## 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.