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Let $A \subseteq B$ be two $k$-algebras, $k$ is a field of characteristic zero. Assume that $\dim(A)=\dim(B) < \infty$.

Is it possible to find a non-maximal ideal $J$ of $B$ such that $J \cap A$ is a maximal ideal of $A$?

Please see this question, in which the following example appears: $A=k[x] \subset k[x,y]=B$. $J = (x,y^2)$ is not a maximal ideal of $k[x,y]$ and $J \cap k[x] = (x)$ is a maximal ideal of $k[x]$; but in this example we have $\dim(A)=1 < \dim(B)=2$.

Thank you very much!

user237522
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3 Answers3

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This can not happen if the extension is integral. This is a special case of the going-up theorem of integral ring extensions. See, for instance, Corollary 5.8 of Atiyah Macdonald.

There are many examples when the extension is not integral. For example, take two zero dimensional rings like a field $A=k$ and $B=k[x,y]/(x,y)^2$. Then the ideal $(x)\subset B$ is not maximal, but $(x)\cap k=(0)$ which is maximal in $k$.

Edit: As pointed out by Badam Baplan, in order for the corollary to apply, the ideal in the larger ring must be prime. This is because the earlier Proposition 5.7 in Atiyah Macdonald only applies when both the rings are domains.

Mike Debellevue
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  • Thank you very much! Nice. Actually, it is not Corollary 5.8 (since here $J$ is not a prime ideal of $B$), but it is Propositions 5.6 (i) and 5.7. – user237522 May 31 '20 at 17:44
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When you don't require $J$ to be prime, things can get weird, even for integral extensions, as Youngsu pointed out (see $k[x^2] \subseteq k[x]$, where $x^2k[x] \cap k[x^2] = x^2k[x^2]$).

The important positive result that works even when $J$ is not prime is the following: if $A \subseteq B$ is an epimorphism of rings and $J$ is an ideal of $B$ such that $J \cap A = \mathfrak{m}$ is maximal, then $J$ is maximal.

Indeed, in this case the pushout $A/\mathfrak{m} \subseteq B/J$ is an epimorphism from a field, so it is a surjection. Thus $B/J$ is a field and $J$ is a maximal ideal [Stacks Lemma 10.106.7, 10.106.8]

Edit: flatness obviously has nothing to do with this property. Consider $k \subseteq k[x]$. Every ideal contained in $xk[x]$ contracts to $0$.

Edit again Also for any field $k$, the finite product $\prod k$ is separable over $k$, and there will be non-maximal ideals which contract to $0$ in $k$.

Badam Baplan
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  • Thank you. Oh I understad now the counterexample of Youngsu for an integral extension. Now I am confused with Mike's answer; even though he may thought of $J$ prime, still my argument about applying 5.6+5.7 seems to work? – user237522 May 31 '20 at 19:06
  • What about other conditions? Flatness? Separability? – user237522 May 31 '20 at 19:09
  • Without looking I guess those are something to the effect that a domain integral over a field is a field. Applying that requires $J$ prime. – Badam Baplan May 31 '20 at 20:05
  • @user237522 updated answer for you to address flatness and separability. Why do you expect that these properties would have anything to do with the property in question? Maybe you should ask a new question where you explain your motivation? – Badam Baplan May 31 '20 at 20:13
  • Thank you for your comments. I am looking for an additional condition that will guarantee that $J$ is maximal or prime (not with $A \subseteq B$ an epimorphism, but an injection). I do not expect flatness and separability to help, they are just properties that I am very slightly familiar with. Other properties are welcome. (Actually, I have asked such a question and deleted it. I am not sure if I know how to undelete it). – user237522 May 31 '20 at 20:32
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    @user237522 Note that epimorphisms of rings aren't necessarily surjections. There are lots of interesting injective epimorphisms. In particular, localizations at multiplicative sets generated by nonzero divisors will give you extensions satisfying your property – Badam Baplan May 31 '20 at 20:44
  • Thank you very much! Sounds nice. (I have not succeeded to undelete my question). – user237522 May 31 '20 at 20:58
  • Please, just to make sure, you meant that for example if $A$ is an integral domain and $S \subseteq A$ is a multiplicative set, then $A \subseteq B=S^{-1}A$ satisfies my property. – user237522 Jun 01 '20 at 22:11
  • @user237522 yes that is one application. you can also show that directly without using any fancy epimorphism arguments. – Badam Baplan Jun 02 '20 at 15:11
  • Thank you very much. Are there additional cases where such $J$ must be maximal? – user237522 Jun 02 '20 at 18:07
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Unless one puts an additional condition on the extension it is possible. Let $R := k[x,xy] \subset S := k[x,y]$, where $k$ is a field. The principal ideal $xS$ contracts to the maximal ideal $(x,xy)R$.

Of course, if one doesn't require such an ideal to be prime, one can take $k[x^2] \subset k[x]$ with $x^2k[x]$.

Youngsu
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  • Thank you very much! Please, what additional conditions on the extension guarantee that $J$ is a maximal ideal of $B$ (or at least prime or radical), under the assumption that $J \cap A$ is a maximal ideal of $A$ (I wish to assume that $J \cap A$ is maximal). – user237522 May 31 '20 at 06:20
  • What if $A \subseteq B$ is flat? integral? separable ring extension? – user237522 May 31 '20 at 06:24
  • You may want to look at the dimension formula, https://stacks.math.columbia.edu/tag/02II. – Youngsu May 31 '20 at 18:16
  • Please, could you elaborate? What additional conditions you had in mind? – user237522 May 31 '20 at 18:43
  • How does the dimension formula help if $J$ is not assumed to be a prime ideal? – user237522 May 31 '20 at 18:50