D+F 7.4.13

Let $\varphi:R \rightarrow S$ be a homomorphism of commutative rings.
(a) Prove that if $P$ is a prime ideal of $S$ then either $\varphi^{-1}(P)=R$ or $\varphi^{-1}(P)$ is a prime ideal of $R$.
(b) Prove that if $M$ is a maximal ideal of $S$ and $\varphi$ is surjective then $\varphi^{-1}(M)$ is a maximal ideal of $R$. Give an example to show that this may not be true if $\varphi$ is not surjective.

$\textit{Proof}$ : Exercise 7.3.24 establishes that the preimage or pullback of an ideal is indeed an ideal. So $\varphi^{-1}(P)$ is an ideal of $R$.

Recall that an ideal of a commutative ring is called a prime ideal if $P \neq R$ and $ab\in{P} \Longrightarrow a\in{P}$ or $b\in{P}$.

Let $r_1r_2 \in{\varphi^{-1}(P)} \Longrightarrow \varphi(r_1r_2)\in{P}$. $\varphi(r_1r_2)=\varphi(r_1)\varphi(r_2)\in{P}$ since $\varphi$ is a ring homomorphism. Since $P$ is a prime ideal in $R$, this means $\varphi(r_1)\in{P}$ or $\varphi(r_2)\in{P}$ which implies
$r_1\in{\varphi^{-1}(P)}$ or $r_2\in{\varphi^{-1}(P)}$ So, $\varphi^{-1}(P)$ is a prime ideal in $R$. Now, if $\varphi^{-1}(P)$ happened to contain a unit (and there is no apparent reason why it cannot), then by proposition 9 . 253, $\varphi^{-1}(P)=R$. In any case, $\varphi^{-1}(P)$ is either a prime ideal or all of $R$.

For part (b), recall that an ideal $M$ of $S$ is a maximal ideal if $M \neq S$ and the only ideals containing $M$ are $M$ and $S$.

Let $M$ be a maximal ideal of $S$. Since $M$ is not all of $S$, then $\exists s \in{S}$ and $s\not \in{M}$. Since $\varphi$ is surjective, $\exists r\in{R}$ s.t. $\varphi(r)=s$ and $r\not \in \varphi^{-1}(M)$. Thus, $\varphi^{-1}(M) \neq R$.
Let $L \subset R$ be an ideal of $R$ containing $\varphi^{-1}(M)$. We want to show that $L=R$ or $L=\varphi^{-1}(M)$.
$L$ contains $\varphi^{-1}(M) \Longrightarrow \varphi(L)$ contains $M$. Then, $\varphi(L)=M$ or $S$ since $M$ is maximal in $S$. Therefore, $L=\varphi^{-1}(M)$ or $R$. $\blacksquare$.

If we relax the condition that $\varphi$ is surjective, then consider the following example:
$\phi: R \rightarrow \mathbb{F}_2$
$\phi: r \mapsto 0$
$(0)=0$ is maximal in $\mathbb{F}_2$. But, $\phi^{-1}(\mathbb{F}_2)=R$, the entire ring, so is not maximal.