Example of $A \subseteq B$, $\dim(A)=\dim(B)$, $J$ non-maximal ideal of $B$ and $J \cap A$ a maximal ideal of $A$

Question

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!

Answer 1

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 $k$.

Answer 2

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 $x^2k[x]$.

Answer 3

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.