2 INTRO TO GROUPS Algebra Study Guide

1 Proofs to Know

Prop: Let R be commutative and let M be a left R-module. Suppose M has an R-basis B of cardinality n and anR-basis E of cardinality m. Then n = m.

Proof. By the remark from class (which states that if M is a free R-module for a commutative ring R with a basis ofn elements, M ∼=Rn), Rm∼=M ∼=Rn (can use the map ϕ : Rn → M given by ϕ(r1, · · · , rn) =

∑n1 risi, where {si}n1

is a basis for M–it’s surjective as it generates M and injective since kerϕ = (0)).Let I be a maximal ideal of R, which exists as R is commutative. Then F = R/I is a field. By an exercise,Rn/IRn∼=R/IR× · · · ×R/IR︸ ︷︷ ︸

n times

= Fn∼=Fm. By vector space theory, n = m.

Prop: Assume the set A = {v1, · · · , vn} spans the vector space V but no proper subset of A spans V . Then A is a basisof V .

Proof. Suffices to show vi are LI. If a1v1 + · · · + anvn = 0 and WLOG a1 6= 0, write v1 as a linear combination ofthe others. So {v2, · · · , vn} spans V . (⇒⇐)

Prop: Let F be a field and V an F -vector space. Let T ⊆ S be subsets of V such that T is LI and S spans V . Thenthere exists a basis B such that T ⊆ B ⊆ S.

Proof. Let Λ = {A : T ⊆ A ⊆ S, A LI}. Note Λ 6= ∅ as T ∈ A. Λ is a poset under ⊆. Let C be a totally orderedsubset of Λ. Let A =

⋃A∈C

A. Show that A ∈ Λ.

By Z-L there exists a maximal element of Λ, say B.Claim: B is a basis for V .Proof: It is enough to show that S ⊆ spanF B. If s ∈ S such that s /∈ spanF B, then B∪{s} is linearly independent.Thus B ∪ {s} ∈ Λ. Since B is maximal, B ⊆ B ∪ {s} =⇒ B = B ∪ {s}.

Thm: Let V be an F -vector space and let W be a subspace of V . Then V/W is a vector space with dimV = dimW +dimV/W (where if one side is infinite then both are).

Proof. Let dimW = m and dimV = n and let w1, w2, · · · , wm be a basis for W . These are LI and can be built upto a basis w1, w2, · · · , wm, vm+1, · · · , vn of V . The natural surjective projection of V into V/W sends all of the wito 0 and none of the vi to 0 (as this would imply vi ∈ W ). Hence, V/W of the projection map is isomorphic to thesubspace of V spanned by the vi, giving that dimV/W = n−m.

2 Intro to Groups

Prop: 1 (Z/nZ)× = {a ∈ Z/nZ : gcd(a, n) = 1}.

Proof. gcd(a, n) = 1 =⇒ ∃x, y ∈ Z such that ax+ ny = 1. Then n|ax− 1 =⇒ ax = 1 so a is a unit. Conversely, ifax = 1 in Z/nZ then n|ax− 1 =⇒ ax− 1 = ny =⇒ gcd(a, n) = 1.

Def: Let F be a field. Then the general linear group of F of degree n is defined by GLn(F ) := Mn(F )× ={A ∈Mn(F ) : detA 6= 0}.

Def: Let G be a group and g ∈ G. Then the order of g is the least positive integer n (if it exists) s.t. gn = 1. If nosuch n exists, then |g| =∞.

Def: Define the dihedral group of order 2n to be the group of rigid motions of the regular n-gon. Then |D2n| = 2n.Let r be the counterclockwise rotation by 2π/n radians and s be the reflection about the x-axis. Note: |r| = nand |s| = 2. Then sri 6= srj for 0 ≤ i < j ≤ n − 1 (if sri = srj =⇒ ri = rj =⇒ 1 = rj−i). ThenD2n =

⟨r, s|rn, s2 = 1, rs = sr−1

⟩.

Prop: 2 Let G be a group and x ∈ G. Let d = |x| <∞. Then for all n ∈ Z, xn = 1⇔ d|n. (For the forward direction,use the division algorithm.)

Prop: Let G be a group, x ∈ G. Then |x| ≤ |G|. If G is a finite group, then every element of G has finite order.

Proof. Nothing to show if |G| = ∞. Assume m = |G| < ∞. Consider{

1, x, x2, · · · , xm}⊆ G. Since G has only m

distinct elements, there exists xi = xj where 0 ≤ i < j ≤ m Then xi−j = 1 and 0 < j − i ≤ m.

Mike Janssen

2 INTRO TO GROUPS Algebra Study Guide 2.1 Generating Groups

Def: Let T be a set, and let ST = {f : T → T : f is a bijection}. This is a group under composition. In the caseT = {1, 2, · · · , n}, denote this group ST by Sn, and it’s called the symmetric group of degree n. Note: |Sn| = n!.

Def: A k-cycle in Sn is a permutation σ such that there exists {a1, a2, · · · , ak} ⊆ {1, 2, · · · , n} such that σ(a1) = a2,σ(a2) = a3, · · · , σ(ak) = a1 and σ(x) = x for all x ∈ {1, · · · , n}\{a1, · · · , ak}. We write this k-cycle as (a1 a2 · · · ak).

Rmrk: Any two disjoint cycles commute, and any permutation can be uniquely written as a product of disjoint cycles.Prop: Let a, b ∈ G and |a| = m < ∞ and |b| = n < ∞. Suppose ab = ba and ∀i, j ∈ Z, ai = bj ⇔ ai = bj = 1. Then|ab| = lcm(m,n).

Cor: Suppose σ, τ are disjoint cycles in Sn. Then |στ | = lcm(|σ|, |τ |).Thm: Let σ = c1c2 · · · ct, where this is a product of disjoint cycles. Let ci be an li-cycle. Then |σ| = lcm(l1, l2, · · · , lt).Prop: 3 Let σ ∈ Sn be a k-cycle. Then |σ| = k.Def: Let σ ∈ Sn \ {1}. Suppose σ is a product of disjoint cycles σ = p1 · · · pl where p1, · · · , pl are of lengths 1 < k1 ≤

k2 ≤ · · · ≤ kl. We say σ has cycle type {k1, k2, · · · , kl}.Def: The quaternion group, Q8, is defined by Q8 = {1,−1, i,−i, j,−j, k,−k}.Def: Let S ⊆ G. The centralizer of S in G is CG(S) := {g ∈ S : gx = xg ∀x ∈ S}.Def: Let H ≤ G. The normalizer of H in G is defined to be NG(H) :=

{g ∈ G : gHg−1 = H

}.

Rmrk: CG(H) ≤ NG(H).Prop: Let G be a group and {Hα}α∈I be a family of subgroups of G. Then ∩α∈IHα ≤ G.

2.1 Generating Groups

Prop: Let S ⊆ G with S 6= ∅. Then 〈S〉 = {se11 · · · senn : si ∈ S, ei = ±1∀i}. If G is finite, 〈S〉 = {s1 · · · , sn : si ∈ S}.

E.g.: 4 Q = 〈{1/pm : p prime, m ∈ N}.Prop: Sn = 〈{(i j) : 1 ≤ i < j ≤ n}〉 = 〈{(1 2), (1 3), · · · , (1n)}〉 = 〈{(1 2), (2 3), · · · , (n− 1n)}〉 = 〈{(1 2), (1 2 3 · · ·n)}〉.Def: If G = 〈a〉, G is said to be a cyclic group (the cyclic group generated by a).Prop: | 〈a〉 | = |a|. If |a| = n, then 〈a〉 =

{1, a, · · · , an−1

}.

Prop: Any subgroup of a cyclic group is cyclic.Thm: 5 Let G = 〈a〉 be a cyclic group with |G| = n <∞.

1. For all i ∈ Z,⟨ai⟩

=⟨ad⟩

where d = gcd(i, n). Thus,{⟨ad⟩

: d > 0, d|n}

is the set of all subgroups of G (theseare all distinct since |

⟨ad⟩| = |ad| = n/d).

2. |ai| = n/gcd(i, n).

3. Every subgroup of G has order dividing n.

Cor: Let G = 〈a〉 be a cyclic group of order n. Then ai is a cyclic generator for G if and only if gcd(i, n) = 1.Cor: Let G = 〈a〉, |G| = n, d a positive divisor of n. Then G has ϕ(d) elements of order d (where ϕ is Euler’s ϕ

function).Cor: Let n be a positive integer. Then

∑d≥1d|n

ϕ(d) = n.

Prop: Let G = 〈a〉 be a cyclic group. Then if |G| = n then G∼=Z/nZ. If |G| =∞ then G∼=Z.Def: 6 The alternating group of order n, denoted by An, is the subgroup of Sn consisting of all even permutations.Prop: If n > 2, An is generated by the set of all 3-cycles.Def: Let G be a group, H ≤ G. Then H is called normal, denoted H C G if gHg−1 = H for all g ∈ G.Rmrk: Let G be a finite group and H ≤ G such that 2|H| = |G|. Then H C G.

Proof. Let σ ∈ G \H. Note |H| = |σH| and H ∩ σH = ∅. Then |H| = |G|/2 = |σH| so H ∪ σH = G. Similarly,H ∪Hσ = G and H ∩Hσ = ∅. Therefore G \H = Hσ = σH, which implies σHσ−1 = H.

Cor:An C Sn.Prop: |A4| = 12. A4 has no subgroup of order 6.

Proof. The number of 3-cycles in A4 (or S4) is(

43

)· 2 = 8. Suppose ∃H ≤ A4 such that |H| = 6 = |A4|/2, which

implies H C A4. So H contains some 3-cycle. WLOG, assume the 3-cycle (1 2 3) ∈ H. Then (1 3 2) ∈ H (as(1 3 2)−1 = (1 2 3). Since H is normal in A4, setting σ = (1 2) implies σ(1 2 3)σ−1 = (1 4 2) ∈ H, so (1 2 4) ∈ H.Doing this again with, say, (1 3)(2 4) yields two more 3-cycles. Therefore H contains at least 6 3-cycles. (⇒⇐)

Mike Janssen

2 INTRO TO GROUPS Algebra Study Guide 2.2 Cosets

2.2 Cosets

Def: Let H ≤ G, g ∈ G. Then the set gH = {gh : h ∈ H} is called a left coset of H.Prop: Let H ≤ G. (1) For all g1, g2 ∈ G, |g1H| = |g2H|. (2) For all g1, g2 ∈ G, g1H = g2H or g1H ∩ g2H = ∅. (3)

G = ∪g∈GgH.

Proof. (1) The maps f : g1H → g2H defined by x 7→ g2g−11 x and f ′ : g2H → g1H defined by y 7→ g1g

−12 y are

mutual inverses. (2) Suppose g1H ∩ g2H 6= ∅. Then x ∈ g1H ∩ g2H, so x = g1h1 = g2h2. Let g1h3 ∈ g1H. Theng1h3 = g1h1h

−11 h3 = g2h2h

−11 h3 ∈ g2H.

Def: If H ≤ G, let [G : H], called the index of H in G be defined to be the number of distinct left cosets of H in G.Thm: (Lagrange) Let G be a finite group, H ≤ G. Then |G| = |H|[G : H].

Proof. Let m = [G : H], say {g1H, · · · , gmH} is the set of distinct left cosets of H. By (3) above, G = ∪mi=1giH. By(2), giH ∩ gjH = ∅ for i 6= j. So |G| = |g1H|+ · · ·+ |gmH| = m|H| by (1).

Cor: 7 Let G be a finite group. Then |a| | |G| for all a ∈ G.

2.3 Group Actions

Def: Let G be a group and A a set. An action of G on A is a function G×A→ A ((g, a) 7→ g · a) such that (1) forall a ∈ A, g1, g2 ∈ G, (g1g2)a = g1(g2a) and (2) for all a ∈ A, 1 · a = a.

Def: Let G be a group acting on a set A. For a ∈ A, the stabilizer Ga of a is Ga := {g ∈ G : ga = a}. Note thatGa ≤ G. The kernel of the action is ker = ∩a∈AGa. The action is called faithful if ker = {1}. The action is freeif Ga = {1} for all a ∈ A. For a ∈ A, the orbit Oa of a is Oa := {ga : g ∈ G}. The action is called transitive ifOa = A for some a ∈ A.

Lma: 8 Let ϕ : G→ G be a group homomorphism. Then ϕ is 1-1 iff kerϕ = 1.Prop: Let A and B be sets such that |A| = |B|. Then SA∼=SB .Def: A homomorphism ϕ : G→ SA is called a permutation representation for ϕ.Rmrk:G acts on A iff for ϕ : G→ SA the kernel of the action = kerϕ.Rmrk:G acts faithfully on A if and only if ϕ : G→ SA is injective. In this case G∼=ϕ(G) ≤ SA.Thm: (Cayley’s Theorem) Let G be a group. Then G is isomorphic to a subgroup of SG. In particular, if |G| = n, then

G∼=H ≤ Sn.E.g.: Examples of Group Actions:

Mike Janssen

2 INTRO TO GROUPS Algebra Study Guide2.4 Quotient Groups and Homomorphisms

Prop: Let H ≤ G, A = {xH : x ∈ G}. Then G acting on A by left multiplication gives a group homomorphismϕ : G→ SA∼=Sn (where n = [G : H]) given by ϕ(g) = σg : A→ A (where σg(xH) = gxH) having kernel⋂

x∈GxHx−1.

If ∩x∈GxHx−1 = 1 then G is isomorphic to a subgroup of Sn, where n = [G : H].Prop: 9 Let G act on A and let a, b ∈ A. Then Oa = Ob or Oa ∩ Ob = ∅. Thus, A is the disjoint union of the distinct

orbits. If |A| <∞ then |A| =n∑i=1

|Oai |, where Oa1 , · · · ,Oan are the distinct orbits of A.

Cor: If G acts transitively on A, then Oa = A for all a ∈ A. Therefore, G acts transitively on A iff ∀a, b ∈ A thereexists g ∈ G such that ga = b.

Prop: Let G act on A. Let a ∈ A. Then there exists a bijection ϕ from {gGa : g ∈ G} → Oa given by gGa 7→ ga.Hence |Oa| = [G : Ga].

Cor:G acts on A, where |A| <∞. Let Oa1 , · · · ,Oanbe the distinct orbits. Then |A| =

n∑i=1

[G : Gai].

E.g.: (The Class Equation) Let G be a finite group. Let G act on itself by conjugation, i.e., G = A, g · a = gag−1. LetOa1 , · · · ,Oan

be the distinct orbits of G under this action. Oa1 ={ga1g

−1 : g ∈ G}

is called the conjugacy class

of a1. Then Ga ={g ∈ G : gag−1 = a

}= CG(a). Therefore, |G| =

n∑i=1

[G : CG(ai)]. Notice that [G : CG(a)] = 1 if

and only if CG(a) = G if and only if a ∈ Z(G). Therefore,

|G| = |Z(G)|+∑

[G : CG(a)],

where a runs through the distinct conjugacy classes and a /∈ Z(G).Prop: Let G be a group with |G| = pn, where p is prime. Then Z(G) 6= 1 (follows from the fact that [G : CG(a)] = pl,

l ≥ 1).

2.4 Quotient Groups and Homomorphisms

Def: A subgroup H of G is normal in G if xHx−1 = H for all x ∈ G.Rmrk:H is normal in G iff xHx−1 ⊆ H for all x ∈ G.Rmrk: Suppose H = 〈S〉. Then H C G iff gsg−1 ∈ H for all s ∈ S and for all g ∈ G.Rmrk: If H ⊆ Z(G) then H C G.Prop: Let ϕ : G1 → G2 be a group homomorphism. Then kerϕ C G1.Rmrk: Let H ≤ G, A = {gH : g ∈ G}. Let G act on A as usual. Recall, GxH = xHx−1. Therefore

⋂x∈G xHx

−1 C G(this is the largest normal subgroup of G contained in H). In particular, if [G : H] = n then there exists a grouphomomorphism ϕ : G→ Sn∼=SA where kernel = ∩x∈GxHx−1.

Prop: 10 Let ϕ : G1 → G2 be a group homomorphism, let K = kerϕ. Then there exists a bijection {xK : x ∈ G1} ↔imϕ (xK ↔ ϕ(x)). In particular, if |G2| <∞, [G : K] = |imϕ| divides |G2|.

Prop: Let G be a finite group and p the smallest prime divisor of |G|. Suppose there exists a subgroup H of G suchthat [G : H] = p. Then H C G. STUDY THIS PROOF

Cor: If [G : H] = 2. Then H C G.Cor: If |G| is odd and [G : H] = 3 then H C G.Def: A group G 6= 1 is called simple if it has no nontrivial normal subgroups.Prop: Let G be a group and H ≤ G. Define an “operation” on the left cosets {xH : x ∈ G} by (xH) · (yH) := (xy)H.

This operation is well-defined if and only if H C G.Prop: Suppose H C G. Then the set of cosets of H is a group under the operation defined above. This group is denoted

G/H and is referred to as a quotient group.Rmrk: If |G| <∞ then |G/H| = |G|/|H| (Lagrange).Prop: 11 Let ϕ : G → A be a surjective group homomorphisms. Then there exists a bijective correspondence between{subgroups of G containing kerϕ} ↔ {subgroups of A} given by K ↔ ϕ(K). Furthermore, K C G⇔ ϕ(K) C A.

Prop: Let G be a group and H C G. Consider the function ϕ : G→ G/H given by ϕ(g) = gH. Then ϕ is a surjectivegroup homomorphism with kerϕ = H.

Cor: Let H C G, ϕ : G → G/H be the canonical map. Then there exists a bijection between {K : K ≤ G, H ⊆ K}and {A : A ≤ G/H} given by K ↔ K/H. Furthermore, if H ⊆ K ⊆ G, K C G⇔ K/H C G/H.

Mike Janssen

2 INTRO TO GROUPS Algebra Study Guide 2.5 Direct Products

Thm: Let ϕ : G→ A be a group homomorphism and let H C G with H ≤ kerϕ. Then there exists a group homomor-phism ϕ : G/H → A given by gH 7→ ϕ(g). Furthermore, imϕ = imϕ and kerϕ = kerϕ/H.

Cor: (1st Isomorphism Theorem) Let ϕ : G→ A be a group homomorphism. Then G/ kerϕ∼= imϕ.Cor: Let H ≤ K ≤ G and H C G, K C G. We already know that K/H C G/H. Then (G/H)/(K/H)∼=G/K via the

map (gH)/(K/H) 7→ gK. (We have kerϕ = K/H in the induced map.)Rmrk: 12 Let ϕ : G→ A be a group homomorphism. Let x ∈ G with |x| <∞. Then |ϕ(x)| | |x|.Thm: (Cauchy’s Theorem) Let G be a finite group and p a prime such that p | |G|. Then G has an element of order p.

Proof. Case 1. G is abelian. Use induction on |G| = n. If n = p, then G is cyclic of order p. If n > p, choosex ∈ G, x 6= 1. Let H = 〈x〉. If p||H|, p||x|, say |x| = ps. Then |xs| = p. If p - |H|, then p|[G : H] = |G|/|H|. Since Gis abelian, H C G. Therefore, p | |G/H| = [G : H]. As H 6= 1, |G/H| < |G|. By induction, G/H has an element oforder p, say yH. Therefore p | |y|, and hence |y| = tp, then |yt| = p.Case 2. G arbitrary. If Z(G) = G, then G is abelian and we’re done. So assume Z(G) � G. If p | |Z(G)|, then Z(G)has an element of order p (as Z(G) is abelian). If p - |Z(G)|, use the class formula: |G| = |Z(G)|+

∑x/∈Z(G)

[G : CG(x)].

Since p||G|, p - |Z(G)|, there exists x /∈ Z(G) such that p - [G : CG(x)] = |G|/|CG(x)|. Therefore, p | |CG(x)|. Asx /∈ Z(G), |CG(x)| < |G|. By induction on |G|, CG(x) has an element of order p. Hence, G does.

Lma: Let a, b ∈ G, |a| <∞, |b| <∞. Suppose (1) 〈a〉 ∩ 〈b〉 = {1} and (2) ab = ba. Then |ab| = lcm(|a|, |b|).Lma: Let H1, H2 C G. Suppose H1 ∩H2 = 1. Then ab = ba for all a ∈ H1, b ∈ H2.Cor: Suppose |G| = pq and let a be an element of order p, b be an element of order q, p < q, p, q primes. Suppose〈a〉 C G and 〈b〉 C G. Then G is cyclic.

Proof. 〈a〉 ∩ 〈b〉 ≤ 〈a〉 , 〈b〉 and 〈a〉 ∩ 〈b〉 = 1 by Lagrange. So ab = ba and hence |ab| = |a| · |b| = pq.

E.g.: Let G be a group, |G| = 6. Then G∼=Z/6Z or G∼=S3.

Proof. Suppose G is not cyclic. By Cauchy there exists a, b ∈ G such that |a| = 2, |b| = 3. Note if ab = ba then|ab| = 6 which implies G is cyclic (⇒⇐). Let H = 〈a〉, K = 〈b〉. As [G : K] = 2, K is normal. Also, H ∩K = 1 (asgcd(|H|, |K|) = 1). By the lemma, A = {xH : x ∈ G}. |A| = [G : H] = 3. Let G act on SA as usual. This gives agroup homomorphism ϕ : G→ S3 and kerϕ is the largest normal subgroup of G contained in H. As |H| = 2 is notnormal, kerϕ = 1. Therefore ϕ is 1-1. As |G| = |S3| = 6, ϕ is onto.

2.5 Direct Products

Def: 13 Let G be a group, A,B ⊆ G (nonempty). Define AB := {ab : a ∈ A, b ∈ B}.Prop: Let H,K ≤ G. Then |HK| · |H ∩K| = |H| · |K|.Prop:H,K ≤ G. Then HK ≤ G⇔ HK = KH.Cor: If H,K ≤ G and either H or K is normal in G, then HK ≤ G.Def: Let G be a group, H1, · · · , Ht ≤ G. Suppose (1) Hi C G for all i, (2) G = H1H2 · · ·Ht = {h1h2 · · ·ht : hi ∈ Hi},

and (3) Hi ∩H1 · · · Hi · · ·Ht = 1 for all i. We say that G is the (internal) direct product of H1, · · · , Ht.Prop: Suppose G is the internal direct product of H1, · · · , Ht. Then G∼=H1 × · · · ×Ht (externally).Prop: Let G be a group of order p2, where p is a prime. Then G∼=Cp2 or G∼=Cp × Cp. (For proof, suppose not cyclic

and find two distinct subgroups of order p.)Thm: 14 Let H,K ≤ G, K C G. Then (1) H ∩K C H and (2) HK/K ∼=H/H ∩K.

2.6 Fund. Thm. of FG Abelian Groups

Thm: 15 Let G be a finitely-generated abelian group. Then there exists unique integers r ≥ 0 and p1, · · · , pk primes,m1, · · · ,mk > 0 such that p1 ≤ · · · ≤ pk and G∼=Cr∞ × Cpm1

1× · · · × Cpmk

k, so G is finite iff r = 0. If |G| <∞ then

|G| = pm11 · · · pmk

k . Special Case: Suppose |G| = pn (p a prime) is abelian. Then G∼=Cpm1 × · · · × Cpmk , where∑mi = n.

Mike Janssen

2 INTRO TO GROUPS Algebra Study Guide 2.7 Automorphisms

2.7 Automorphisms

Def: Let G be a group. An isomorphism ϕ : G→ G is called an automorphism of G. Let Aut(G) denote the setof automorphisms, which is a group under function composition.

Def: Let G be abelian, m ∈ Z. Then the map ϕm : G→ G given by ϕm(g) = gm is a group homomorphism. Note thatϕm ◦ ϕl = ϕml.

Prop: Let G = Cn, n ≥ 1. Then Aut(Cn) = {ϕm : Cn → Cn : gcd(m,n) = 1}∼= (Z/nZ)×.Prop: 16 Cm × Cn∼=Cmn ⇔ gcd(m,n) = 1.Cor: If m1, · · · ,mk are pairwise relatively prime positive integers then Cm1 × · · · × Cmk

= Cm1···mk.

Def: Let G be a group. Define the exponent of G by e(G) = inf{d ≥ 1 : gd = 1∀g ∈ G

}. If |G| < ∞, then by

Lagrange, e(G) ≤ |G|.Notation: Let G be a group, x ∈ G. Define ψx : G→ G by ψx(g) = xgx−1.Def: The map ψx is a homomorphism for all x ∈ G, and is called a inner automorphism of G. Define the group of

inner automorphisms to be Inn(G) = {ψx : x ∈ G}.Exercise: Inn(G) E Aut(G).Note: Aut(G)/ Inn(G) is called the group of automorphisms.Exercise: Inn(G)∼=G/Z(G) (isomorphism given by ψx 7→ xZ(G)).Def: Let H ≤ G. H is called a characteristic subgroup of G, denoted H charG if ϕ(H) = H for all ϕ ∈ Aut(G).Rmrk:H charG⇔ ϕ(H) ⊆ H for all ϕ ∈ Aut(G).Rmrk:H charG =⇒ H E G.Rmrk: If H is the unique subgroup of order |H| then H charG.Rmrk: Z(G) charG.Rmrk: In general, K C H and H C G 6 =⇒ K C G.Rmrk:K charH and H charG implies K charG.Rmrk:K charH and H C G =⇒ K C G.Def: Let G be a group, x, y ∈ G. The element denoted [x, y] := xyx−1y−1 is called the commutator of x and y. The

subgroup of G generated by all commutators is called the commutator subgroup (sometimes denoted G′ or [G,G]).So G′ =

⟨xyx−1y−1|x, y ∈ G

⟩.

Rmrk:G′ charG.Rmrk: 17 Let G be a group, H ≤ G. (1) If H ⊇ G′ then H C G and G/H is abelian. (2) The converse is also true. If

H C G and G/H is abelian then H ⊇ G′.

2.8 Sylow’s Theorems

Def: Let G be a finite group, p a prime such that p | |G|. Suppose pn | |G| but pn+1 | |G|. A p-subgroup of G is asubgroup of G of order pα for some α > 0. Let Sylp(G) denote the set of Sylow p-subgroups of G. Let np = |Sylp(G)|.

Thm: (Sylow I) If pα | |G| then G has a subgroup of order pα.

Proof. Induction on |G| = n. If n = 1, the statement is vacuously true. Then the class equation gives |G| =

|Z(G)|+l∑i=1

[G : CG(gi)] where CG(gi) 6= G.

Case 1. p | |Z(G)|. By Cauchy’s Theorem there exists a ∈ Z(G) where |a| = p. Let K = 〈a〉 C G (as a ∈ Z(G)).Consider G = G/K. Then |G|/p < |G|. Also, pα−1 | |G|. By the induction hypothesis G has a subgroup of orderpα−1. Such a subgroup has the form H/K, where K ⊆ H ⊆ G. Then |H/K| = |H|/|K| = pα−1, which implies|K| = p =⇒ |H| = pα.

Case 2. p - |Z(G)|. Then the class equation implies there exists some i for which p - [G : CG(gi)]. Then p - |G|/|CG(gi)|implies pα | |CG(gi)|. But |CG(gi)| < |G|. Since [G : CG(gi)] > 1, CG(gi) has a subgroup of order pα (byinduction).

Thm: (Sylow II) If G is a finite group, pn | |G|, pn+1 - |G|. Let P ∈ Sylp(G), np = |Sylp(G)|. Then

1. np = [G : NG(P )] for all P ∈ Sylp(G). In particular, Sylp(G) = {P} if and only if P C G (i.e., G = NG(P )).

2. np ≡ 1 mod p

3. Any two Sylow p-subgroups of G are conjugate.

4. Any p-subgroup of G is contained in a Sylow p-subgroup.

Mike Janssen

2 INTRO TO GROUPS Algebra Study Guide 2.9 Solvable Groups

2.8.1 Applications

Prop: 18 Let |G| = pq where p < q primes and p - q − 1. Then G is cyclic.

Proof. Let np | |G|/p = q implies np = 1 or q and nq|p implies nq = 1 or p. And np ≡ 1 mod p and nq ≡ 1 mod q.As p < q, p 6≡ 1 mod q. Therefore nq = 1. As q 6≡ 1 mod p, np = 1. Therefore both the Sylow p-subgroup P and

the Sylow q-subgroup Q are normal. Therefore |PQ| = |P | · |Q||P ∩Q|

= pq. So G = PQ, P ∩Q = 1, P C G, Q C G,and

hence G∼=P ×Q∼=Cp × Cq ∼=Cpq.

Exercise: Suppose |G| = p2q where p, q are primes. Prove G is not simple.E.g.: 19 No group of order 144 is simple.Prop: Let G be a finite group. G is cyclic if and only if G has a unique subgroup of order d for every positive divisor d

of |G|. Study Proof!

2.9 Solvable Groups

Def: Let G be a group. A normal series for G is a sequence of subgroups

1 = Hn E Hn−1 E · · · E H0 = G.

Note: G, nontrivial, is simple iff its only normal series is the trivial normal series. The groups Hi/Hi−1 are calledthe factor groups of the series. The length of the series is the number of nontrivial factor groups.

Def: A composition series for G is a normal series such that all the factor groups are simple.E.g.: Easy composition series: {1} E A5.E.g.: {1} E {1, (1 2)(3 4)} E V4 E A4 E S4 (A4/V4

∼=C3, S4/A4∼=C2, V4/H ∼=C2).

Def: 20 A subgroup H of G is called a maximal normal subgroup if H 6= G, H E G and there are no normalsubgroups properly between H and G (equivalently, H E G and G/H is simple).

Prop: Any finite group has a composition series.Thm: (Jordan-Holder Theorem) Suppose G has a composition series. Then any two composition series for G have the

same length, and the set of factor groups are the same up to isomorphism.Def: A refinement of a normal series for G is just another normal series for G obtained by inserting more subgroups

between the terms of the original.Rmrk: If G has a composition series, any normal series has a refinement which is a composition series.Note: A finite abelian group is simple iff G has prime order.Note: The factor groups for any composition series for an abelian group have prime order.Def: A solvable series for a group G is a normal series in which the factor groups are all abelian, i.e., 1 = Hn C · · · C

H0 = G such that Hi/Hi+1 is abelian. A group is solvable if it has a solvable series.Prop: Suppose |G| <∞. Then G is solvable iff all the factor groups in the composition series have prime order.Exercise: Prove this!

Def: 21 Define G(n) := G(n−1)′ where G(0) = G (recall G′ is the commutator subgroup of G). Recall that G′ charG.Since “char” is transitive, G(n) charG for all n ≥ 0. In particular, G(n−1)/G(n) is abelian. The series · · · C G(2) CG(1) C G(0) = G is called the derived series.

Prop:G is solvable iff G(n) = 1 for some n ≥ 1 (i.e., the derived series is normal).

Mike Janssen

3 RING THEORY Algebra Study Guide

Lma: Let ϕ : G→ A be a surjective group homomorphism. Then ϕ(G(n)) = A(n) for all n ≥ 0.Cor: Suppose H C G. Then (G/H)(n) = G(n)H/H.Thm: Let G be a group, H ≤ G. (1) If G is solvable, so is H. (2) If H C G then G is solvable iff H and G/H are

solvable.Prop: Any p-group is solvable. (For proof, use induction and note that Z(G) and G/Z(G) is solvable.)

3 Ring Theory

3.1 Intro to Ring Theory

Def: 22 Let R be a ring. Let a ∈ R. Then a is called a zero-divisor if there exists nonzero b ∈ R such that eitherab = 0 or ba = 0. If R is commutative and has no nonzero zero divisors (nonzero-divisors or NZDs) then R is calledan integral domain (or just a domain).

Def: Let R be a ring with 1. Let a ∈ R. An element b ∈ R is called a left-inverse for a if ba = 1 and b is aright-inverse for a if ab = 1.

Rmrk: Let a ∈ R. Suppose b is a left-inverse for a and c is a right-inverse. Then b = c.E.g.: Let (G,+) be an abelian group. Let Hom(G,G) = {f : G→ G : f a group homom.}(= End(G)). Define +

and ◦ on Hom(G,G) as follows: given f, g ∈ Hom(G,G), (f + g)(x) = f(x) + g(x) and (fg)(x) = (f ◦ g)(x) for allx ∈ G. Then Hom(G,G) is a ring with 1.

E.g.: LetG = Z∞, andR = Hom(G,G). Let f, g ∈ R be given by f(a1, a2, a3, · · · ) = (0, a1, a2, · · · ) and g(a1, a2, a3, · · · ) =(a2, a3, · · · ). Then gf = 1 but fg((1, 0, 0, · · · )) = (0, 0, · · · ) so fg 6= 1. Thus, f is left-invertible but not right-invertibleand vice versa for g. (And actually, if αn ∈ R is given by αn((a0, a1, · · · )) = (na0, 0, 0, · · · ), (g + αn)f = 1, so f hasinfinitely many left inverses.)

Rmrk: 23 A unit in a ring cannot be a zero-divisor.Def: A division ring is a ring R 6= {0} with 1 such that every nonzero element is a unit. A field is a commutative

division ring.Prop: Let n ≥ 2. Then n is prime iff Z/nZ is a field iff Z/nZ is a domain.

3.2 Polynomial Rings

Rmrk: If R is a domain, then R[x] is a domain.Thm: 24 (Division Theorem) Let R be commutative with 1, and x a variable. Let f(x), d(x) ∈ R[x], where the

leading coefficient of d(x) is a unit in R. Then there exist unique polynomials q(x), r(x) ∈ R[x] such that f(x) =d(x)q(x) + r(x) and deg r < deg d.

Def: A polynomial is called monic if its leading coefficient is a unit. So the division theorem “works” whenever youdivide by a monic polynomial.

Cor: Let f(x) ∈ R[x], r ∈ R. Then f(r) = 0 iff (x− r)|f(x).

3.3 Ideals and Ring Homomorphisms

Def: A map ϕ : R→ S (R,S rings) is called a ring homomorphism if ϕ(a+b) = ϕ(a)+ϕ(b) and ϕ(ab) = ϕ(a)ϕ(b)for all a, b ∈ R.

E.g.: 25 Suppose ϕ : R → S is a nonzero ring homomorphism. Suppose that R,S have identity and S is a domain.Then ϕ(1R) = 1S .

Def: 26 Let R be commutative with 1. A nonempty subset I of R is an ideal if (I,+) ≤ (R,+) and for all r ∈ R andi ∈ I, ri ∈ I.

Def: Let R be commutative with 1. Let A ⊂ R. Define the ideal of R generated by A by (A) :=⋂

I idealA⊂R

I.

Exercise: (A) = {r1a1 + · · ·+ rnan : ri ∈ R, ai ∈ A, n ∈ N}.Exercise: Consider the ideal (2, x) ⊂ Z[x] is not principal.Def: Let R be commutative with identity, I ⊆ R an ideal. Let R/I denote the set of left cosets of (I,+) in (R,+).

Define +, · on R/I by (1) (a+ I) + (b+ I) = (a+ b) + I and (2) (a+ I) · (b+ I) = (a · b) + I.Prop: 27 Let ϕ : R→ S be a homomorphism of commutative rings. Then R/ kerϕ∼= imϕ (as rings).Prop: Suppose I ⊂ J are ideals of R. Then J/I is an ideal of R/I.

Mike Janssen

3 RING THEORY Algebra Study Guide 3.4 Operations on Ideals

Prop: (R/I)/(J/I)∼=R/J .Prop: Let I be an ideal ofR. Then there exists a bijection, inclusion-preserving correspondence between {J : J an ideal of R}

and {ideals of R/I} given by J 7→ J/I = ϕ(J) and K 7→ ϕ−1(K), where ϕ : R → R/I is the canonical surjectivehomomorphism.

3.4 Operations on Ideals

Def: Let I, J be ideals of R. Then we define I+J := {a+ b : a ∈ I, b ∈ J}. This is the smallest ideal of R containing

I ∪ J (which is not an ideal). Further, IJ = 〈{ij} i ∈ I, jıJ〉 ={

n∑k=1

ikjk : ik ∈ I, jk ∈ J}

, which is the smallest

ideal containing {ij} i ∈ I, j ∈ J .Thm: (Chinese Remainder Theorem) Let R be commutative with identity, I1, · · · , In ideals of R such that Ii + Ij = R

for all i 6= j. Then (1) I1 ∩ · · · ∩ In = I1I2 · · · In and (2) there is an isomorphism of rings ϕ : R/(I1 · · · In) →R/I1 × · · · ×R/In given by r + I1 · · · In 7→ (r + I1, · · · , r + In).

Thm: 28 Let d1, · · · , dn be pairwise relatively prime positive integers and a1, · · · , an any integers. Then there existsx ∈ Z such that x ≡ ai mod di for i = 1, · · · , n.

Rmrk: If R,S are rings with 1, then (R× S)×∼=R× × S× as groups.Def: Let R be commutative with 1. An ideal m of R is called maximal if m 6= R and the only ideals containing m are

m and R.Prop: An ideal I of R is maximal iff R/I is a field.Rmrk: Let ϕ : R→ F is a surjective ring homomorphism such that ϕ 6≡ 0, F a field. Then kerϕ is a maximal ideal, as

R/ kerϕ is isomorphic to a field.Rmrk: Let m, n be distinct maximal ideals of a commutative ring R. Then m + n = R.Def: An element r ∈ R is called nilpotent if rn = 0 for some n ∈ N. R is called reduced if there are no nonzero

nilpotents.

3.5 Zorn’s Lemma

Def: 29 Let A be a set. A partial order ≤ on A is a binary relation with the following properties: for all a, b, c ∈ A,(1) a ≤ a; (2) a ≤ b and b ≤ a iff a = b; (3) a ≤ b, b ≤ c implies a ≤ c. Additionally, ≤ is a total order on A if for alla, b ∈ A, a ≤ b or b ≤ a.

Def: Let (A,≤) be a poset, and B a subset of A. An element x ∈ A is an upper bound for B if b ≤ x for all b ∈ B.An element b ∈ B is maximal if for every y ∈ B with y ≤ b implies y = b.

Thm: (Zorn’s Lemma) Let (A,≤) be a nonempty poset and suppose every totally ordered subset of (A,≤) has an upperbound in A. Then A is a maximal element.

Prop: Let R 6= 0 be commutative with 1 and I 6= R an ideal. Then there exists a maximal ideal m of R containing I.Def: Let R be commutative. An ideal P of R is called prime if P 6= R and whenever ab ∈ P , then either a ∈ P or

b ∈ P .Rmrk: P is prime iff R/P 6= 0 is a domain.Rmrk: Let ϕ : R→ S be a commutative ring homomorphism. Suppose ϕ 6= 0 and S is a domain. Then kerϕ is a prime

ideal of R.Rmrk:R is a domain iff {0} is prime. R is a field iff {0} is maximal.Rmrk: If P is prime and P ⊇ IJ then P ⊇ I or P ⊇ J .

3.6 Quotient Fields

Def: Given a domain R, the smallest field containing R is called the quotient field (or field of fractions) of R,denoted by Q(R).

Prop: Let R be a domain and suppose R ⊆ F , F a field. Then there exists an injective field homomorphism f : Q(R)→F that fixes R.

3.7 Euclidean Domains

Def: 30 Let R be a domain. R is called a Euclidean domain if there exists a function δ : R\{0} → N0 = {0, 1, 2, · · ·}such that for all a, b ∈ R with b 6= 0 there exists q, r ∈ R such that a = bq + r with r = 0 or δ(r) < δ(b).

Mike Janssen

3 RING THEORY Algebra Study Guide 3.8 Factorization Conditions

Thm: Z[i] is a Euclidean domain with δ : Z[i]→ N0 by a+ bi 7→ a2 + b2.Thm: Z[i] = {a+ bi : a, b ∈ Z} is a Euclidean Domain with δ : Z[i]→ N given by a+ bi 7→ a2 + b2 = (a+ bi)(a− bi).

Proof. Note that, for all α, β ∈ Z[i], δ(αβ) = δ(α)δ(β). Suppose β 6= 0. Then

α

β=αβ

ββ= x+ yi, for x, y ∈ Q.

Let a, b ∈ Z and u, v ∈ Q such that x = a + u and y = b + v for |u|, |v| ≤ 1/2. Then observe α = β(x + yi) =β(a+ bi) + β(u+ vi). Let γ = β(u+ vi) and note γ = α− β(a+ bi) implies γ ∈ Z[i].

Claim: If we show δ(γ) < δ(β), we’re done: α = β(a+ bi) + γ.Notice that δ can be extended to a function on Q[i]→ Q≥0 by x+yi 7→ x2+y2. Then δ(γ) = δ(β(u+vi)) = δ(β)δ(u+vi),

and δ(u+ vi) = u2 + v2 ≤ 14 + 1

4 = 12 . Thus, δ(γ) ≤ 1

2δ(β) < δ(β).

Def: An integral domain in which every ideal is principal is a PID.Thm: Euclidean Domains are PIDs.

Proof. Let R be a ED under δ and let I ⊂ R be an ideal. If I = {0}, we’re done. Assume I 6= {0}. Choose d ∈ I \{0}such that δ(d) ≤ δ(a) for all a ∈ I \ {0}. Claim: I = (d). Pick a ∈ I. Then a = dq + r for some q, r ∈ R with r = 0or δ(r) < δ(d). Since r = a − dq ∈ I, we may not have δ(r) < δ(d), as δ(d) ≤ δ(b) for all b ∈ I. Thus, r = 0, anda ∈ (d).

Rmrk: The domain R = Z[√−5] is not a PID (the ideal (3, 2 +

√−5) is not principal).

Rmrk: 31 The domain R = Z[

1+√−19

2

]is a PID that is not a Euclidean Domain.

Prop: Let R be a PID and I = (p) 6= 0 a prime ideal. Then I is maximal.

3.8 Factorization Conditions

Def: Let R be a domain. An element r ∈ R is called irreducible if r is not a unit and whenever r = ab, then oneof a or b is a unit. An element p ∈ R \ {0} is called prime if (p) is a prime ideal (i.e. if p|ab then p|a and p|b).

Prop: Let R be a domain. Then any prime element is irreducible.Rmrk: 2 ∈ R = Z[

√−5] is irreducible but not prime.

Def: 32 Let R be a domain. Then R is a UFD if (1) every nonunit of R is a product of irreducibles and (2) ifa1a2 · · · am = b1 · · · bn, where ai, bj is irreducible, then m = n and after reindexing ai = uibi for ui ∈ R×.

Def: Let R be a commutative ring. R is said to satisfy the ascending chain condition (ACC) on principal ideals ifgiven any ascending chain of principal ideals

(a1) ⊆ (a2) ⊆ · · ·

there exists n ≥ 1 such that for all k ≥ n, (ak) = (ak+1) i.e.,

(a1) ⊆ (a2) ⊆ · · · ⊆ (an) = (an+1) = (an+2) = · · · .

Rmrk: Let R be a commutative ring. Then R satisfies ACC on principal ideals iff every nonempty set Λ of principalideals of R has a maximal element in Λ.

Thm: Let R be a domain. Then R is a UFD if and only if R satisfies ACC on principal ideals and every irreducibleelement is prime.

Prop: 33 Let R be a PID. Then R is a UFD.Def: Let R be a domain, a, b ∈ R. An element d ∈ R \ {0} is a greatest common divisor (gcd) of a, b if (1) d|a, d|b

and (2) if e|a, e|b, then e|d.Rmrk: If d1, d2 are gcds of a and b then (d1) = (d2).Prop: Let R be a UFD. Then gcd(a, b) exists for all a, b ∈ R.Def: Let R be a domain, a, b ∈ R both nonzero. The least common multiple (lcm) of a and b is an element

m ∈ R \ {0} such that (1) a|m, b|m, and if a|m and b|n, then m|n.Rmrk: Any two lcms of a and b are associate.Exercise: Let R be a UFD, a, b ∈ R \ {0}. Then (a) ∩ (b) = (lcm(a, b)).Exercise: Find an example of a domain R and nonzero elements a, b such that gcd(a, b) does not exist. (R = Z[

√−5],

a = 2(1 +√−5), b = 6.)

Mike Janssen

4 MODULES Algebra Study Guide

Def: 34 Let R be a UFD, x a variable. Let f(x) = anxn + · · · + a0 ∈ R[x] \ {0}. The content of f(x) is defined by

c(f) = gcd(an, · · · , a1, a0). f(x) is called primitive if c(f) = 1.Rmrk: c = c(f) implies f = cf1, where f1 is primitive.Rmrk: If d ∈ R, d|c(f), then f/d ∈ R[x] and c(f/d) = c(f)/d.Rmrk: If f = cg where c ∈ R, then c|c(f).Rmrk: Suppose f = gh, f, g, h ∈ R[x] then c(g)c(h)|c(f).Thm: (Gauss’ Lemma) Let R be a UFD, f = gh ∈ R[x] \ {0}. Then c(f) = c(g)c(h).Cor: 35 The product of primitive polynomials is primitive.Cor: Let R be a UFD, F = Q(R). Let f(x) ∈ R[x], g(x) ∈ R[x] primitive. Suppose g|f in F [x] (i.e. f = gh for

h ∈ F [x]). Then g|f in R[x].Cor: Let R be a UFD and F = Q(R), f(x) ∈ R[x]. Suppose f(x) = g(x)h(x), where g, h ∈ F [x]. Then there exist

a, b ∈ F such that ag, bh ∈ R[x] and f = (ag)(bh).Cor: Let R be a UFD and F = Q(R). Let f(x) ∈ R[x] be primitive. Then f(x) is irreducible in R[x] if and only if f(x)

is irreducible in F [x].Thm: Let R be a UFD. Then R[x] is a UFD.Cor: If k is a field (or even just a UFD), k[x1, · · · , xn] is a UFD for all n.Rmrk: 36 If deg f(x) > 1 and f(x) is irreducible then f(x) has no roots in F .Rmrk: (x2 + 1)2 ∈ Q[x] is reducible but has no roots.Rmrk: If deg f = 2 or 3 then f is irreducible if and only if f has no roots in F .Thm: (Rational Root Theorem) Let R be a UFD and f(x) = anx

n + · · · a0 ∈ R[x]. Suppose α ∈ F = Q(R) is a root off . Then α = c/d where c|a0 and d|an, for c, d ∈ R.

Thm: (Eisenstein’s Criterion) Let R be a UFD and f(x) = anxn+ · · ·+a0 ∈ R[x] and suppose there exists an irreducible

element π ∈ R such that (1) π - an; (2) π|ai for 0 ≤ i ≤ n − 1, and (3) π2 - a0. Then f(x) is irreducible in F [x],where F = Q(R).

4 Modules

4.1 Intro to Module Theory

Def: 37 Let R be a ring. A left R-module is an abelian group (M,+) together with an R-action, i.e., a map fromR×M →M ((r,m) 7→ rm) with the following properties: For all r, s ∈ R and m,n ∈M ,

• r(m+ n) = rm+ rn

• (r + s)m = rm+ sm

• (rs)m = r(sm)

• 1m = m.

E.g.: Let R be a ring

• {0}, R

• Let R be commutative. Any ideal I of R is a left (or right) R-module.

• R a ring, n ∈ N. Define the set Rn := {(x1, · · · , xn) : xi ∈ R}. Rn is a left R-module, and is called a free leftR-module of rank n.

• Let R = Z. Any abelian group (G,+) is a Z-module via the action of adding elements up m or −m times.

• Let R be a ring and n ∈ N. Let S = Mn(R). Then S acts on Rn via matrix multiplication.

• Let ϕ : R → S be a ring homomorphism such that ϕ(1R) = 1S . Then S is a left R module via the actionr · s = ϕ(r)s.

• If R ⊂ S and 1R = 1S , then S is a left R-module via r · s = rs.

Let R be commutative and I and ideal of R. There is a canonical surjective ring homomorphism ϕ : R→ R/Igiven by ϕ(r) = r + I. Thus, R/I is an R-module.

Mike Janssen

4 MODULES Algebra Study Guide 4.1 Intro to Module Theory

Def: Let F be a field. An F -vector space is a left F -module.Def: Let R be a ring and let M be a left R-module. A left R-submodule of M is a subgroup N of M such that N is

a left R-module via the action of R on M restricted to N .Prop: Let M be a left R-module and N a nonempty subset of M . Then N is a left R-submodule of M if for all r ∈ R

and for all x, y ∈ N , x+ ry ∈ N .E.g.: Let R be a ring. Then N =

{(r1, r2, r3) ∈ R3 : r1 + 2r2 − r3 = 0

}is a left R-submodule of R3.

E.g.: Let R be a commutative ring and M a left R-module. Let r ∈ R and N = {m ∈M : rm = 0}. Then N is a leftR-module.

Rmrk: 38 Let ϕ : R→ S be a ring homomorphism with ϕ(1R) = 1S . Then any left S-module M is also a left R-modulein a natural way: ∀r ∈ R, m ∈M , r ·m = ϕ(r) = m.

E.g.: ϕ : Z→ Q. Then Q2 is a Z-module in a natural way.E.g.: Let E be a field and F a subfield. Then ϕ : F → E is a field homomorphism via the inclusion map. Then E is an

F -vector space and F is an F -subspace of E.Def: Let M,N be R-modules. A function f : M → N is an R-module homomorphism (or an R-linear map) if

• f(x+ y) = f(x) + f(y) for all x, y ∈M .

• f(rx) = rf(x) for all r ∈ R, x ∈M .

Rmrk: Given an R-linear map f from M to N , ker(f) := {m ∈M : f(m) = 0} and im(f) := {f(m) : m ∈M}. ker(f)and im(f) are submodules of M and N , respectively.

Def: Let M be an R-module and {Mi}i∈I a collection of submodules of M . Define∑i∈I

Mi :={mi1 + · · ·+min : n ∈ N, mij ∈Mij , for j = 1, · · · , n

}.

Then∑i∈I

Mi is the smallest submodule of M containing∑i∈I

Mi. In the case I = {1, · · · , n}, we write M1 + M2 +

· · ·+Mn =∑i∈I

Mi.

Def: If S is a nonempty subset of M , then RS :=∑s∈S

Rs.

Def: Let M be an R-module. The annihilator of M is defined by

AnnR(M) := {r ∈ R : rm = 0∀m ∈M}.

Note: AnnR(M) is a two-sided ideal of R.Rmrk: Let R be commutative and M an R-module and I an ideal of R with I ⊂ AnnR(M). Then M is an R/I module.Def: 39 Let {M1, · · · ,Mn} be a collection of R-modules. The external direct product is defined to be

M1 ⊕M2 ⊕ · · · ⊕Mn := {(m1, · · · ,mn) : mi ∈Mi}.

The general case: Let {Mi}i∈I be a collection of R-modules. Then⊕i∈I

Mi := {(mi)i∈I : mi ∈Mi, mi = 0 for all but finitely many i}.

Def: Let {Mi}i∈I be a set of R-modules. We define the external direct product to be∏i∈I

Mi := {(mi) : mi ∈Mi∀i}.

Thus,⊕i∈I

Mi ⊆∏i∈I

Mi, with equality if |I| <∞.

Prop: Let M be an R-module and {Mi}i∈I a collection of submodules of M . Then M =⊕i∈I

Mi if and only if (1)

M =∑i∈I

Mi and (2) for all i ∈ I, Mi ∩

(∑j 6=i

Mj

)= (0).

Prop: Let {Mi}i∈I be submodules of M . If M = ⊕i∈IMi (internal) then M ∼=⊕i∈IMi (external).Def: Let M be an R-module. A subset S of M is called an R-basis for M if

Mike Janssen

5 VECTOR SPACES Algebra Study Guide 4.2 Module Isomorphism Theorems

1. M = RS (i.e., S generates M)

2. S is R-linear independent (i.e., if r1s1 + · · ·+ rnsn = 0 (with ri ∈ R, si ∈ S and distinct) then ri = 0 for all i).

If M has an R-basis, then M is called a free R-module.Rmrk: If M has a basis of n elements, then M ∼=Rn.Rmrk: If R is not commutative, then a free module may have bases of different cardinalities. If R is commutative,

then any two bases for a given free module have the same cardinality. This cardinality is called the rank of the freemodule.

Rmrk: Suppose R is commutative. Let I be an ideal. Then I is free if and only if I = (0) or I = Rs = (s), where s is anon-zero divisor.

Rmrk: Let R be commutative and I an ideal of R. Then R/I is free if and only if I = R or I = (0).

4.2 Module Isomorphism Theorems

1. Let ϕ : M → N be an R-module homomorphism. Then M/ kerϕ∼=ϕ(M).

2. Let A,B be submodules of M . Then (A+B)/B∼=A/(A ∩B).

3. Let L ⊆M ⊆ N be R-modules. Then(N/L)/(M/L)∼=N/M.

4. Let N ⊆M be R-modules. Then there exists a one-to-one correspondence between the set of R-submodules ofM containing N and the R-submodules of M/N (i.e., Q↔ Q/N).

5 Vector Spaces

Let R = F a field. Then a set V is an F -module if and only if V is an F -vector space40.Def: If V is an F -vector space and S ⊂ V , then FS = spanF (S). If spanF (S) = V , we say S spans V .Prop: Let F be a field and V and F -vector space. Let T ⊆ S be subsets of V such that (1) T is linearly independent

and (2) S spans V . Then there exists a basis β with T ⊆ β ⊆ S.Cor: Let V be an F -vector space for a field F . Then

1. V has a basis.

2. Every linearly independent subset of V is contained in a basis.

3. Every spanning set for V contains a basis.

Thm: (Replacement Theorem) Let V be an F -vector space. Let S be a spanning set for V and T a linearly independentsubset of V . For every (distinct) t1, · · · , tn ∈ T there exists s1, · · · , sn ∈ S (distinct) such that (S \ {s1, · · · , sn}) ∪{t1, · · · , tn} spans V .

Cor: Let V be a finitely-generated F -vector space. Let S be a finite spanning set for V and T a linearly independentsubset of V . Then |T | ≤ |S|.

Cor: Let V be a finitely-generated F -vector space. Then (1) V has a finite basis and (2) any two bases have the samenumber of elements.

Def: Let V be an F -vector space and dimV = |β| for any basis β of V (finite or infinite).Cor: Let V be a finite-dimensional vector space and W a subspace of V . Then dimW ≤ dimV and W = V if and only

if dimW = dimV .Prop: 41 Let F be a finite field. Then |F | = pn for p ∈ Z a prime and n ≥ 1.Prop: Let ϕ : V → W be a linear transformation of F -vector spaces. Suppose dimV = n < ∞. Let K = kerϕ and

C = imϕ. Then dimF V = dimF K + dimF C.Def: The rank of ϕ is rank(ϕ) := dimC and the nullity of ϕ is nullity(ϕ) := dimK.Cor: Let ϕ : V →W be an F -linear transformation. Suppose dimV = dimW <∞. Then the following are equivalent:

1. ϕ is an isomorphism.

2. ϕ is injective.

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6 MATRIX ALGEBRA Algebra Study Guide

3. ϕ is surjective.

4. ϕ takes a basis of V to a basis of W .

The proof follows from the rank + nullity theorem.

6 Matrix Algebra

6.1 Intro to Matrices

Def: Let R be a commutative ring and m,n ∈ N. An m × n matrix over R is an array (aij) with m rows, ncolumns, and mnth entry amn.

Let Mm×n(R) denote the set of m × n matrices over R. We add them in the usual way. Then (Mm×n(R),+) is anabelian group.

Define an R-action on Mm×n(R) by r · (aij) = (raij) for all r ∈ R. Thus, Mm×n(R) is an R-module.So for 1 ≤ i ≤ m and 1 ≤ j ≤ n let Eij denote the m× n matrix with 1 = aij and 0 elsewhere.Then {Eij : 1 ≤ i ≤ m, 1 ≤ j ≤ n} is an R-basis for Mm×n(R). Then Mm×n(R)∼=Rmn.Let A = (aij) be an m× n matrix and b = (bj1) be an n× 1 matrix. Define

A · b =

c1...cm

,

where ci =n∑j=1

aijbj1.

Prop: Let R be a commutative ring, and M an R-module with B = {e1, · · · , cn} ⊆M . The following are equivalent:

1. B is a basis for M .

2. Given any R-module N and x1, · · · , xn ∈ N , there exists a unique R-module homomorphism ϕ : M → N suchthat ϕ(ei) = xi for all i.

6.2 Transition Matrices

Def: LetR be a commutative ring, and letM andN beR-modules. Define HomR(M,N) = {f : M → N : f R-linear}.Rmrk: HomR(M,N) is an R-module with R-action: for r ∈ R, f : M → N , define r · f : M → N by m 7→ rf(m).Rmrk: If M = N , then HomR(M,M) is a ring with identity (and multiplication given by composition). HomR(M,M)

is usually denoted EndR(M) (the endomorphism ring).Prop: Let R be a commutative ring, M an R-module, β = {e1, · · · , en} ⊆M . The following are equivalent:

1. β is a basis for M

2. Given any R-module N and x1, · · · , xn ∈ N there exists a unique R-homomorphism ϕ : M → N such thatϕ(ei) = xi for all i.

Def: Let R be commutative, M,N free R-modules of ranks m and n, respectively. Let β = {β1, · · · , βm} and E ={E1, · · · , En} be bases for M,N , respectively. Given an R-homomorphism ϕ : M → N we have for 1 ≤ j ≤ m, there

exist unique elements aij ∈ R such that ϕ(βj) =m∑i=1

aijEi. The n×m matrix (aij) is called the matrix of ϕ with

respect to β and E and denoted by MEβ (ϕ).Prop: Let R be commutative, and M and N free R-modules of ranks m,n respectively. Fix bases B, E for M,N . Then

the map from HomR(M,N)→Mn×m(R) taking ϕ 7→MEB (ϕ) is an R-module homomorphism.Prop: Let ϕ : M → N , ψ : L → M be R-module homomorphisms. Let δ,B, E be bases for L, M , and N respectively.

Then MEδ (ϕ ◦ ψ) = MEB (ϕ)MBδ (ψ).Cor: If M is free of rank n, then the map EndR(M)→Mn×n(R) given by ϕ 7→MBB (ϕ) is a ring isomorphism.Cor: Let F be a field and A ∈Mn(F ). TFAE:

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6 MATRIX ALGEBRA Algebra Study Guide 6.3 Hom and Duality

1. A is invertible.

2. Ax = 0 if and only if x = 0 for all x ∈Mn×1(F ).

3. For all y ∈Mn×1(F ) there exists x ∈Mn×1(F ) such that Ax = y.

Def: 42 Let A,B ∈Mn×n(R). Then A and B are called similar if there exists an invertible n× n matrix P such thatA = P−1BP .

E.g.: Let F be a field and let V = {f ∈ F [x] : deg f ≤ 2}. Let B ={

1, x, x2}

and E ={

1, 1 + x, 1 + x+ x2}

. Then

MEB (I) =

1 −1 00 1 −10 0 1

and

MBE (I) =

1 1 10 1 10 0 1

6.3 Hom and Duality

Def: Let R be commutative, M an R-module. The dual module (or dual space) of M is

M∗ := HomR(M,R).

In the case that R is a field, the elements of M∗ are called linear functionals.Prop: Let F be a free R-module of rank n. Let {e1, · · · , en} be a basis for F . For each i ∈ {1, · · · , n}, define maps

e∗i := F → R

by e∗i (ej) = δij (the Dirac delta function). Then {e∗1, · · · , e∗n} is a basis for F ∗ called the dual basis for {e1, · · · , en}.In particular, F ∗ is free of rank n.

Def: Let M be an R-module, m ∈ M . Define a map evm : M∗ → R by f 7→ f(m). Note: evm is R-linear. Therefore,evm ∈ HomR(M∗, R) = M∗∗. This gives a natural homomorphism from M to M∗∗, i.e., ϕ : M → M∗∗ given byϕ(m) = evm.

Rmrk: For finite-dimensional vector spaces, V ∼=V ∗∼=V ∗∗.

6.4 Determinants

Def: 43 Let R be a commutative ring and M an R-module with n ∈ N. Then an n-multilinear form on M is afunction f : Mn → R such that for all i and for all m1, · · · , mi, · · · ,mn ∈M (fixed) the function ϕi : M → R givenby

x 7→ f(m1, · · · ,mi−1, x,mi+1, · · · ,mn)

is an R-module homomorphism. We say f is alternating if for all m1, · · · ,mn ∈ M there exists i such thatmi = mi+1 im plies f(m1, · · · ,mn) = 0.

Prop: Let f be an alternating n-multilinear form on M . Let m1, · · · ,mn ∈M . Then

1. f(m1, · · · ,mi−1,mi+1,mi,mi+2, · · · ,mn) = −f(m1, · · · ,mn) (i.e., the value of ϕ on an n-tuple is negated iftwo adjacent components are interchanged).

2. ∀σ ∈ Sn, f(mσ(1), · · · ,mσ(n)) = sign(σ)f(m1, · · · ,mn).

3. If mi = mj for some i 6= j, f(m1, · · · ,mn) = 0.

4. ∀α ∈ R and i 6= j, f(m1, · · · ,mi + αmj , · · · ,mn) = f(m1, · · · ,mn) (where mi + αmj is in the ith position).

Proof. 1. Let ψ : M×M → R be given by ψ(x, y) = f(m1, · · · ,mi−1, x, y,mi+2, · · · ,mn). then ψ(x+y, x+y) = 0as f is alternating. By the fact that ψ is an R-linear map, ψ(x + y, x + y) = ψ(x, x + y) + ψ(y, x + y) =ψ(x, x) + ψ(x, y) + ψ(y, x) + ψ(y, y) = 0 = ψ(x, y) + ψ(y, x).

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6 MATRIX ALGEBRA Algebra Study Guide 6.4 Determinants

2. Recall: every σ ∈ Sn can be written as a product of transpositions of the form (i i + 1). We do induc-tion on the number of transpositions in such a decomposition. If σ = (i i + 1), we’re done by part 1.Now let σ = τ(i i + 1), where τ is a product of n − 1 such transpositions. Then f(mσ(1), · · · ,mσ(n)) =f(mτ(1), · · · ,mτ(i−1),mτ(i+1),mτ(i), · · · ,mτ(n)) = −f(mτ(1), · · · ,mτ(n)) (by 1) = − sign(τ)f(m1, · · · ,mn) =sign(σ)f(m1, · · · ,mn).

3. Let σ be any permutation which fixes i and sends j to i+1. Then f(m1, · · · ,mn) = sign(σ)f(mσ(1), · · · ,mσ(n)) =0, as mj is now in the i+1st component, giving that f equals 0 by the definition of an alternating n-multilinearform.

4. f(m1, · · · ,mi + αmj , · · · ,mn) = f(m1, · · · ,mn) + αf(m1, · · · ,mj , · · ·mn) = f(m1, · · · ,mn) + 0(by 3, as theith and jth components are equal in the second summand).

Prop: Let f be an alternating n-multilinear form on M . Suppose for j = 1, · · · , n and wj =n∑i=1

αijvi, where wi ∈ M ,

vi ∈M , αij ∈ R. Then

f(w1, · · · , wn) =

(∑σ∈Sn

sign(σ)ασ(1)1 · · ·ασ(n)n

)f(v1, · · · , vn).

Def: Let R be a commutative ring and n ∈ N. A determinant on R is a function det : Mn×n(R)→ R such that detis an alternating n-multilinear function on the columns of R (M = Rn) and such that det(In×n) = 1.

Notation: If A ∈Mn×n(R), let A1, · · · , An denote the columns of A, i.e., det(A1, · · · , An) is alternating multilinear.Thm: There exists a unique determinant function on R. If A = [aij ] then

det(A) =∑σ∈Sn

sign(σ)aσ(1)1 · · · aσ(n)n.

Prop: 44 Let A,B ∈Mn×n(R). Then det(AB) = det(A) det(B).Def: Let A ∈Mm×n(R). Say A = [aij ]. The transpose of A, denoted At, is the n×m matrix whose ijth entry is aji.Prop: Let A ∈Mn×n(R). Then det(A) = det(At).Thm: (Cramer’s Rule) Let A ∈ Mn×n(R) with columns A1, · · · , An. Let [x1, · · · , xn] ∈ Mn×1(R). Let [y1, · · · , yn] =

x1A1 + · · ·+ xnAn (so AX = Y ). Then xi detA = det(A1, · · · , Ai−1, Y, Ai+1, · · · , An).Def: Let A = [aij ] ∈ Mn×n(R). Then the ijth cofactor of A is (−1)i+j det(Aij), where Aij is the n − 1 × n − 1

submatrix of A obtained by deleting the ith row and jth column.

Prop: (Cofactor Expansion) Let A = [aij ]. Fix i ∈ {1, · · · , n}. Then detA =n∑i=1

(−1)i+j det(Aij).

Def: Let A ∈ Mn×n(R). Define the adjoint of A, denoted adjA, to be the n × n matrix whose ijth entry is(−1)i+j det(Aji).

Thm: Let A ∈Mn×n(R). Thenadj(A) ·A = A · adj(A) = det(A)I.

Cor: 45 Let A ∈Mn×n(R). Then A is invertible IFF det(A) ∈ R×.Def: Let R be a domain, A ∈Mn×n(R). Then the column rank of A is the maximum number of linearly independent

columns of A, (and similarly for row rank).Prop: Let R be a domain, A ∈Mn×n(R). TFAE:

1. col rank(A) = n

2. row rank(A) = n

3. det(A) 6= 0.

Def: Let R be commutative, A ∈ Mm×n(R). Let 1 ≤ r ≤ min {m,n}. An r × r minor of A is the determinant of anr × r submatrix of A obtained by deleting any m− r rows and n = r columns.

Def: Define the rth fitting ideal of A, denoted Ir(A), to be the ideal of R generated by all the r × r minors of A.Rmrk:

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7 MODULES OVER A PID Algebra Study Guide

1. I0(A) = R and Ir(A) = 0 for r ≥ min {m,n} by convention.

2. Ir(A) = Ir(At) for all r ≥ 0.

3. Ir+1(A) ⊆ Ir(A) for all r ≥ 0.

4. For any A,B, Ir(AB) ⊆ Ir(A) ∩ Ir(B).

Thm: Let R be a domain and A an arbitrary m× n matrix over R. Then col rank(A) = sup {r : Ir(A) 6= (0)}.

7 Modules over a PID

7.1 Basic Theory

Def: 46 Let M be an R-module. M is called Noetherian if M satisfies the ascending chain condition on submodules,i.e., given any ascending chain

N0 ⊆ N1 ⊆ N2 ⊆ · · ·

of submodules of M there exists t such that for all n ≥ 0, Nt = Nt+n, Nt = Nt+1 = Nt+2 = · · · .We say R is Notherian as a ring if it is Noetherian as an R-modules (i.e., R satisfies ACC on ideals).Prop: Let M be an R-module. The following are equivalent:

1. M is Noetherian

2. Every submodule of M is finitely generated.

3. Every nonempty set of submodules of M has a maximal element.

Cor:R is a Noetherian ring if and only if for all ideals I of R, I is finitely generated.Prop: Let M ⊆ N be R-modules. Then N is Noetherian if and only if M and N/M are Noetherian.Cor: Let M1, · · · ,Mn be R-modules. Then M1 ⊕M2 ⊕ · · · ⊕Mn is Noetherian if and only if each Mi is Noetherian.

Proof. By induction on n (M1 ⊕M2)/M1∼=M2.

Prop: Let R be a Noetherian ring. Then every finitely-generated R-module is Noetherian.Prop: Let R be a domain and M a free R-module of rank n. Then every set of n+1 elements of M is linearly dependent.Def: Let R be a domain and M an R-module. A set x1, · · · , xn ∈ M is said to be a maximal linearly independent set

if {x1, · · · , xn} is linearly independent and {x1, · · · , xn, z} is linearly dependent for all z ∈ M . The rank of M isdefined to be

rank M = sup {n : ∃x1, · · · , xn ∈M forming a maximal LI set}.

Prop: 47 Let R be a domain and let L ⊆M be R-modules. Then

rank M = rank L+ rank M/L.

Cor: rank M1 ⊕ · · · ⊕ rank Mn =n∑i=1

rank Mi.

Proof. Induction and (M1 ⊕M2)/M1∼=M2.

Def: Let R be a domain and M an R-module. Define the torsion submodule of M to be

TorR(M) := {m ∈M : rm = 0 for some r ∈ R \ {0}}.

If M = TorR(M), M is called torsion. If TorR(M) = 0, M is said to be torsion-free.Rmrk: Let R be a domain.

1. If M is free, M is torsion-free.

2. Submodules of torsion-free modules are torsion-free.

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7 MODULES OVER A PID Algebra Study Guide 7.1 Basic Theory

3. R = Z, Z/2 is a torsion Z-module. (2) is torsion-free, as it is isomorphic to Z (as groups).

4. R = Q[x], I = (x, y). Then I is torsion-free but not free (as I is not principal).

5. If M is a finitely-generated torsion module then there exists r ∈ R \ {0} such that rM = 0.

Def: Let R be a commutative ring and M an R-module. Define the annihilator of M to be

AnnR(M) := {r ∈ R : rM = 0}.

It is clear that AnnR(M) is an ideal of R.To restate (5), if M is finitely-generated module and R is a domain, M is torsion if and only if AnnR(M) 6= (0).Thm: Let R be a PID and M a free R-module of rank m <∞ and L a submodule of M . Then L is free of rank ` 6 m.

In fact, there exists a basis y1, · · · , yn of M and a1, · · · , a` ∈ R such that

1. {a1y1, · · · , a`y`} is a basis for L.

2. ai|ai+1 for all i > 1.

Thm: 48 Let R be a PID, M a finitely-generated R-module of rank r. Then there exist nonzero nonunits a1, · · · , at ∈ Rsuch that ai|ai+1 for all i > 1 and

M ∼=Rr ⊕R/(a1)⊕ · · · ⊕R/(at).

The ideals (a1), · · · , (at) are called the invariant factors of m.Cor: Let R be a PID and M a finitely-generated R-module of rank m. Then

1. M is free iff M is torsion-free.

2. M = F ⊕ TorR(M), where F is free of rank m.

Thm: Let R be a PID and M a finitely-generated R-module of rank r. Then there exist prime elements p1, · · · , pr ∈ Rnot necessarily distinct and non-negative integers α1, · · · , αr such that

M ∼=Rr ⊕R/(pα11 )⊕ · · · ⊕R/(pαr

r ).

The ideals (pα11 ), · · · , (pαr

r ) are called the elementary divisors of M .Thm: 49 The elementary divisors and invariant factors are unique.Cor: Let R be a PID and M1,M2 are finitely-generated R-modules.

1. M1∼=M2

2. rank(M1) = rank(M2) and M1 and M2 have the same set of elementary divisors.

3. rank(M1) = rank(M2) and M1 and M2 have the same invariant factors.

Rmrk: Let M be a finitely-generated torsion R-module, where R is a PID. Then AnnR(M) = (at), where (at) is thelargest invariant factor of M , i.e., M ∼=R/(at).

Def: 50 Let A,B ∈Mn(F ). We say A and B are similar if and only if A = PBP−1.Prop: Let S, T ∈ EndF (V ). TFAE:

1. S, T are similar

2. For all bases B of V , MB(S) is similar to MB(T )

3. For all bases B of V there exists a basis B′ of V such that MB′(S) = MB(T ).

Prop: If A,B ∈Mn(F ) are similar, that det(A) = det(B).Def: Let V be a finite-dimensional F -vector space. Let T ∈ EndF (V ). Define

detT := det(MB(T ))

for all bases B of V .Rmrk:

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7 MODULES OVER A PID Algebra Study Guide 7.2 Rational Canonical Form

1. detST = detS detT for all S, T ∈ EndF (V ).

2. detS 6= 0⇔ S is an isomorphism

3. If S, T are similar, then detS = detT .

Def: Let V be a finite-dimensional F -vector space. A nonzero v ∈ V is called a eigenvector of T ∈ EndF (V ) if

Tv = λv

for some λ ∈ F .If such a v exists, λ is called an eigenvalue of T . If λ is an eigenvalue, Eλ(T ) := {v ∈ V : Tv = λv} is called the

eigenspace of T corresponding to λ.Prop: Let T ∈ EndF (V ) and λ ∈ F . Then λ is an eigenvalue for T if and only if det(λI − T ) = 0.Rmrk: The eigenvalues of a linear transformation T are the roots of cT (x).

7.2 Rational Canonical Form

Rmrk: 51 Let F be a field and x a variable.

1. dimF F [x] =∞, with{

1, x, x2, · · ·}

a basis.

2. dimF (F [x]/(f(x))) = n, where deg f = n, with{

1, x, · · ·xn−1}

a basis.

3. dimF EndF (V ) = (dimF V )2 (as EndF (V ) = HomF (V, V )∼=Mn(F )).

Let T ∈ EndF (V ). For all c0, c1, · · · , cn ∈ F , c0 + c1T + · · · + cnTn ∈ EndF (V ), where c ∈ F is identified with

cI ∈ EndF (V ).For a fixed T ∈ EndF (V ) there exists a map

ψT : F [x]→ EndF (V )

given by ψT (f(x)) = f(T ). Note that ψT is a ring homomorphism and an F -linear transformation.By comparing dimensions, kerψT 6= (0). Hence, kerψT = (f(x)) for some nonzero f ∈ F [x]. Therefore, deg f(x) ≤ n2.Def: Let T ∈ EndF (V ). Define the minimal polynomial of T to be the unique monic polynomial mT (x) such that

kerψT = (mT (x)).Let T ∈ EndF (V ) and ψT : F [x] → EndF (V ) the ring homomorphism from before. Thus, V is a left F [x]-module via

ψT , i.e., for v ∈ V ,x · v := T (v),

i.e., as x 7→ T , x · v 7→ T (v).Notation: Let VT denote the F [x]=module V via ψT .Rmrk: Let {v1, · · · , vn} be an F -basis for V . In particular, V = Fv1 + · · · + Fvn. Then VT = F [x]v1 + · · · + F [x]vn.

Therefore, VT is a finitely-generated F [x]-module. Note: V = VT as F -vector spaces.Also note mT (x)VT = 0, so VT is torsion (mT (x) · v = mT (T ) · v = 0(v) = 0 for all v ∈ V ). By the structure theorem,

VT ∼=F [x]/(a1(x))⊕ · · · ⊕ F [x]/(at(x)),

where the ai(x)’s are monic (and hence unique), with ai|ai+1 for all i. The ai’s are called the invariant factors of T .Rmrk: Let f(x) ∈ F [x]. Then f(T ) = 0 ⇔ f ∈ (mT (x)) ⇔ f(x)VT = 0 ⇔ f ∈ AnnF [x] VT ⇔ f(x) ∈ (at(x)) ⇔

(mT (x)) = (at(x))⇔ mT (x) = at(x) .Prop: Let S, T ∈ EndF (V ). TFAE:

1. S and T are similar.

2. VS ∼=VT as F [x]-modules.

3. S and T have the same invariant factors.

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7 MODULES OVER A PID Algebra Study Guide 7.2 Rational Canonical Form

7.2.1 Rational Canonical Form and Minimal/Characteristic Polynomials

Def: 52 Let x be an indeterminate over F . The polynomial det(xI − T ) is the characteristic polynomial of Tand will be denoted cT (x). If A is an n× n matrix with coefficients in F , det(xI − A) is called the characteristicpolynomial of A and will be denoted cA(x).

Def: The unique monic polynomial which generates the ideal Ann(V ) in F [x] is called the minimal polynomial of Tand will be denoted mT (x). The unique monic polynomial of smallest degree which when evaluated at the matrix Ais the zero matrix is called the minimal polynomial of A and will be denoted mA(x).

Prop: The minimal polynomial mT (x) is the largest invariant factor of V . All the invariant factors of V divide mT (x).Def: Let a(x) = xk + bk−1x

k−1 + · · ·+ b1x+ b0 be any monic polynomial in F [x]. The companion matrix of a(x) isthe k×k matrix with 1’s down the first subdiagonal, −b0,−b1, · · · ,−bk−1 down the last column and zeros elsewhere.The companion matrix of a(x) will be denoted by Ca(x). I.e.,

Ca(x) =

0 0 · · · · · · · · · −b01 0 · · · · · · · · · −b10 1 · · · · · · · · · −b2

0 0. . .

......

.... . .

...0 0 · · · · · · 1 −bk−1

Given f ∈ F [x]/(a(x)), Ca(x) is the matrix corresponding to T , which is the linear transformation given by multiplication

of f by x.Thus, if V ∼=F [x]/(a1(x))⊕ · · · ⊕ F [x]/(am(x)), the matrix for the linear transformation T (given by multiplication by

x) has as matrix the direct sum of the companion matrix for the invariant factors, i.e.,Ca1(x)

Ca2(x)

. . .Cam(x)

Def: A matrix is said to be in rational canonical form if it is the direct sum of companion matrices for monic

polynomials a1(x), · · · , am(x) of degree at least one with a1(x)|a2(x)| · · · |am(x). The polynomials ai(x) are calledthe invariant factors of the matrix. Such a matrix is also said to be a block diagonal matrix with blocks thecompanion matrices for the ai(x).

Thm: Let A and B be n× n matrices over the field F . Then A and B are similar if and only if A and B have the samerational canonical form.

Cor: Let A and B be two n× n matrices over a field F and suppose F is a subfield of the field K.

1. The rational canonical form of A is the same whether it is computed over K or over F . The minimal andcharacteristic polynomials and the invariant factors of A are the same whether A is considered as a matrix overF or as a matrix over K.

2. The matrices A and B are similar over K if and only if they are similar over F .

Lma: Let a(x) ∈ F [x] be any monic polynomial.

1. The characteristic polynomial of the companion matrix of a(x) is a(x).

2. If M is the block diagonal matrix

M =

A1 0 · · · 00 A2 · · · 0...

.... . .

...0 0 · · · Ak

given by the direct sum of the matrices A1, · · · , Ak, then the characteristic polynomial of M is the product ofthe characteristic polynomials of A1, · · · , Ak.

Mike Janssen

7 MODULES OVER A PID Algebra Study Guide 7.3 Jordan Canonical Form

Prop: Let A be an n× n matrix over the field F .

1. The characteristic polynomial of A is the product of all the invariant factors of A.

2. The minimal polynomial of A divides the characteristic polynomial of A.

3. The characteristic polynomial of A divides some power of the minimal polynomial of A. In particular thesepolynomials have the same roots, not counting multiplicities.

7.3 Jordan Canonical Form

Rmrk: 53 Let T ∈ EndF (V ) and dimV <∞. We assume F contains all eigenvalues of T . Equivalently, cT (x) factorscompletely into linear factors in F [x]. Then cT (x) = (x− λ1) · · · (x− λn) ∈ F [x], where the λi’s are eigenvalues, notnecessarily distinct. Then the elementary divisor form for T is

VT ∼=F [x]/(p1(x)m1)⊕ · · · ⊕ F [x]/(ps(x)ms).

Note that the pi(x)’s are irreducible and monic, and hence unique.Rmrk: Note that cT (x) = a1(x) · · · at(x) = p1(x)m1 · · · ps(x)ms . Note that pi(x) = x− λi, as the roots of the pi’s are in

F and are the eigenvalues of T . Hence,

VT = F [x]/((x− λ1)m1)⊕ · · · ⊕ F [x]/((x− λs)ms),

where the λi’s are not necessarily distinct.Rmrk: Consider the special case VT ∼=F [x]/((x − λ)m). Then B =

{1, x− λ, · · · , (x− λ)m−1

}is a basis for VT as an

F -v.s. Note dimF VT = m. In general, T ((x − λ)i) = x(x − λ)i = (λ + x − λ)(x − λ)i = λ(x − λ)i + 1 · (x − λ)i+1.Also, T ((x− λ)m−1) = λ(x− λ)m−1. Thus, the Jordan block corresponding to (x− λ)m is

J(x−λ)m = [T ]B =

λ 0 0 · · · 01 λ 0 · · · 00 1 λ · · · 0...

. . . λ 00 0 0 1 λ

In general, if

VT = F [x]/((x− λ1)m1)⊕ · · · ⊕ F [x]/((x− λs)ms),

with Bi ={

1, x− λi, · · · , (x− λi)mi−1}

and B = ∪n1Bi, then

JCF (T ) =

J(x−λ1)m1 0 · · · 0

0 J(x−λ2)m2 0...

0 0. . . 0

0 · · · 0 J(x−λs)ms

E.g.: Let A = [1] ∈M3(Q). Then cA(x) = x2(x− 3) and mA(x) = x(x− 3). The EDs are x, x, x− 3, so

JCF =

0 0 00 0 00 0 3

Prop: Suppose A ∈ Mn(F ) and cA splits over F . Then A is similar to a diagonal matrix if and only if mA has no

repeated roots.Prop: If F contains all eigenvalues for A ∈Mn(F ), then A is similar to a lower (upper) triangular matrix.

Mike Janssen

8 FIELD THEORY Algebra Study Guide

8 Field Theory

8.1 Introduction

Def: 54 Let R be a commutative ring with 1. There is a natural ring homomorphism ϕ : Z → R given byϕ(m) = m · 1R. Then kerϕ = (n) ⊆ Z for a unique n ≥ 0. The characteristic of R is n, where kerϕ = (n), and wewrite charR := n.

Rmrk: If R is a domain, then charR = 0 or p for some prime p > 0.Def: Note that, with definitions as above, imϕ ⊆ R. Further, imϕ is contained in every subring of R containing 1R.

We call imϕ the prime subring of R. If charR = n > 0, then imϕ∼=Z/(n), and if charR = 0, imϕ∼=Z.Def: Let F be a field. The prime subfield of F is defined to be the quotient field of the prime subring of F . The

prime subfield is contained in any subfield of F .Rmrk: If CharF = p > 0, Zp is the prime subfield. If charF = 0, the prime subfield of F is (isomorphic to) Q.Def: Let E be a field and F a subfield. We call E/F a field extension. The degree of E/F , denoted [E : F ], is

defined to be dimF E. E/F is finite if [E : F ] <∞.Prop: If E/F is finite, then E/F is algebraic.Def: Let E/F be a f.e. and α ∈ E. Define F [α] := {f(α) : f(x) ∈ F [x]}. F [α] is the smallest subring of E containing

both F and α. Define F (α) := Q(F [α]).Prop: Let E/F be a field extension and α ∈ E. TFAE:

1. α is algebraic over F .

2. F (α) = F [α].

3. [F (α) : F ] <∞.

Prop: Let E/F be a f.e. and α1, · · · , αn ∈ E. TFAE:

1. α1, · · · , αn are all algebraic.

2. F (α1, · · · , αn) = F [α1, · · · , αn].

3. [F (α1, · · · , αn) : F ] <∞.

Cor: Let E/F be a f.e. and α, β ∈ E. If α, β are algebraic over F , then α ± β, αβ, α/β are all algebraic over F(assuming β 6= 0).

Prop: Let E/F be a f.e. Let L = {α ∈ E : α algebraic over F}. Then L is a subfield of E with F ⊆ L ⊆ E (L is thealgebraic closure of F in E).

Prop: 55 If F ⊆ E ⊆ L are fields, then [L : F ] = [L : E][E : F ].E.g.: [C : R] = 2.Def: Let E/F be a f.e. and let α ∈ E be algebraic over F . Consider the natural ring homomorphism ψ : F [x]→ F [α]

given by f(x) = f(α). Then kerψ = (p(x)) for some non-constant monic p ∈ F [x]. Then p is uniquely-determinedby α. Denote p(x) = Min(α, F ).

Prop: Let E/F be a f.e. and α ∈ E algebraic over F . Let p(x) ∈ F [x] be monic and p(α) = 0. TFAE:

1. p(x) = Min(α, F )

2. p(x) ∈ Irr(F [x])

3. deg p(x) = F [(α) : F ]

4. If f(x) ∈ F [x] \ {0} and f(α) = 0 then deg f(x) ≥ deg p(x).

Rmrk: If f(α) = 0 and f(x) ∈ F [x], then Min(α, F ) = p(x)|f(x).Rmrk: If [F (α) : F ] = n, then F (α) = F [α]∼=F [x]/(p(x)) with deg p = n,

{1, x, · · · , xn−1

}is an F -basis of F [x]/(p(x)),

which implies{

1, α, · · · , αn−1}

is an F -basis of F [x]/(p(x)).E.g.: Find [Q(i, 3

√2) : Q]. Then [Q(i) : Q] = 2 and [Q( 3

√2) : Q] = 3; further [Q(i, 3

√2) : Q(i)] ≤ 3 and [Q(i, 3

√2) :

Q( 3√

2)] ≤ 2. Since i /∈ Q( 3√

2) ⊆ R, [Q(i, 3√

2) : Q( 3√

2)] > 1, so [Q(i, 3√

2) : Q] = 6.Rmrk: 56 If E/F is algebraic, E/F is not necessarily finite. Example: Q = {α ∈ C : α algebraic over Q} and n

√2 ∈ Q.

Rmrk: Let E/F be an alg. extension. Then E/F is finite if and only if E is f.g. as a field over F (IFF E = F (α1, · · · , αn)for some α1, · · · , αn ∈ F .

Prop: Let F ⊆ E ⊆ L be fields. Then L/F is algebraic if and only if L/E and E/F are algebraic.

Mike Janssen

8 FIELD THEORY Algebra Study Guide 8.2 Composite Fields

8.2 Composite Fields

Def: Let E and F be subfields of a field L. The composite field EF of E and F is⋂E∪F⊆T⊆LT a field

T

Suppose:

E

��EF

??

K

??F

��

Then we have the following remarks.Rmrk: EF = E(F ) = F (E).Rmrk: If E = K(α1, · · · , αn), then EF = F (α1, · · · , αn). If additionally F = K(β1, · · · , βl), then EF = K(α1, · · · , αn, β1, · · · , βl).Rmrk: If E/K is algebraic then EF/F is algebraic.Rmrk: If E/F is algebraic then EF = {e1f1 + · · ·+ enfn : ei ∈ E, fi ∈ F, n ∈ N}.Rmrk: [EF : F ] ≤ [E : K].

8.3 Splitting Fields

Prop: 57 Let F be a field and f(x) ∈ F [x] irreducible. Then there exists a field E containing F such that f(x) hasa root in E. If α ∈ E is a root of f(x) then

F (α)∼=F [x]/(f(x)).

Cor: Let F be a field and f(x) ∈ F [x] with deg f ≥ 1. There exists a field E containing a root of F .Def: Let f(x) ∈ F [x] and E an extension field of F . We say f(x) splits in E if f(x) is the product of linear factors in

E[x].Prop: Let f(x) ∈ F [x] \ F . Then there exists a field E containing F such that f(x) splits over E.Def: Let f(x) ∈ F [x] with deg f > 0. A splitting field for f(x) (over F ) is a field E containing F such that f(x)

splits in E but f(x) does not split in any proper subfield of E containing F .Prop: Let f(x) ∈ F [x] \ F . Then a splitting field for f(x) over F exists. In particular, let L be a field containing F in

which f(x) splits. Say f(x) = c(x− α1) · · · (x− αn) for some c ∈ F , αi ∈ L. Then E = F (α1, · · · , αn) is a splittingfield for f(x) over F .

Lma: Let ω be a primitive pth root of unity where p is prime. Then [Q(ω) : Q] = p− 1.

8.4 Algebraic Closure

Def: 58 A field F is algebraically closed if every nonconstant polynomial in F [x] splits into linear factors in F [x].Rmrk: Let F be a field. TFAE:

(1) F is algebraically closed

(2) Every nonconstant polynomial in F [x] has a root in F

(3) 6 ∃ a field properly containing F which is algebraic over F

Def: Let F be a field. An algebraic closure of F is an algebraically closed field containing F which is algebraic overF .

Prop: Let F ⊆ L be fields where L is algebraically closed. Let F = {α ∈ L : α is algebraic over F}. Then F is analgebraic closure of F .

Thm: Let F be a field. Then there exists an algebraic closure of F .Def: 59 Let E/F and L/K be field extensions and τ : E → L and σ : F → K be field homomorphisms. We say τ

extends σ (or “τ is defined over σ”) if τ |F = σ.

Mike Janssen

8 FIELD THEORY Algebra Study Guide 8.5 Splitting Fields

Rmrk: : Let σ : F → K be a field isomorphism. For f(x) = cnxn + · · · + c1x + c0 ∈ F [x], let fσ(x) = σ(cn)xn +

σ(cn−1)xn−1 + · · · + σ(c0) ∈ K[x]. Note: (fg)σ = fσgσ and (f + g)σ = fσ + gσ. Thus, σ : F [x] → K[x] is anisomorphism of polynomial rings sending f to fσ.

Lma: Let σ : F → K be an isomorphism of fields. Let F and K be algebraic closures of F and K respectively.Let f(x) ∈ F [x] be an irreducible polynomial and β ∈ K be a root of fσ(x). Then there exists an isomorphismτ : F (α)→ K(β) which extends σ such that τ(α) = β, i.e.,

F (α) τ

α7→β// K(β)

F σ// K

Thm: Let E/F and L/K be field extensions. Let σ : F → K be a nonzero field map. Suppose E/F is algebraic and Lis algebraically closed. Then there exists τ : E → L which extends σ.

Cor: 60 Let F be a field and say E1, E2 are two algebraic closures of F . Then there exists an isomorphism τ : E1 → E2

which fixes F (i.e., τ(x) = x for all x ∈ F ).

8.5 Splitting Fields

Def: Let F be a field and S ⊂ F [x] \F . A splitting field for S (over F ) is a field E containing F such that everyf ∈ S splits in E[x] and E is minimal with respect to this property.

Rmrk: Let S ⊂ F [x]\F as above and let F be any algebraic closure of F . Let T ={α ∈ F : α is a root of some f ∈ S

}.

Then E = F (T ) is a splitting field for S.Exercise: Suppose E is a splitting field for a subset S ⊂ F [x] \F . Then E is a splitting field for f ∈ F [x] if and only if

E/F is finite.Prop: Let F be a field and S ⊂ F [x] \ F . Let E1 and E2 be splitting fields for S over F . Then there exists an

isomorphism τ : E1 → E2 fixing F .Prop: Let E/F be an algebraic extension and σ : E → E be a field homomorphism fixing F . Then σ is an isomorphism.Thm: Let E/F be an algebraic extension. The following are equivalent:

1. E be a splitting field for some set S ⊆ F [x] \ F .

2. Every irreducible f(x) ∈ F [x] which has a root in E splits completely over E.

3. Let E be an algebraic closure of E and σ : E → E be any field homomorphism fixing F . Then σ(E) = E, i.e.,σ is an automorphism of E.

Def: An algebraic extension E/F satisfying 1-3 above is called a normal extension.Def: 61 Let F be a field, f(x) ∈ F [x], and α ∈ F a root of f(x). We say α is a repeated root of f(x) if (x− α)2|f(x)

in F [x].Def: Let F be a field and x an indeterminate over F . Define an F -linear transformation Dx : F [x] → F [x] by

Dx(xn) = nxn−1 for n ≥ 1 and Dx(1) = 0.Exercise: Let E/F be a field extension and f, g ∈ F [x]. Then gcdF [x](f, g) = gcdE[x](f, g).Prop: Let F be a field and f(x) ∈ F [x] \ F . Then f(x) has a repeated root if and only if gcd(f, f ′) 6= 1.Def: Let F be a field. A polynomial f(x) ∈ F [x] \ F is separable if f(x) has no repeated roots. Let α ∈ F . We say α

is separable over F if Irr(α, F ) is separable. An algebraic extension E/F is called separable if every element ofE is separable over F .

Def: F is called a perfect field if every algebraic extension of F is separable.Prop: Any field of characteristic 0 is perfect.Prop: Any finite field is perfect.Prop: Let F be a finite field. Then (F×, ·) is a cyclic group.Rmrk: 62 If F is a finite field then |F | = pn for p = charF .Thm: Let p be a prime and n > 0. Let E be a splitting field for xp

n − x over Fp. Then |E| = pn. Moreover, any otherfield of order pn is also a splitting field for xp

n − x over Fp. Therefore, any two fields of order pn are isomorphic.Prop: Let F be a finite field. Then every algebraic extension of F is normal.

Mike Janssen

9 GALOIS THEORY Algebra Study Guide 8.6 Cyclotomic Stuff

8.6 Cyclotomic Stuff

Def: Let n > 0 be an integer, w = e2πi/n ∈ C a primitive nth root of unity. Define the nth cyclotomic polynomial

Φn(x) :=∏

1≤i≤ngcd(i,n)=1

(x− wi) =∏

τ∈C annth root of unity

(x− τ) ∈ C[x].

Rmrk: Φ1(x) = x− 1, Φ2(x) = x+ 1Rmrk: xn − 1 =

∏d∈Nd|n

Φd(x)

Rmrk: x3 − 1 = Φ1(x) = Φ3(x) so Φ3(x) =x3 − 1x− 1

= x2 + x + 1, Φ4(x) =x4 − 1Φ1Φ2

= x2 + 1, Φ5(x) =x5 − 1x− 1

=

x4 + x3 + x2 + x+ 1, etc.Rmrk: deg Φn = ϕ(n), where ϕ is Euler’s ϕ-function.Rmrk: 63 Let F be a field and R a subring of F . Let f, g ∈ R[x] with f monic. Suppose f |g in F [x]. Then f |g in R[x].Lma: Φn(x) ∈ Z[x] for all n ≥ 1.Thm: Φn(x) is irreducible over Q for all n.Cor: If w ∈ C is a primitive nth root of unity, [Q(w) : Q] = ϕ(n) = deg Φn(x) and Φn = Irr(w,Q).

9 Galois Theory

9.1 Basic Definitions

Def: Let E/F be a field extension. Define Aut(E) := {σ : E → E : σ is a field isomorphism} to be the automor-phism group of E (which is a group under composition). Define Aut(E/F ) := {σ ∈ Aut(E) : σ|F = 1F }.

Rmrk: If L is the prime subfield of E then Aut(E) = Aut(E/L).Rmrk: Let σ ∈ Aut(E/F ) and f(x) ∈ F [x]. If α ∈ E is a root of f(x) then σ(α) is a root of f(x).E.g.: 64 Aut(Q( 3

√3)) = 1, as the other roots of x3 − 2 are non-real.

E.g.: Let E be the splitting field of x6 + 3 over Q. Then Aut(E) = S3 or Aut(E) = C6.Lma: Any algebraic extension of a perfect field is perfect.Thm: Let F be a perfect field and σ : F → F ′ a field isomorphism (so F ′ is also perfect). Let f(x) ∈ F [x] and E be a

splitting field for f(x) over F . Let E′ by a splitting field for fσ(x) over F ′.Let ΛEσ = {τ : E → E′ : τ is an isomorphism such that τ |F = σ}. Then |ΛEσ | = [E : F ].

Cor: Let F be perfect and f(x) ∈ F [x] and E a splitting field for f over F . Then |Aut(E/F )| = [E : F ].Def: Let E/F be algebraic. Then E/F is called Galois if E/F is normal and separable. If E/F is Galois, then the

Galois group of E/F is Aut(E/F ), usually denoted Gal(E/F ).Rmrk: We’ve proved if F is perfect and E/F is a finite normal extension then |Gal(E/F )| = [E : F ].Rmrk: Suppose F ⊆ L ⊆ E and E/F is Galois. Then E/L is also Galois but L/F may not be.E.g.: 65 Let E = Q(

√2,√

3). Note that E is a splitting field for (x2− 2)(x2− 3) which implies E/Q is normal. Further,|Gal(E/Q)| = [E : Q] = 4. If σ ∈ Gal(E/Q) then σ is determined by σ(

√2) and σ(

√3). There are only 4 possibilities,

and hence they are all elements of Gal(E/Q).Prop: Let ω ∈ C be a primitive nth root of unity. Then Q(ω)/Q is Galois and Gal(Q(ω)/Q)∼=Z×n . In particular, if n

is prime, then Gal(Q(ω)/Q) is cyclic of order n− 1.

9.2 The Fundamental Theorem of Galois Theory

Thm: (Fundamental Theorem of Galois Theory) Let K/F be a Galois extension and set G = Gal(K/F ). Then thereis a bijection between the subfields E of K containing F and the subgroups H of G given by the correspondencesE 7→ {the elements of G fixing E} and H 7→ {the fixed field of H} which are inverse to each other.

Prop: Let E/F be a finite Galois extension and letG = Gal(E/F ). Then EG = F (where EG := {u ∈ E : σ(u) = u∀σ ∈ G}).Cor: Let E/F be a finite Galois extension and L an intermediate field. Then EGal(E/L) = L.Cor: Let E/F be a finite Galois extension. The map ψ(L) = Gal(E/L) for L ∈ {intermediate fields of E/F} is 1-1 and

hence there are only finitely many intermediate fields.

Mike Janssen

9 GALOIS THEORY Algebra Study Guide 9.3 Radical Extensions

Thm: 66 (Primitive Element Theorem) Let E/F be a finite extension with only finitely many intermediate fields. Thenthere exists α ∈ E such that E = F (α). Then α is called a primitive element for E/F .

Cor: Let E/F be a finite Galois extension and L an intermediate field. Then L = F (α) for some α ∈ L.Cor: Let F be a perfect field, and E/F a finite extension. Then E = F (α) for some α ∈ E.Lma: Suppose E/F is a Galois extension (not necessarily finite). Suppose there exists n ≥ 1 such that [F (α) : F ] ≤ n

for all α ∈ E. Then E/F is finite and [E : F ] ≤ n.Thm: (Artin) Let E be a field and G a finite subgroup of Aut(E). Let F be the fixed field of G, i.e., F = EG.

1. E/F is finite Galois, and

2. Gal(E/F ) = G.

Rmrk: On FTGT: Gal(E/EH) = H for some H ≤ Gal(E/F ) (and EH an intermediate field).Rmrk: On FTGT: L1 ⊆ L2 if and only if Gal(E/L1) ≥ Gal(E/L2).Rmrk: On FTGT: H1 ≤ H2 if and only if EH1 ⊇ EH2 .Rmrk: On FTGT: [E : EH ] = |H|Rmrk: On FTGT: [EH : F ] = [G : H].Thm: 67 Let E/F be a finite Galois extension and G = Gal(E/F ). Let L be an intermediate field and H = Gal(E/L).

Then

1. L/F is normal if and only if H C G.

2. If L/F is normal, then Gal(L/F )∼=G/H.

Prop: 68 Let F be a perfect field, f(x) ∈ F [x] of degree n. Assume f(x) has distinct roots (e.g., if f is irreducible).Let E be a splitting field for f(x) over F . Then E/F is finite Galois, and Gal(E/F )∼=H ≤ Sn.

Cor: [E : F ] ≤ n!.

9.3 Radical Extensions

Def: 69 A finite extension of fields E/F is called radical if there exist elements u1, · · · , un ∈ E such that E =F (u1, · · · , un) and for each i, there exists mi such that umi

i ∈ F (u1, · · · , ui−1), i.e., there exists a root towerF = F0 ⊆ F1 ⊆ · · · ⊆ Fn = E, where Fi = Fi−1(ui) and umi

i ∈ Fi−1. A polynomial f(x) ∈ F [x] is called solvableby radicals if f(x) splits in some radical extension of F .

Prop: Suppose E = F (u), where charF = 0 and un ∈ F , where F contains a primitive nth root of unity. Then E/F isGalois and Gal(E/F ) is cyclic.

Lma: 70 Let E/F be a finite radical extension and charF = 0. Then E ⊆ E′ such that E′/F is finite, Galois, andradical.

Thm: 71 Let E/F be Galois, radical with charF = 0. Then the Galois group is solvable.

Proof. Let F = F0 ⊂ F1 ⊂ · · · ⊂ Fn = E, Fi = Fi−1(αi) and αmii ∈ Fi−1. Let m = m1 · · ·mn, and ω a primitive

mth root of 1. Then Gal(E/F )∼= Gal(E(ω)/F )/Gal(E(ω)/E). Hence if Gal(E(ω)/F ) is solvable, so is Gal(E/F ),as quotients of solvable groups are solvable. Now Gal(E(ω)/F ) is solvable if and only if Gal(E(ω)/F (ω)) is solvableand Gal(F (ω)/F ) is solvable. By the homework exercise, Gal(F (ω)/F ) is abelian and hence solvable. It is enoughto show that Gal(E(ω)/F ) is solvable. (Hence we may assume F (in our original setup) contains a primitive mithroot of unity.) By a previous proposition, Fi/Fi−1 is Galois and has cyclic Galois group. let Hi = Gal(E/Fi). Thenwe have {1} = Hn C Hn−1 C · · · C H0 = Gal(E/F ). The quotients Hi−1/Hi

∼= Gal(Fi/Fi−1), which is cyclic. Thisgives a solvable series for G.

Thm: (Galois) Let F be a field of characteristic 0, f(x) ∈ F [x]. Let E be a splitting field for f over F . Then if f(x) issolvable by radicals over F , then Gal(E/F ) is solvable.

Proof. If f(x) is solvable by radicals, there exists a finite radical extension L/F such that E ⊆ L. By a lemma, thereexists a Galois radical extension L′/F with L ⊆ L′. So by the theorem, Gal(L′/F ) is solvable, and Gal(E/F ) =Gal(L′/F )/Gal(L′/E).

Mike Janssen

9 GALOIS THEORY Algebra Study Guide 9.3 Radical Extensions

Prop: Let charF = 0, and let f(x) ∈ F [x] be irreducible of degree p, where p is a prime. Let E be the splitting fieldfor f(x) over F . Suppose there exists an automorphism σ : E → E fixing F and all but two roots of f . ThenGal(E/F )∼=Sp.

E.g.: Let f(x) = x5−2x3−8x−2 ∈ Q[x]. Then f is not solvable by radicals, as complex conjugation is an automorphismof C fixing 3 real roots and transposing 2 others. Then Gal(E/F )∼=S5, which is not solvable by group theory.

Def: Let F be a field, x, t1, · · · , tn indeterminates over F . Then the polynomial fn(x) = xn + t1xn−1 + · · · tn−1x+ tn ∈

F (t1, · · · , tn)[x] is called the general equation of degree n over F . We say fn(x) is solvable by radicals if theroots of fn(x) lie in a radical extension of F (t1, · · · , tn).

Thm: Let E be a splitting field for fn(x) over F (t1, · · · , tn). Then Gal(E/F (t1, · · · , tn))∼=Sn.Thm: 72 (Fundamental Theorem of Algebra) C is algebraically closed.

Mike Janssen

10 OLD EXAMS Algebra Study Guide

10 Old Exams

10.1 Exam 1, Fall 2007

1. Let G be a group and Z the center of G. Provethat if G/Z is cyclic then G is abelian. Also, givean example (no proof necessary) where this resultfails if the hypothesis is changed to assume G/Z isabelian.

2. Let H be a subgroup of a finite group G and con-sider the set

Λ :={xHx−1 : x ∈ G

}.

(a) Prove that |Λ| = [G : NG(H)] where NG(H) ={g ∈ G : gHg−1 = H

}.

(b) Prove that⋃x∈G xHx

−1 = G if and only ifH = G. (Hint: Use part (a) to show that ifH 6= G then the number of elements int heunion of the conjugates of H is less than |G|.)

3. Let G be a finite group of whose order is square-free.(That is, no square integer greater than 1 divides|G|.) Let p be the smallest prime dividing |G| andsuppose G contains a subgroup H of index p. Provethat H is the only subgroup of G of index p.

4. Let G be a group with exactly three conjugacyclasses. Prove that G∼=C3 or G∼=S3.

10.2 Exam 2, Fall 2007

1. Let G be a finite group and suppose G has a uniquesubgroup of order d for each positive divisor d of|G|. Prove that G is cyclic. (October 24, 2007)

2. Prove that if |G| = 3·5·7 then G has a cyclic normalsubgroup of order 35.

3. Let G be a finite group and suppose pn divides |G|for some prime p and positive integer n. Prove thatG has a subgroup of order pn.

4. Let G be a finite group and suppose G contains asimple subgroup H 6= 1 such that [G : H] = 2. Sup-pose further that |Z(G)| is odd. Prove that H is theonly nontrivial normal subgroup of G.

5. Let G be a group and K a finite cyclic normal sub-group of G. Prove that G′ ⊆ CG(K), where G′

is the commutator subgroup of G and CG(K) ={g ∈ G : gk = kg for all k ∈ K}. (Hint: Let g ∈ Gand consider the inner automorphism ψg of G.Since K is normal, ψg restricted to K is an au-tomorphism of K. This gives a natural homomor-phism ϕ : G → Aut(K). Examine the kernel andthe image of this map.)

6. Let G be a group of order 3 · 7 · 11. Suppose Gcontains an element of order 21. Prove that G iscyclic.

10.3 Final Exam, Fall 2007

1. Let R be a ring and I1, · · · , In ideals of R which arepairwise co-prime (i.e., Ii + Ij = R for all i 6= j).Prove that

I1I2 · · · In = I1 ∩ I2 · · · In.

2. Let I be an ideal of R and define the radical√I of

I by√I := {r ∈ R : rn ∈ I for some n}.

Prove that√I is an ideal of R and that R/

√I has

no nonzero nilpotent elements.

3. Prove that every Euclidean domain is a PID.

4. Prove that Z[√−5] is a UFD. (Show all details.)

5. Prove that the following polynomials are irreducibleover the given field:

(a) x5 − 2x+ 2 over Q.(b) x3 + x2 + 2 over F3.(c) x4 + x2 + 6 over Q.

6. Let R be a ring and r ∈ R such that r is not nilpo-tent. Let Λ be the set of all ideals I of R such thatrn /∈ I for all n ≥ 1. Prove that Λ is a maximal ele-ment and that any maximal element of Λ is a primeideal.

10.4 Exam 1, Spring 2008

1. Let M be an R-module and ϕ : M →M a surjectiveR-module homomorphism.

(a) Suppose M is Noetherian. Prove that ϕ is anisomorphism.

(b) Give an example of such an M and ϕ such thatϕ is not an isomorphism.

2. Let R be a domain and M an R-module. Recallthat a subset S of M is called a maximal linearlyindependent set of M if S is linearly independentand any subset of M properly containing S is lin-early dependent.

(a) Let T be a linearly independent subset of M .Prove that T is contained in some maximallinearly independent subset of M .

(b) Let T be a linearly independent subset of Mand N the R-submodule of M generated byT . Prove that T is a maximal linearly inde-pendent subset if and only if M/N is torsion.

Mike Janssen

10 OLD EXAMS Algebra Study Guide 10.5 Exam 2, Spring 2008

3. Let I be an ideal of R.

(a) Prove that I is a free R-module if and only ifI = (0) or I = (a) for some a ∈ R which is nota zero-divisor.

(b) Give an example of a submodule of a free R-module which is not free.

4. Let I be an ideal of R and M a finitely-generatedR-module.

(a) Suppose IM = M . Prove there exists an ele-ment a ∈ I such that (1− a)M − 0.

(b) Suppose I is finitely generated and I2 = I.Prove that I = (e) for some e ∈ R such thate2 = e.

5. Let F be a field and V a finite dimensional F -vectorspace. Let W be a subspace of V . Prove thatdimV = dimW + dimV/W .

6. Let A = [aij ] be an n × n matrix over a field F .Define the trace of A, denoted tr(A), by tr(A) =∑ni=1 aii.

(a) Prove that for n × n matrices A, tr(AB) =tr(BA).

(b) If A and B are similar, prove that tr(A) =tr(B).

(c) Let A be a 2 × 2 matrix which is not a scalarmatrix (i.e., not a scalar multiple of the iden-tity matrix.) Prove that A is similar to aunique matrix of the form:[

0 −det(A)1 tr(A)

].

7. Let A be a matrix with rational entries whose char-acteristic polynomial is c(x) = x2(x− 1)2(x+ 1).

(a) Find all possible sets of invariant factors for A.

(b) Find a specific matrix A as above whose min-imal polynomial has degree 3.

8. Find the RCF of the following matrix in M3(Q):1 1 11 1 11 1 1

10.5 Exam 2, Spring 2008

1. Let E be a splitting field of x6 + 5 over Q. Find[E : Q]. Justify all details.

2. Let f(x) be an irreducible polynomial of degree nover a field F . Let g(x) be any nonconstant poly-nomial in F [x] and set h(x) = f(g(x)). Prove thatif q(x) ∈ F [x] is any irreducible factor of h(x) thendeg q is divisible by n. (Hint: If α is a root of h(x)then g(α) is a root of f(x).)

3. Let E be a splitting field for x4 − 4x2 − 1 over Q.Find the Galois group of E/Q. That is, explicitlydescribe all automorphisms of E and identify thegroup structure (e.g., by showing it is isomorphicto some well-known group).

4. Let E/F be a finite Galois field extension and Kand L intermediate fields. Suppose E = KL andF = K ∩ L.

(a) Give an example of fields E, F , K, and L asabove such that [E : F ] 6= [K : F ][L : F ].

(b) If either K or L is normal over F , prove that[E : F ] = [K : F ][L : F ].

5. Let ω be a primitive 7th root of unity and letE = Q(ω). Find all subfields of E and primitiveelements (over Q) for each.

6. Let E/F be a finite Galois extension and K anintermediate field. Let G = Gal(E/F ) and H =Gal(E/K). Let NG(H) denote the normalizer of Hin G. Prove that NG(H) = {g ∈ G : g(K) = K}and NG(H)/H ∼= Aut(K/F ).

7. Let Q be the algebraic closure of Q in C and F amaximal subfield of Q not containing

√5 (such a

field exists by Zorn’s Lemma). Let E be a finitenormal extension of F . Prove that Gal(E/Q) iscyclic.

10.6 Final Exam, Spring 2008

1. Let R be a ring and M an R-module and N a sub-module of M . Prove that M is Noetherian iff Nand M/N are Noetherian.

2. Let R be a domain and M a finitely-generated tor-sion R-module. Prove there exists a nonzero r ∈ Rsuch that rM = 0, i.e., r annihilates M . Givean example of a torsion Z-module M such that nononzero element of Z annihilates M .

3. Consider the matrix A below with rational coeffi-

Mike Janssen

10 OLD EXAMS Algebra Study Guide 10.6 Final Exam, Spring 2008

cients.

A =

1 0 0 0 0 0 00 0 0 0 0 0 00 1 1 0 0 0 00 0 0 0 0 0 00 0 0 1 0 0 00 0 0 0 1 0 −10 0 0 0 0 1 2

.

(a) Find the rank of A.

(b) Find the invariant factors and the characteris-tic polynomial of A.

(c) Find the eigenvalues of A.

(d) Find the JCF of A.

4. Let F be a field and V an F -vector space (but notnecessarily of finite dimension). Let S be a subsetof V which spans V . Prove that S contains a basisfor V .

5. Let E/F be a finite field extension. Suppose f :E → E is a field homomorphism which fixes F .Prove that f is an automorphism of E.

6. Prove that the group of units of a finite field iscyclic.

7. Let E be the splitting field for x5 − 2 over Q andG = Gal(E/Q). Find |G| and a set of generators forG. Also, find generators for the normal subgroup ofH such that G/H is cyclic of order 4.

8. Let F be a finite field. Prove that F [x] has an irre-ducible polynomial of degree m for every m ≥ 1.

9. Let R be a commutative ring and M an R-module.Let M∗ denote the R-module HomR(M,R). Form ∈ M , let evm : M∗ → R be the map defined byevm(f) = f(m) for all f ∈M∗.

(a) Prove evm ∈ HomR(M∗, R) = M∗∗.

(b) Prove the map ϕ : M → M∗∗ defined byϕ(m) = evm is an R-module homomorphism.

(c) Assume R is a domain and ϕ : M → M∗∗ isinjective. Prove that M is torsion-free.

10. Let F be a field and A a square matrix with en-tries from F . Prove that A is similar to AT , thetranspose of A.

Mike Janssen

NOTES Algebra Study Guide NOTES

Notes1Math 817 Class notes, August 29, 20072Math 817 Class notes, August 31, 20073Math 817 Class notes, September 5, 20074Math 817 Class notes, September 10, 20075Math 817 Class notes, September 12, 20076Math 817 Class notes, September 14, 20077Math 817 Class notes, September 17, 20078Math 817 Class notes, September 19, 20079Math 817 Class notes, September 21, 2007

10Math 817 Class notes, September 24, 200711Math 817 Class notes, September 26, 200712Math 817 Class notes, September 28, 200713Math 817 Class notes, October 1, 200714Math 817 Class notes, October 3, 200715Math 817 Class notes, October 10, 200716Math 817 Class notes, October 12, 200717Math 817 Class notes, October 17, 200718Math 817 Class notes, October 19, 200719Math 817 Class notes, October 24, 200720Math 817 Class notes, October 26, 200721Math 817 Class notes, October 29, 200722Math 817 Class notes, November 7, 200723Math 817 Class notes, November 9, 200724Math 817 Class notes, November 12, 200725Math 817 Class notes, November 14, 200726Math 817 Class notes, November 14, 200727Math 817 Class notes, November 16, 200728Math 817 Class notes, November 19, 200729Math 817 Class notes, November 26, 200730Math 817 Class notes, November 30, 200731Math 817 Class notes, December 3, 200732Math 817 Class notes, December 5, 200733Math 817 Class notes, December 7, 200734Math 817 Class notes, December 10, 200735Math 817 Class notes, December 12, 200736Math 817 Class notes, December 14, 200737Math 818 Class notes, January 14, 200838Math 818 Class notes, January 16, 200839Math 818 Class notes, January 18, 200840Math 818 Class notes, January 23, 200841Math 818 Class notes, January 25, 200842Math 818 Class notes, January 30, 200843Math 818 Class notes, February 1, 200844Math 818 Class notes, February 4, 200845Math 818 Class notes, February 6, 200846Math 818 Class notes, February 8, 200847Math 818 Class notes, February 11, 200848Math 818 Class notes, February 13, 200849Math 818 Class notes, February 15, 200850Math 818 Class notes, February 18, 200851Math 818 Class notes, February 20, 200852Adapted from Section 12.2 of Dummit and Foote53Math 818 Class notes, February 29, 200854Math 818 Class notes, March 3, 200855Math 818 Class notes, March 5, 200856Math 818 Class notes, March 7, 200857Math 818 Class notes, March 14, 200858Math 818 Class notes, March 24, 200859Math 818 Class notes, March 26, 200860Math 818 Class notes, March 28, 200861Math 818 Class notes, March 31, 200862Math 818 Class notes, April 2, 200863Math 818 Class notes, April 4, 200864Math 818 Class notes, April 7, 200865Math 818 Class notes, April 9, 200866Math 818 Class notes, April 11, 2008

67Math 818 Class notes, April 14, 200868Math 818 Class notes, April 16, 200869Math 818 Class notes, April 21, 200870Math 818 Class notes, April 23, 200871Math 818 Class notes, April 25, 200872Math 818 Class notes, April 28, 2008

Mike Janssen

11 NOTES Algebra Study Guide

11 Notes

Mike Janssen

11 NOTES Algebra Study Guide

Mike Janssen

11 NOTES Algebra Study Guide

Mike Janssen

11 NOTES Algebra Study Guide

Mike Janssen

11 NOTES Algebra Study Guide

Mike Janssen

11 NOTES Algebra Study Guide

Mike Janssen

11 NOTES Algebra Study Guide

Mike Janssen