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Recall that a Jacobson ring $A$ is a ring such that all prime ideals are an intersection of maximal ideals. Is the following true? If $f:A\to R$ is a ring map with $A$ Jacobson, and $\mathfrak m$ is a maximal ideal of $R$, then $f^{-1}\mathfrak m$ is maximal?

Proof attempt: Let $\mathfrak p = f^{-1}(\mathfrak m)$. We know $\mathfrak p$ is prime, hence $\mathfrak p = \bigcap_{\mathfrak M\supset\mathfrak p}\mathfrak M$. So $A/\mathfrak p\to R/\mathfrak m$ is injective, the target is a field. By the Chinese remainder theorem, $A/\bigcap{\mathfrak M} = \prod A/\mathfrak M$ which is a subring of a field, and in particular a domain. But a product of fields is a domain if and only if the product consists of a single field, so $\mathfrak p$ is maximal.

I think this is wrong though, because this seems to imply that the inverse of any prime ideal is maximal. The proof didn't really use maximality of $\mathfrak m$ in an essential way. Is there an obvious mistake/easy fix or is the argument or statement completely wrong?

Nico
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1 Answers1

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The statement is wrong. Let $k$ be a field and consider the natural inclusion $f:k[x,y]\to k[x,y]_{(x)}$. The ideal $(x)\subset k[x,y]_{(x)}$ is maximal, as $k[x,y]_{(x)}/(x)\cong k(y)$, but the preimage of $(x)$ along $f$ is just $(x)$, which is not a maximal ideal of $k[x,y]$.

KReiser
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  • What's wrong with the proof? Can the statement be salvaged? – Nico May 12 '23 at 23:21
  • You can't salvage the statement in any interesting way. The part with the Chinese Remainder Theorem is wrong - you're potentially using an infinite collection of ideals there, but the theorem is only true for a collection of finitely many ideals. – KReiser May 12 '23 at 23:28