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.

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