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

Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.4 Exercise 7.4.13


(1) By Exercise 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 $M$.


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