## The complete homomorphic preimage of a prime ideal is a prime ideal

Let $R$ and $S$ be commutative rings and let $\varphi : R \rightarrow S$ be a ring homomorphism.

1. Prove that if $P \subseteq S$ is a prime ideal then $\varphi^\ast[P] \subseteq R$ is a prime ideal. Apply this to the special case when $R \subseteq S$ to deduce that if $P$ is prime in $S$ then $P \cap R$ is prime in $R$.
2. Prove that if $M \subseteq S$ is a maximal ideal and if $\varphi$ is surjective, then $\varphi^\ast[M] \subseteq R$ is maximal. Give an example to show that this need not be the case if $\varphi$ is not maximal.

1. By §7.3 #24, $\varphi^\ast[P]$ is an ideal of $R$. Now suppose $ab \in \varphi^\ast[P]$. Then $\varphi(ab) = \varphi(a)\varphi(b) \in P$, so that since $P$ is prime, either $\varphi(a) \in P$ or $\varphi(b) \in P$. Thus either $a \in \varphi^\ast[P]$ or $b \in \varphi^\ast[P]$. Hence $\varphi^\ast[P]$ is a prime ideal of $R$.

Note that if $\iota : R \rightarrow S$ is the inclusion map, then $\iota^\ast[P] = R \cap P$.

2. Let $M \subseteq S$ be maximal, and note that $\varphi^\ast[M] \subseteq R$ is an ideal. Note that $\varphi^\ast[M] \neq R$ since $\varphi$ is surjective. Let $\pi : S \rightarrow S/M$ denote the natural projection. Since $\varphi$ is surjective, $\pi \circ \varphi : R \rightarrow S/M$ is a surjective ring homomorphism and $S/M$ is a field. Moreover, $\varphi^\ast[M] \subseteq \mathsf{ker}\ \pi \circ \varphi$. Now $R/\mathsf{ker}\ (\pi \circ \varphi) \cong S/M$ is a field, and thus has only the trivial ideals. Using the Lattice Isomorphism Theorem and since $\varphi^\ast[M] \neq R$, we have $\varphi^\ast[M] = \mathsf{ker}\ \pi \circ \varphi$. Since $R/\varphi^\ast[M]$ is a field, $\varphi^\ast[M]$ is maximal in $R$.

Now let $M \subseteq R$ be a maximal ideal and consider the inclusion map $\iota : M \rightarrow R$. Then $\iota^\ast[M] = M$ is not maximal in $M$.