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After receiving interesting answers to this question (see also this question), I wish to concentrate on the following special case: Let $A \subseteq B=k[x,y]$, $k$ is a field of characteristic zero. Assume that $\dim(A)=\dim(B)$ (of course, $\dim(B)=2$).

Question 1: Is it possible to find a non-maximal ideal $J$ of $B=[x,y]$ such that $J \cap A$ is a maximal ideal of $A$?

The answer to question 1 is yes, with the following being an example of Youngsu:

$A=k[x,xy]$, $J=xk[x,y]=(x)_{k[x,y]}$ contracts to the maximal ideal $(x,xy)A=(x,xy)_{k[x,xy]}$.

So now I add the condition that $J$ is generated by two algebraically independent elements of $B$, but still gets an example: $A=k[x,y(y-1)]$, $J=xk[x,y]+y(y-1)k[x,y]=(x,y(y-1))_{k[x,y]}$ contracts to $xA+y(y-1)A=(x,y(y-1))_A$, which is clearly maximal in $A$. ($J$ is non-maximal in $B$ and also non-prime in $B$).

So now I ask:

Question 2: 'When' such $J$ (= two generated by two alg. ind. elements of $B$) must be maximal (or at least prime) in $B$, given its contraction $J \cap A$ is maximal in $A$?

Ideas for 'When':

(1) $A \subseteq B$ is integral: Not a good idea, since in my last example $A \subseteq B$ is integral, but $J$ is not a maximal ideal of $B$ (and also not a prime ideal of $B$).

(2) $A \subseteq B$ is flat: (I have not yet tried to figure out if $k[x,y(y-1)] \subseteq k[x,y]$ is flat or not). For faithfully flat extension $A \subseteq k[x,y]$, we have for every ideal $I$ of $A$: $Ik[x,y]\cap A=I$.

(3) $Q(A)=Q(B)$: I do not know if the fraction fields are relevant here.

(4) $A \subseteq B$ is separable: I think that in my last example the extension is not separable, but do not see any connection between my question and being separable.

Any hints and comments are welcome!

user237522
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  • As for (2), your example is already flat. One way to see this: for any univariate polynomial $f(y)$, $k[f]$ is a PID. Torsion-free modules over PIDs are flat, so $k[f] \subseteq k[y]$ is flat. Also $k[f] \subseteq k[f,x]$ is flat, so changing base ring we get $k[y] \otimes_{k[f]} k[f,x] = k[x,y]$ is flat over $k[f,x]$. – Badam Baplan Jul 22 '20 at 13:24
  • Also I have to ask: since you say that you "do not see any connection between [your] question and being separable," why are you asking about separability? – Badam Baplan Jul 22 '20 at 14:17
  • Thank you for your comments. Concerning your question: I named a few properties, hoping that one of them will be relevant. – user237522 Jul 22 '20 at 19:40
  • Btw, when you say algebraically independent, do you mean algebraically independent over $k$? Over $A$ – Badam Baplan Jul 22 '20 at 19:57
  • Yes, I meant algebraically independent over $k$. Please, do you have any ideas relevant for my above question? – user237522 Jul 22 '20 at 20:19
  • Not really, besides quoting the answer I posted to the linked question. Again if $A \subseteq B$ is a ring epimorphism and $J \cap A $ is maximal then $J$ is maximal. This is not at all specific to your assumptions. I'm sorry if this isn't relevant to what you're looking for. It might help if you have context into where this question is coming from. – Badam Baplan Jul 22 '20 at 22:17
  • Ah, one more thing. Now that we've clarified what you mean by algebraically independent, i can help that (3) also is insufficient. You can reuse Youngsu's example of $k[x,xy]$ but instead take $J = (x,y^2)$. – Badam Baplan Jul 22 '20 at 22:22
  • Thank you for helping me.Your previous idea (epimorphism) a month ago in the other question was very nice. (You are right concerning the context). Thanks for (3). – user237522 Jul 23 '20 at 11:06

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