Algebra Solutions: Hungerford III.2

1) The set of all nilpotent elements in a commutative ring forms an ideal.
Proof-
Let R be a commutative ring and let N = \{a \in R | a^n = 0 \text{ for some natural number n} \}. Since 0 is in N we know that N is nonempty. From exercise 3.1.12 we know that the sum of two nilpotent elements is nilpotent. Let a,b \in N, then say that a^m = 0, b^n=0 \Rightarrow (-b)^n = \pm b^n = \pm 0 = 0. Therefore -b is nilpotent. Thus for a,b \in N we have that a-b \in N being a sum of nilpotent elements. Lastly, let r\in R. Then we must show that ra \in N. Note that (ra)^m = r^ma^m = r^m0 = 0. Thus ra= ar \in N so that N is an ideal by the ideal test theorem.

2) Let I be an ideal in a commutative ring R and let Rad(I) = \{ r \in R | r^n \in I \text{ for some n} \}. Show that Rad(I) is an ideal.
Proof-
0 is in Rad(I) trivially so that Rad(I) is nonempty. Let a,b \in Rad(I) and r\in R with a^n,b^m\in I. We want to show that a-b \in Rad(I). Consider then:
(a-b)^{n+m} = \sum_{k=0}^{n+m} \dbinom{n+m}{k} a^{n+m-k}(-b)^k. If k \leq m \Rightarrow n+m-k > n \Rightarrow a^{n+m-k} \in I. On the other hand if k\geq m then (-b)^k \in I. Thus \forall 0 \leq k \leq n+m we have that \dbinom{n+m}{k} a^{n+m-k}(-b)^k\in I. Thus (a-b)^{n+m} \in I \Rightarrow a-b \in Rad(I). Lastly (ra)^n =r^na^n = \in I since I is an ideal. Thus ra=ar \in Rad(I). So once again by the ideal test, Rad(I) is an ideal.

3) If R is a ring and a\in R, then J = \{r\in R | ra = 0\} is a left ideal and K = \{r\in R | ar = 0\} is a right ideal in R.
Proof-
0 is obviously in J so J is nonempty. Let b,c\in J and r\in R. Then ba = ca = 0. Then (b-c)a = ba-ca = 0-0 = 0 so that b-c \in J. Also, (rb)a = r(ba) = r0 = 0 \Rightarrow rb \in J. Thus J is a left ideal. The proof for K is exactly the same but on the right side.

4) If I is a left ideal of R, then A(I) = \{r\in R | rx = 0 \forall x\in I\} is an ideal in R.
Proof-
Since 0x=0 \forall x\in I we have that 0\in A(I) so that A(I) is nonempty. Now let a,b\in A(I) and r\in R. Then ax=bx=0 \forall x \in I. Thus (a-b)x = ax-bx=0-0=0 \forall x \in I. Therefore a-b\in A(I). Lastly, (ra)x = r(ax) = r0=0 \Rightarrow ra \in A(I) and (ar)x = a(rx)= a0 = 0 since rx\in I because x\in I and I is an ideal. Thus ra,ar \in A(I) so that A(I) is an ideal in R.

5) If I is an ideal in a ring R, let [R:I] = \{ r\in R | xr \in I \forall x\in R \}. Prove that [R:I] is an ideal of R which contains I.
Proof-
First note that x0=0 \in I \forall x\in R so that 0 \in [R:I]. Thus [R:I] is nonempty. Now let a,b\in [R:I] and r\in R. Then xa, xb \in I \forall x\in R. Then x(a-b) = xa-xb \in I \forall x \in R. Therefore a-b \in [R:I]. Also x(ra) = (xr)a \in I because xr\in R and a\in [R:I]. Thus ra \in [R:I]. Similarly ar \in [R:I] so that [R:I] is an ideal. Lastly, let c \in I. Then c\in R \Rightarrow xc \in I \forall x\in R because I is an ideal. Thus c \in [R:I] so that I \subset [R:I].

10) (a) Show that \mathbb{Z} is a principle ideal ring.
(b) Every homomorphic image of a principle ideal ring is also a principle ideal ring.
(c) \mathbb{Z}_m is a principle ideal ring for every m>0.
Proof- a)
Let I be an ideal in \mathbb{Z}. Then being a subring of the integers it must have a lowest positive integer, call it a. Claim that I = (a). Obviously (a) \subset I so we must show that I \subset (a). Let b \in I. Then we can write by the division algorithm, b = aq + r, where 0\leq r < a. Then r = b-aq. We have that b \in I, and since I is an ideal and a\in I \Rightarrow aq \in I. This gives that b-aq = r\in I. But a was the least positive integer in I, so that r must be 0. Thus b = aq \Rightarrow b \in (a). Thus I = (a) and since I was arbitrary, any ideal of \mathbb{Z} must be principle.

b) Let R be a principle ideal ring and f:R \rightarrow S be a homomorphism. Want to show that Im(f) is a principle ideal ring, i.e. that every ideal in Im(f) is principle. Let J be an ideal in Im(f). Then consider I = f^{-1}(J) = \{ r\in R | f(r) \in J\}. Since R is principle, and the inverse image of an ideal is an ideal, we have that I is a principle ideal, so we can write, I = (a). Since f is onto the image we know that f((a)) is an ideal in S. We have that (a) = \{ ra | r\in R\} so that:
f((a)) = \{ f(ra) | r \in R\} = \{ f(r)f(a) | r \in R\} = \{ sf(a) | s \in S\} = (f(a)) since f is surjective onto the image.
Want to show that f((a)) = (f(a)) = J. Since f(a) \in J we have that (f(a)) \subset J. Now let y\in J. Then since f is surjective onto the image, there exist x\in R such that f(x) = y. In fact, x\in f^{-1}(J) = (a). Thus we can write x = r'a. Then y = f(x) = f(r'a) = f(r') f(a) \in (f(a)). Therefore (f(a)) = J so that Im(f) is a principle ideal ring.

c) Consider the canonical homomorphism \pi : \mathbb{Z} \rightarrow \mathbb{Z}_m by a\mapsto \bar{a}. Since this is a surjective homomorphism and since the integers are a principle ideal ring, from part b we have that \mathbb{Z}_m is a principle ideal ring.

13) Let f:R \rightarrow S be a homomorphism of rings, I an ideal in R, and J an ideal in S.
(a)f^{-1}(J) is an ideal in R that contains Kerf.
(b) If f is an epimorphism then f(I) is an ideal in S. If f is not surjective then f(I) need not be an ideal in S.
Proof- a
) Let a,b\in f^{-1}(J) and r\in R. Then we have that f(a),f(b) \in J \Rightarrow f(a)-f(b) = f(a-b) \in J \Rightarrow a-b \in f^{-1}(J). Also, f(ra) = f(r)f(a) \in J because J is an ideal. Therefore ra \in f^{-1}(J), and so f^{-1}(J) is an ideal. Now let x\in Kerf. This gives us that f(x) = 0 \in J \Rightarrow x\in f^{-1}(J). So that the Kernel is contained in f^{-1}(J).

b) Now we assume that f is onto. Let c,d \in f(I), and s\in S. Since f is onto there exists, a,b\in I such that f(a) = c, f(b) = d. Since I is an ideal, a-b \in I and then f(a-b) = f(a)-f(b) = c-d \Rightarrow c-d \in f(I). Also, since f is onto, there exists an r\in R such that f(r) = s. Since I is an ideal, ra \in I. Then f(ra) = f(r)f(a) = sc \Rightarrow sc \in f(I). Similarly, cs \in f(I), so that f(I) is an ideal in S.

17) Let f: R\rightarrow S be an epimorphism of rings with kernel K.
(a) If P is a prime ideal in R that contains K, then f(P) is a prime ideal in S.
(b) If Q is a prime ideal in S, then f^{-1}(Q) is a prime ideal in R that contains K.
(c) There is a 1-1 correspondence between the set of all prime ideals in R that contain K and the set of all prime ideals in S, given by P\mapsto f(P).
(d) If I is an ideal in a ring R, then every prime ideal in R/I is of the form P/I where P is a prime ideal in R that contains I.
Proof- a)
Let P be a prime ideal in R that contains K. Then since f is an epimorphism we know that f(P) is an ideal in S. Let c,d\in S and assume that cd\in f(P). Since f is onto there exist a,b \in R such that f(a) = c, f(b) =d . Then f(ab) = f(a)f(b) = cd \in f(P). Since f(P) is an ideal this also tells us that -f(ab) \in f(P) so that there exists p\in P such that f(p) = -f(ab). Then 0 = f(ab) + f(p) = f(ab+p) \Rightarrow ab+p \in K \subset P, by assumption. Therefore there is some p'\in P such that ab+p = p' \Rightarrow ab = p'-p\in P. Since P is a prime ideal this tells us that a\in P or b\in P. Therefore f(a) = c \in f(P) or f(b) = d \in f(P) so that f(P) is prime.

b) Let Q be a prime ideal in S. Then we have that f^{-1}(Q) is an ideal in R that contains K. Let a,b\in R and assume that ab\in f^{-1}(Q). Then f(ab) = f(a) f(b) \in Q \Rightarrow f(a) \in Q or f(b) \in Q since Q is a prime ideal. Therefore a\in f^{-1}(Q) or b\in f^{-1}(Q) which tells us that f^{-1}(Q) is prime.

c) Let X be the set of all prime ideals in R that contain K and Y be the set of all prime ideals in S. Defined \phi :X \rightarrow Y by \phi (P) = f(P). Claim that \phi is bijective. Let P,Q \in X. and assume that \phi (P) = \phi (Q). Then, f(P) = f(Q). Let p\in P. Then f(p) \in f(P) = f(Q) gives us that there exists a q\in Q such that f(q) = f(p). Then 0 = f(p-q) \Rightarrow p-q \in K \subset Q. Which means that p\in Q. So then P \subset Q, and we can similarly show that Q \subset P so that P = Q. Therefore \phi is one to one. Next, let J \in Y. From part b we know that f^{-1}(J) is a prime ideal so it is in the set X. Then \phi (f^{-1}(J)) = f(f^{-1}(J)) = J so that \phi is surjective. Thus we have a one to one correspondence.

d) Let \bar{P} be a prime ideal in R/I. Consider \pi : R \rightarrow R/I given by \pi (r) = r+ I. Since \pi is an epimorphism with Kernel I we have that \pi^{-1} (\bar{P}) is a prime ideal in R which contains I by part (b). Then \bar{P} = \pi ( \pi^{-1} (\bar{P} )) =  \{ q+I | q\in \pi^{-1} (\bar{P})\} = \pi^{-1} (\bar{P})/I . Then call \pi^{-1} (\bar{P}) = P so that \bar{P} = P/I, where P has the wanted properties.

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