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I suspect this is easy, but that I'm missing something obvious: If a ring homomorphism $f:R\rightarrow S$ is such that a maximal ideal $m$ in $S$ does not have $f^{-1}(m)$ maximal in $R$, must there be some non-unit element $x\in R$ whose image under $f$ is a unit in $S$?

My intuition is just that some "localization" must have occurred. In the classic example of $\mathbb{Z}\hookrightarrow\mathbb{Q}$, the zero ideal becomes maximal because 2, 3, 5, ... all become units.

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2 Answers 2

up vote 8 down vote accepted

I'm not sure that I properly understand the question, especially in light of your acceptance of Bill Dubuque's answer and your accompanying comment.

Still, maybe these remarks will be useful:

The map $R/f^{-1}(m) \to S/m$ is a map from an integral domain to a field, and if the source is not also a field (i.e. if $f^{-1}(m)$ is not itself maximal), then there are certainly non-units in $R/f^{-1}(m)$ that map to units in $S/m$.

One interpretation of your question is whether, in this context, there must necessarily be a non-unit of $R$ that maps to a unit in $S$?

The answer is "no" in general. Here is a counterexample:

Let $R$ be a domain which is not a field, and let $I$ by any subset of $R\setminus \{0\}$ which generates $R\setminus \{0\}$ as a monoid under multiplication. Let $S$ be the polynomial ring $R[\{x_i\}_{i \in I}].$

If $Q$ denotes the fraction field of $R$, then there is a natural surjection $S \to Q$ given by mapping $x_i$ to $1/i$. (This is where we use the assumption that $I$ generates $R\setminus \{0\}$, so as to get a surjection.) The kernel of this surjection is a maximal ideal $m$ of $S$, whose preimage in $R$ is the zero ideal (hence not maximal).

On the other hand, since $S$ is a polynomial ring over $R$, any non-unit in $R$ remains a non-unit in $S$.

Some additional comments: if $R$ is Jacobson (e.g. $\mathbb Z$, or a polynomial ring over a field), then any map between finite type $R$-algebras preserves the corresponding MaxSpecs. On the other hand, for such domains, the set $I$ is necessarily infinite, and so $S$ is necessarily infintely generated as an $R$-algebra (and hence no contradiction ensues!).

But we can get finite type examples by taking $R$ to be e.g. a DVR. Then $I$ can be taken to consist of a single element (the uniformizer), and so $S = R[x]$ is a polynomial ring in a single variable. For example, if $R = \mathbb Z_p$ (the $p$-adic integers) then we have $$\mathbb Z_p \hookrightarrow \mathbb Z_p[x] \mapsto \mathbb Z_p[x]/(p x - 1) = \mathbb Q_p,$$ and so $p x - 1$ generates a maximal ideal in $\mathbb Z_p[x]$ whose preimage is not maximal, but every non-unit in $\mathbb Z_p$ remains a non-unit in $\mathbb Z_p[x]$.

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Ah, now I see my error (and Bill's): in general, we have only $f(f^{-1}(m))\subseteq m$, not $f(f^{-1}(m))=m$. Thus, just because a maximal ideal $M\supset f^{-1}(m)$ in $R$ does not mean that $f(M)\supset m$, only $f(M)\supset f(f^{-1}(m))$. I'd seen the "localizing without localizing" trick in Eisenbud, but hadn't thought to apply it here. It does an excellent job of teasing apart where the failure to preserve non-units / maximal ideals actually does occur. Thanks for your answer. –  Zev Chonoles Oct 5 '10 at 18:04

HINT Consider the image of a maximal ideal containing $f^{-1}(m)$.

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hits head Of course. And it's also obvious that all maps that send non-units to units must force there to be a maximal ideal in the codomain whose preimage is not maximal. I was toying with the idea that MaxSpec would be a functor on the category of rings and homomorphisms that preserve non-units, though I doubt this is terribly useful. –  Zev Chonoles Oct 3 '10 at 19:49
1  
This was a wonderful hint! –  Jason DeVito Oct 3 '10 at 20:03

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