Number Theory - Notesgpagi/PDF/numTh.pdfRDedekind, K= Frac(R). The set of Fractional ... Number...

Number Theory - Notes

Gilad Pagi

Las update - October 28, 2016

1 Reminders

1. Primitive Element Theorem: Every finite and separable extension F/K (over Q for example) issimple, F = K(α). Thus algebraic.

2. Gauss’s Lemma: A polynomial in Z[x] ⊂ Q[x] is irreducible in Z[x], then it is irreducible in Q[x]. Theconverse is true for monic polynomials.

3. Every extension of Q is separable: every minimal polynomial has no multiple roots.

• L/K separable of degree n, then there are n embeddings L→ K, fixing K.

4. K : Q is Galois if |AutQ(K)| = [K : Q].

5. K : Q is Galois ⇐⇒ K is a splitting field over Q.

6. Let K : Q be Galois. Given p(x) ∈ Q[x], the elements of the Galois group permute the roots. If K = Q(α)and g(x) is α’s min poly with d roots not in Q, then the Galois group G injects into Sd.

7. Def: the splitting filed of xn − 1 over Q is denoted Q(ζn) - cyclotomic extension.

8. Let R be a ring:

• an irreducible element is a ∈ R, not a unit, such that if a = bc, one of them is a unit.

• a prime element is p ∈ R, not a unit, such that p|ab⇒ p|a or p|b.• In a domain: prime is irreducible. The converse is true in a UFD.

9. Euclidian domain ⇒ PID ⇒ UFD.

10. Let K/Q. The norm of α ∈ K, NK/Q(α), is the determinant of the linear map K → K over Q, x 7→ αx.

• If Glaois with group G, NK/Q(α) =∏G g(α)

11. Let K/Q. The trace of α ∈ K, TrK/Q(α), is the trace of the linear map K → K over Q,x 7→ αx.Tr : K → Q, a linear map.

• If Glaois with group G, TrK/Q(α) =∑G g(α).

• It is an additive K-linear map.

12. Def: Let V be a K-vector space. Let B be a bilinear form V × V → K. It is non-degenerate ifB(x, y) = 0∀y ⇒ x = 0 and same for y. Equivalently, fix x 6= 0. Then B(x, ) : V ] ∼= HomK(V,K) =V ∗, x 7→ [y 7→ B(x, y)].

13. Structure of finitely generated modules over a PID For every finitely generated module M over aprincipal ideal domain R, there is a unique decreasing sequence of proper ideals (d1) ⊇ (d2) ⊇ · · · ⊇ (dn)such that M is isomorphic to the sum of cyclic modules:

M ∼=⊕i

R/(di) = R/(d1)⊕R/(d2)⊕ · · · ⊕R/(dn)

Some of those ideals may be (0) which constitutes the free part.

14. Finite extension of Finite fields has a cyclic Galois group.

15. G(L/K galois. H,H ′ ≤ G fixing F, F ′, Then H ∩H ′ fixes FF ′.

1

Number Theory Notes - Gilad Pagi 2

16. Euler function:

φ(pk) = pk − pk−1 = pk−1(p− 1) = pk(

1− 1

p

)17. Let ζ be a pth root of unity. Then:

xpr − 1

xpr−1−1 = 1 + xpr−1

+ (xpr−1

)2 + ...+ (xpr−1

)p−1 =∏

(k,pr)=1

(x− ζk)

18. Let R be a normal domain and let S be a domain generated over R by one integral element, s. Thenthe minimal monic polynomial f(x) of s over the fraction field of R has coefficients in R and S ∼=R[x]/f(x)R[x].

19. Let R be a PID. Let M be a free R module of rank n. Let N be a submodule. Then N is free of rankm ≤ n. There exists x1, ..., xn is a free basis for M , and r1, ..., rm ∈ R such that r1x1, ..., rmxm is a freebasis for M .

2 Main Results

1. 2016-Sep-6,p11

2. Def:

• A number field is a finite field extension of Q (thus, algebraic).

• An algebraic number is a number is a number field.

• An Algebraic integer is an algebraic number α that satisfies a monic polynomial in Z[x]. For afield K, the set of algebraic integers is denoted OK .

3. OQ = Z.

4. K = Q(√d) then:

• OK = Z + Z√d if d = 2, 3 mod 4, and then disc(K) = 4d

• OK = Z + Z(1 +√d)/2 if d = 1 mod 4. disc(K) = d

5. OQ(ζn) = Z[ζn].

6. If α ∈ OK then its min poly over Q is in Z[x].

7. Let K be a number field. Then OK is a ring (subring of K, thus a domain).

8. Usually OK is not a UFD.

9. Let R be a domain with factorization: every element can be written as a finite product of a unit andirreducibles. Then R is a UFD ⇐⇒ every irreducible is a prime.

10. A finitely generated module over a noetherian ring R is noetherian.

11. 2016-Sep-8, p4

12. Def: If R is a domain, we say that R is integral closed (normal) is the set of integral elements of Frac(R)over R is in R.

13. R domain. S an R algebra. s ∈ S satisfies sI ⊂ I for some ideal in R. Then s is integral over R.

14. R ↪→ S -an R algebra. S is module finite over R, then it is integral over R.

15. Let A - a domain, K = Frac(A), L some finite separable extension of K, B the integral closure of A inL. Let A be noetherian and integrally closed. Then B is a finitely generated A-module, and a noetherianring as well.

B

int

L

fin.sep

A K1What is MAGMA?

Last Updated: October 28, 2016 2


16. For L/K degree n and separable, then we have σ1, ..., σn embedding of L ↪→ K fixing K (which are infact automorphism of the galois closure of L over K), and TrL/K(α) =

∑i σi(α).

17. L/K finite and sep, then TrL/K 6= 0. If not sep, Tr = 0.

18. L/K finite and sep. TrL/K : L×L→ K is a non-degenerate bilinear form. So L ∼= HomK(L,K) = L∗byx 7→ [y 7→ Tr(xy)]. Fix a basis x1, .., xr of L/K. Then there exists a dual basis with respect to the trace:y1, ...yr such that Tr(xiyj) = δij .

19. For a number field K, OK is a noetherian domain, finitely generated as an Z-module. It is integrallyclosed, and every non-zero prime is maximal. Thus OK is a Dedekind domain ( normal noeth domain ofdim 1)

20. OK is the integral closure of Z in K, and K = Frac(OK)

OK

int

K

fin.Sep

Z Q

21. a finite domain is a field.

22. Def: R domain, K = Frac(R). A Fraction Ideal of R in K is an R-submodule of K, M , such that thereexists α ∈ K,α 6= 0 with αM ⊂ R. α induces isomorphism M ∼= αM an (integral) ideal of R.

23. R noeth domain, then every ideal contains a product of prime ideals.

24. Let R be a Dedekind domain. Then every non-zero ideal in R factors uniquely as a produce of nonzeroprime ideals.

25. Let R be Dedekind, P non zero prime. Define P−1 = {x ∈ K | xP ⊂ R}. Then P−1 is a fractional idealand P−1P = R.

26. R Dedekind, K = Frac(R). The set of Fractional (and integral) ideals form a group with multiplication.I−1 = {x ∈ K | xI ⊂ R.

27. Def: I, J fractional ideals in R Dedekind, K = Frac(R). We say that I divides J if equivalently:

• Exsits integral ideal Q such that IQ = J .

• I−1J is an integral ideal

• I ⊃ J .

• In the prime decompositions, every power of a prime in I occurs also in J (with higher power).

28. Def: Ideal Class Group Cl(R) =(fraction ideals)/(principle fractional ideals).

29. K is a number field, then Cl(OK) is finite.

30. R Dedekind. Then R UFD ⇐⇒ PID ⇐⇒ Cl(R) = 0.

31. 2016-Sep-15,p11

32. Chinese Reminder Thm: A1, ..., Ak ideals in R, Ai + Aj = R, then the map M → M/A1M × ... ×M/AkM (with kernel ∩AiM) is surjective and M/(A1...Ak)M ∼= M/A1M × ... ×M/AkM , so ∩AiM =(A1...Ak)M . Usually it is applied with M = R.

• With M = R, For any a1, ..., an ∈ R we can find x ∈ R such that x = ai mod Ai.

33. DEF: Let R be Dedekind. We define a Valuation with respect to to a prime P . vP (a) is the power of Pin (a).

• vP (a) = r ⇒ (a) ∈ P r − P r+1.

34. R Dedekind. Then every ideal is generated by 2 elements. If a ∈ I then there exists b ∈ I such that(a, b) = I.

35. Let W be a multiplicative system in R. Denote S = W−1R, and consider the map R→ S.



(a) Every ideal in S is an extended ideal of the form W−1a for some ideal a ⊂ R.

(b) a is a contracted ideal in R if and only if no element of W is a zero divisor in R/a.

(c) The prime ideals of S are in one-to-one correspondence with the prime ideals of R which does notmeet W .

(d) Let Q be a prime in S, and P = Qc. Then P is a prime and W−1P = Q.

(e) Let P be a prime disjoint from W . Let Q = W−1P . Then Q is a prime and Qc = P .

36. 2016-Sep-20, p14

37. A is integrally closed, S mul system, then so is S−1A.

38. A dedekind, S mul system, then S−1A dedekind.

39. A dedekind, then AP is a PID of dim 1, so DVR. Gen by x ∈ P − P 2. Moreover, P i/P i+1 is dim 1A/P -vector space.

40. A ↪→ B integral extension of domain. Then A is a field ⇐⇒ B is.

41. A ↪→ B integral extension. Q prime in B. Then Qc maximal in A ⇐⇒ Q is maximal in B.

42. (HW1) LetA ↪→ B ↪→ C be rings. Suppose that B is integral over A and C is integral over B. Then C isintegral over A.

43. A ↪→ B integral extension. Then S−1A ↪→ S−1B is integral.

44. Lying over A ↪→ B integral. P prime in A. Then there are finitely many primes Q1, ..., Qn in B lyingover A (from CA: they are incomparable). The correspond to the max ideals in BP /PBP .

45. K a field, R finitely generated over K as a K-vector space, then R is Artin (every prime is maximal, andthere are finitely many of those because they are minimal).

46. (See p65, thm21, Marcus - Number fields”) Let L be a field over a number field K, [L : K] = n. Fix aprime in A = OK , p. Let q1, ...qt be the primes lying over p in B = OL. Then pOL = qe11 · · · q

ett . OL/qi

is a field extension of OK/p. Let fi denote the degree. Then: n =∑t

1 fiei.

• 2016-Sep-18,p17

• If L/K is Galois, then f1 = ... = ft, e1 = ... = et.

• G acts on {q1, ..., qr} and the action is transitive, so the stabilizer is of size |G|/r.• For a tower of number filed extensions which are Galois, if q overQ over p then e(q/p) = e(q/Q)e(Q/p), f(q/p) =f(q/Q)f(Q/p).

• Def: Denote D(q|p) as the stabilizer of q ∈ {q1, ..., qr} under the G action. |D| = |G|/r = ef.

• Denote k(p) = Ok/p, k(q) = OL/q. Then k(q)/k(p) is a normal extension - every polynomial ink(p)[x] with a root in k(q) splits. It is not necessarily separable (see HW2). Moreover, the naturalmap D(q|p)→ Aut(k(q)/k(p)) is surjective.2

• Oct - 4- 2016, p21.

• Denote Lq as the intermediate field of D(q|p), R = OLq . Let Q lie over q in R/A. Then R/Q = A/p,i.e. f(Q|p) = 1.

• In B/R, Q is a unique prime under q, r(q|Q) = 1. Lq is the smallest intermediate field with thatproperty.

• We get that for (f, e, r) in B/A we get (e, f, 1) in B/R and (1, 1, r) in R/A. Lq is the largest subfieldK ′ with a prime Q′ over p with f(Q′|p) = 1 = e(Q′|p).

47. In the above setup, let k(q)/k(p) be separable, thus galois, so we get D(q|p) � Gal(k(q)/k(p)).

• The kernel of the map I(q|p) is called the Inertia Group. I = {σ ∈ D | σx ≡ x (mod q),∀x ∈ B}meaning it fixes B/q.

• |I| = e. Denote LI as the field fixed by I.

• we get a tower L−−LI −−Lq −−K, with q−−p′′−−Q−−p and (1, e, 1)−−(f, 1, 1)−−(1, 1, r).

2If L/K is purely inseparable, then Aut(L/K) is trivial?



• So e(q|p′′) = e = [L : Li], A galois extension of degree n with (f, e, r) = (1, n, 1) is called totallyramified.

• LI is the largest subfield K with p′ over p and e(p′|p) = 1.

• LI is the smallest subfield K with p′ under q and e(q|p′) = [L : K ′], thus the smallest totally ramifiedextension.

• For any τ ∈ Gal(L/K), D(τq|p) = τD(q|p)τ−1 and I(τq|p) = τI(q|p)τ−1

• If D(q|p) is normal, then we have one field Lq, with Lq/K galois with (f, e, r) = (1, 1, r) so p =Q1 · · ·Qr. We say that p splits completely.

• If further I(q|p) is normal in G and thus in D, then each of the Q′is expends to a prime in LI .

48. Frobenius conjugacy class / element: Let L/K be an extension, q over p in A = OL/B = OK suchthat e(q|p) = 1 and k(p), k(q) are finite. Then D(q|p) ∼= Gal(k(q)/k(p) which is cyclic and generated byx 7→ x#k(p). We denote the corresponding generator in D(q|p) as Frob(q|p).

• For any τ ∈ Gal(L/K), Frob(τq|p) = τFrob(q|p)τ−1. This is the Frobenius Conjugacy class.Denote Frob(p).

• If G is abelian then we have only one element, the Frobenius element.

• OCT-6-2016, worksheet

• Let q be a prime, then the multiplicative group (Z/qZ)× is cyclic. Every finite subgroup of amultiplicative group of a field is cyclic.

• Legandre symbol Let q be a prime and a be an integer relatively prime to q. Then(aq

)is defined

to be 1 if x2 = a (mod q) has a solution, -1 otherwise.

• For any odd primeq:(−1q

)= (−1)

q−12 .

• Q(√d)/Q with G = {±1}, q odd prime, q - d then (q) is unramified (e=1) and Frob(q) =

(dq

)• Gal(Q(ζn)/Q) = (Z/nZ)×. R = OQ(ζn) = Z[ζn]. (q) is unramified in R if and only if q does not

divide n. For any prime q, Frob(q) = [q], the class in (Z/nZ)×.

• In above setup, Let Q be in R, lying over (q). 1 ≤ a ≤ n, 1 ≤ b ≤ n. Then n = (1−ζ)(1−ζ2) · · · (1−ζn−1), and ζa ≡ ζb (mod q)⇒ a = b.

• Let p be an odd prime and let ζ be a primitive pth root of unity. Then L = Q(ζ) contains a uniquequadratic subextension, which happens to be Q(

√p∗) where p∗ = (−1)(p−1)/2p.

• Let q 6= p another odd prime. Then(qp

)=(p∗

q

)and(

q

p

)(p

q

)= (−1)

p−12

q−12

• OCT-11, p23

• Bottom line: L = Q(ζp) contains a unique subquadratic extension L∗ = Q(√±p). For a prime q,

the Frob in L is q (mod p), and in L∗ is(±pq

). Since the Frob elements in L∗ is a restricition from

L, in both cases it is periodic mod p.

49. Def: (Absolute determinant) Let K be a number field. Let L ⊂ K be a lattice, i.e. L = Zα1⊕ ...⊕αj ,a free Z module of rank [K : Q]. (independence over Z implies independence over Q, so ) the generatorsof L span K over Q. Take the n embedding of K in its Galois closure, σ1, ..., σn. The Disc(L) is definedto be, equivalently:

Disc(L) = |det[σi(αj)]2| = |det[TrK/Q(αiαj)]| ∈ Q− 0

Disc in independent of change of basis.

50. K = Q(α). Then Disc(Z[α]) = ±NK/Q(f ′(α)) where f(x) the the irreducible polynomial of α in K.

51. Disc(Z[ζp]) = ±pp−2.

52. Disc(Z[ζm]) divides mφ(m) (φ is the Euler function).

53. L ⊂ OK ⇒ OK ⊂ 1dL with d = Disc(L). Moreover, for any α ∈ OK ,

α = m1α1

d+ ...+mn

αnd, with d | m2

i



54. Let K,L be number fields of degree n,m. Suppose [KL : Q] = nm. Then OKL ⊂ 1dOKOL with

d = gcd(Disc(OK), Disc(OL))

55. Def: The Discriminant is like the absolute discriminant, just without the absolute value:

disc(L) = det[σi(αj)]2 = det[TrK/Q(αiαj)] ∈ Q− 0

56. Let K be a number field. (p) ramifies (e > 1 for one of the primes) in K ⇐⇒ p | disc(OK).

57. Consider the setup in 15. Let A be dedekind, and B = A[α]. Let f(x) = irr(α,K, x) ∈ A[x]. Let p be aprime in A. Let f̄(x) ∈ K(p)[x] the reduction mod p. Suppose f̄(x) = P1(x)e1 · · ·Pn(x)en composition toirreducbiles in K(p)[x]. lift each of the Pi’s to an fi ∈ A[x], monic of the same degree. Then

pB = qe11 · · · qerr , qi = pB + (fi(α))

with f(qi|P ) = degPi = deg fi.

58. If L1 ⊂ L2 are two latices, then disc(L1) = a2 · disc(L2) for some a ∈ Z.

59. Let A be a ring, B = Ax1 ⊕ ...⊕Axn. Then

discA(B) = det[Tr(xixj)] is defined up to a square of a unit.

60. Let A ⊂ B, B ∼= An, p prime in A. Then B/pB is A/p free and discA/p(B/pB) = discA(B) in A/p.

61. A ⊂ B1, B2, then discA(B1 ×B2) = discA(B1)discA(B2)

62. 2016-10-20, p31

63. Def: The Dual of a lattice L ⊂ K is L∗ = {x ∈ K | Tr(xy) ∈ Z,∀y ∈ L}.

64. If L = Zx1 ⊕ ...⊕ Zxn ⇒ L∗ = Zy1 ⊕ ...⊕ Zyn where yi is a dual basis with respect to the trace, see 18.

65. Properties of L∗:

• L1 ⊂ L2 ⇒ L∗1 ⊃ L∗2• L∗∗ = L.

• (L1 + L2)∗ = L∗1 ∩ L∗2

66. For K = Q(α),L = Z[α], f(x) = irr(α,Q, x)⇒

L∗ = Zc0(α)

f ′(α)⊕ ...⊕ Z

cn−1(α)

f ′(α)

withf(x)/(x− α) = c0(α) + c1(α)x1 + ...+ cn−1(α)xn−1

•

f(x) = a0 + a1x+ ...+ anxn ⇒

c0(α)...cn−3(α)cn−2(α)cn−1(α)

=

an an−1 ... ... a0

. . .

an an−1 an−2an an−1

an

αn−1

...α2

α1

• Let α = α1, α2, ..., αn be the roots of f(x). Then

n∑1

1

f ′(αi)

f(x)

(x− αi)= 1

n∑1

αkif ′(αi)

f(x)

(x− αi)= xk

Tr(αkcj(α)

f ′(α)) = δjk



• α ∈ OK ⇒ L∗ = 1f ′(α)L

67. O∗K is a fractional ideal. Define the Different, an ideal in OK :

DK/Q = (O∗K)−1

68. If OK = Z[α]⇒ D = f ′(α)OK

69. Def: Norm of an ideal. In the setup of 15, NL/K : ideals(B) → ideals(A), by sending a prime Q in B

with P = Q ∩A to P f with f = [B/Q : A/P ].

70. For K/Q, Q prime in OK over p, |NK/Q(Q)| = pf = |OK/Q| the number of elements in the FP -vectorspace.

71. In the setup 153, let α ∈ B. Then NL/K(αB) = NL/K(α)A where the right hand side is calculated usingthe original definition of the norm (10). For ideal I ⊂ OK , |N(I)| = |OK/I] = [OK : I]

• If L/K is Galois, For αB = Qe11 · · ·Qerr ⇒ N(Qi)B =∏σ∈G σ(Qi)

72. For K/Q, |N(DK/Q)| = |disc(K/Q)| = |OK/D| = [OK : D]

73. D ⊂ OK ⊂ D−1 = O∗K and [D−1 : OK ] = [OK : D]

74. In Setup 15, if I, J ⊂ A ideals with IB = JB then I = J in A.

75. (background) Let R be a PID, M free R module with x1, ...xn as a basis. N be a submodule. Then Nis free of rank m ≤ n. There exists a basis y1, ..., yn for M such that d1y1, ....dmym is a basis for N withd1 | d2 | ... | dm. If m = n, R = Z, then [M : N ] and subgroups is d1 · · · dn and it is also the determinantof a transformation matrix from a basis of M to a basis of N .

76. 2016-10-25, p35

77. Let K be a number field, Q a prime over (p) ⊂ Z. Then Q is ramified over (p) ⇐⇒ Q | DK/Q. Moreover,if pOK = Qe(others) then

• Qe−1 | D and,

• Qe | D ⇐⇒ p | e.

78. For x ∈ OK , TrK/Q(x) = TrOK/Z(x) ∈ Z.

79. K number field then Cl(OK) is finite.

• There exists an absolute constant c(K) such that every ideal class admits an ideal I such that|N(I)| ≤ c(K).

80. Let R be a region in Rn clsoed, convex, an symmetric. let D be a fundamental domain (generator) of alattice L is Rn. If vol(R) ≥ 2nvol(D) then R contains a lattice point.

81. Let K/Q be a number field, degree n. Let r1=# real embeddings, r2=# complex embeddings (r1 +2r2 =n). In the class group of OK , every class contains an ideal J with:

|N(J)| ≤(

4

π

)r2 n!

nn

√|disc(K)|

3in this setup, are all ideal of A contracted?


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