Solutions to Homework 2 - Math 4400

In this assignment we shall use often the subgroup criterion:

Subgroup criterion: Let (G, +) be an additive group. H ⊂ G is an additive subgroup if(i) H is closed under addition, (ii) 0 ∈ H, (iii) If h ∈ H then −h ∈ H.

1. (# 22. p. 244). Let φ : R → R′ be a ring homomorphism and let N be an ideal ofR.(a). Show φ(N) is an ideal of φ(R).(b). Give an example showing that φ(N) need not be an ideal of R′.(c). Let N ′ be an ideal either of φ(R) or of R′. Show that φ−1(N ′) is an ideal of R.Solution. (a). φ(N) is an additive subgroup of φ(R) since:(i) φ(x) + φ(y) = φ(x + y) ∈ φ(N) since x + y ∈ N for x, y ∈ N ,(ii) φ(0) = 0 ∈ φ(N),(iii)−φ(x) = φ(−x) ∈ φ(N) is the additive inverse of φ(x) ∈ φ(N).Thus φ(N) satisfies the subgroup criterion and is an additive subgroup of φ(N). Nextif φ(r) ∈ φ(R) and φ(x) ∈ φ(N) then φ(r)φ(x) = φ(rx) ∈ φ(N) as rx ∈ N . Likewise,φ(x)φ(r) = φ(xr) ∈ φ(N) as xr ∈ N . Thus φ(N) is closed under left and rightmultiplication and is an ideal of φ(R).(b). Let φ : Z → Z × Z be the diagonal map. That is, φ(x) = (x, x). We saw on thelast homework that this was a ring homomorphism. Note that N = Z is an ideal inR = Z. We have that

φ(N) = {(x, x) | x ∈ Z}.

This is not an ideal in Z×Z. Since (1, 1) ∈ φ(N) and (1, 0) ∈ Z×Z but (1, 0) · (1, 1) =(1, 0) /∈ φ(N).(c). Let N ′ be an ideal of either φ(R) or of R′. (We shall see that the argument isindependent of which case we are in.). We shall show that

N = φ−1(N ′) = {x ∈ R | φ(x) ∈ N ′}

is an ideal in R. Let x, y ∈ φ−1(N ′). Then(i) φ(x + y) = φ(x) + φ(y) ∈ N ′ as N ′ is an additive group and hence x + y ∈ N ,(ii) φ(0) = 0 ∈ N ′ as N ′ is an additive group and 0 ∈ N ,(iii) φ(−x) = −φ(x) ∈ N ′ as N ′ is an additive group and thus −x ∈ N .It follows that N is an additive subgroup of R. Let r ∈ R and x ∈ φ−1(N ′). Thusφ(rx) = φ(r)φ(x) ∈ φ(R)N ′ ⊆ N ′ and hence rx ∈ N . Similarly, we can show xr ∈ N .Thus φ−1(N ′) is an ideal of R.

2. (# 28. p. 245). Let R and R′ be rings with corresponding ideals N and N ′. Supposethat φ : R → R′ is a ring homomorphism and φ(N) ⊆ N ′. Show there is a well-definedhomomorphism from R/N to R′/N ′.Solution. We define φ∗(x + N) = φ(x) + N ′. First we must check that this map iswell-defined. Let x + N = y + N be two representatives of a coset. It follows thatx = y + n for some n ∈ N . Then

φ∗(x + N) = φ(x) + N ′ = φ(y + n) + N ′ = φ(y) + φ(n) + N ′ = φ(y) + N ′

since φ(n) ∈ φ(N) ⊆ N ′. Thus φ∗ is well-defined. We now check that it is a ringhomomorphism. Let x + N, y + N ∈ R/N . Then

φ∗((x + N) + (y + N)) = φ∗(x + y + N) = φ(x + y) + N ′ = φ(x) + φ(y) + N ′

= (φ(x) + N ′) + (φ(y) + N ′) = φ∗(x + N) + φ∗(y + N)

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and

φ∗((x + N)(y + N)) = φ∗(xy + N) = φ(xy) + N ′ = φ(x)φ(y) + N ′

= (φ(x) + N ′)(φ(y) + N ′) = φ∗(x + N)φ∗(y + N).

It follows that φ∗ is a ring homomorphism.

3. (# 30. p. 245). Show that the collection of all nilpotent elements in a commutativering, R, is an ideal, the nilradical of R.Solution. Let N denote the set of nilpotent elements.(i) Suppose a, b ∈ N . Then by the last homework a+ b is nilpotent and thus a+ b ∈ N .(ii) 0 ∈ N since 01 = 0.(iii) Suppose a ∈ N such that an = 0 and let −a denote its additive inverse. We havethe identity

aN = (−a)N (1)

for N even. (This follows from the identity ab = (−a)(−b) and thus may be proven byinduction on N). Thus if n is even then

(−a)n = an = 0

and −a is nilpotent. If n is odd then

(−a)n+1 = an+1 = an · a = 0a = 0

and −a is nilpotent. By the subgroup criterion N is an additive subgroup of R. Nowsuppose r ∈ R and a ∈ N such that an = 0. Then

(ra)n = rnan = rn0 = 0

since R is commutative. It follows that rN ⊆ N . Thus N is an ideal of R. (We neednot check right multiplication as R is commmutative.)

4. (# 34. p. 245). Let R be a commutative ring and N an ideal of R. Let√

N denotethe set of all a ∈ R such that an ∈ N for some n ∈ N.

√N is called the radical of R.

Solution. (i) Let a, b ∈√

N . That is, there exist m,n ∈ N such that am, bn ∈ N . LetJ = n + m. Then

(a + b)J =∑

u+v=J,u,v≥0

(J

u

)aubv

=m−1∑u=0

(J

u

)aubv +

J∑u=m

(J

u

)aubv.

Now as am ∈ N we have au ∈ N for u ≥ m. Thus(Ju

)aubv ∈ N for u ≥ m as N is an

ideal and the second sum is an element of N . Note that in the first sum

v = J − u ≥ J − (m− 1) = nm− (m− 1) = m(n− 1) + 1 ≥ (n− 1) + 1 = n.

Thus bv ∈ N for 0 ≤ u ≤ m − 1 and(Ju

)aubv ∈ N for 0 ≤ u ≤ m − 1 as N is an ideal

and the first sum is an element of N . It follows that a + b ∈√

N .(ii) Note that 01 = 0 ∈ N.(iii) Suppose that a ∈

√N and an ∈ N . If n is even then (−a)n = an ∈ N . If n is odd,

then (−a)n+1 = an+1 = ana ∈ N .It follows that

√N is an additive subgroup of R.

Let r ∈ R and a ∈√

N . That is, an ∈ N . Now (ra)n = rnan ∈ N as N is an ideal.Thus rN ⊆ N and N is an ideal.

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5. (#28 p. 253). Let R be a commutative ring with unity, 1. Prove directly from thedefinitions of maximal and prime ideals that a maximal ideal is a prime ideal.Solution. Let M be a maximal ideal in R. Suppose that ab ∈ M . If a ∈ M then weare finished. Suppose that a /∈ M . Note that

aR = {ar | r ∈ R}

is an ideal of R (this was mentioned in class). By a later exercise, aR + M is an idealof R. But notice that, M ⊂ aR +M since any m ∈ M satisfies m = a0+m ∈ aR +M .Moreover, M 6= aR + M as a /∈ M . As M is maximal it follows that aR + M = R.Thus 1 ∈ aR + M . Thus there exist r ∈ R and m ∈ M such that

1 = ar + m.

Multiplying by b we obtainb = abr + mb.

Since ab ∈ M and m ∈ M , it follows that b = abr + mb ∈ M as M is an ideal. Thusb ∈ M and M is a prime ideal.

6. (#34 p. ) (a). We need to show that A+B is an ideal of R. That is, it’s an additivesubgroup and closed under left and right multiplication. We shall apply the subgroupcriterion.

(i) Let a + b, a′ + b′ ∈ A + B. Then (a + b) + (a′ + b′) = (a + a′) + (b + b′) ∈ A + B.(ii) 0 = 0 + 0 ∈ A + B.(iii) Let a+b ∈ A+B. Then −a ∈ A and −b ∈ B since they each are additive subgroupsof R. Thus (−a) + (−b) ∈ A + B and (a + b) + (−a) + (−b) = 0.

Next we check that A + B is closed under left multiplication. Let r ∈ R and a + b ∈A + B. Then r(a + b) = ra + rb ∈ A + B since A and B are both ideals. Similarly,(a + b)r = ar + br ∈ A + B as A and B are ideals. Therefore A + B is an ideal.

(b). Show that A ⊆ A + B and B ⊆ A + B.Let a ∈ A. Then a = a + 0 ∈ A + B. Thus A ⊆ A + B. Similarly, if b ∈ B thenb = 0 + b ∈ A + B and hence B ⊆ A + B.

7. (# 35 p. 254) Let A,B be ideals in R. (a). Show that AB is an ideal in R.Solution. (i) Note that if x =

∑mi=1 aibi and y =

∑nj=1 a′jb

′j ∈ AB then x + y =∑m+n

i=1 cidi where

ci ={

ai 1 ≤ i ≤ ma′n+i m + 1 ≤ i ≤ m + n

and

di ={

bi 1 ≤ i ≤ mb′n+i m + 1 ≤ i ≤ m + n

.

(ii) Also 0 = 0 · 0 ∈ AB.(iii) Now if x =

∑mi=1 aibi then −x =

∑mi=1(−ai)bi is clearly an element of AB (−ai ∈ A

as A is an additive subgroup.) Thus AB is an additive subgroup of R.Let r ∈ R and x =

∑mi=1 aibi. Then we have

rx = x =m∑

i=1

(rai)bi ∈ AB

as rai ∈ A as A is an ideal and bi ∈ B. Similarly,

xr =m∑

i=1

ai(bir) ∈ AB

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by a similar argument. It follows that AB is closed under left and right multiplicationby R and is thus an ideal.(b). Note that aibi ∈ A as A is an ideal for 1 ≤ i ≤ m. Thus

∑mi=1 aibi ∈ A as A is an

additive subgroup. Also, aibi ∈ B as B is an ideal for 1 ≤ i ≤ m. Thus∑m

i=1 aibi ∈ Bas B is an additive subgroup. It follows that

∑mi=1 aibi ∈ A ∩B. Hence AB ⊆ A ∩B.

8. (#36 p. 254). Let A,B be ideals of R. Show that quotient

A : B = {r ∈ R | rb ∈ A for all b ∈ B}

is an ideal of R.Solution. Let x, y ∈ A : B. That is, xb ∈ A for all b ∈ B and yb ∈ A for all b ∈ B.Thus

(x + y)b = xb + yb ∈ A

for all b ∈ B. Next 0b = 0 ∈ A for all b ∈ B. Finally, if x ∈ A : B then (−x)b =x(−b) ∈ A for all b ∈ B. Thus A : B is closed under additive inverses and A : B is anadditive subgroup of R. Let r ∈ R and x ∈ A : B. Then

(rx)b = x(rb) ∈ A : B

for all b ∈ B since rb ∈ B. Thus r(A : B) ⊆ A : B and A : B is an ideal of R.

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