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Let $A \subseteq B$ be two commutative $\mathbb{C}$-algebras. Let $q$ be an ideal of $B$ and denote $p=q \cap A$.

By [Lemma 3][1], if $A \subseteq B$ is an integral extension, then: $q$ is maximal in $B$ if and only if $p$ is maximal in $A$.

If $A \subseteq B$ is not necessarily integral, when is it possible to deduce the following half claim: If $p$ is maximal (prime) in $A$, then $q$ is maximal (prime) in $B$.

I guess that this does not hold in general. Does one of the following additional conditions help: (i) flatness of $A \subseteq B$; (ii) faithful flatness of $A \subseteq B$, see [this][2]; (iii) separability of $A \subseteq B$; (iv) $A$ and $B$ being UFD's.

Thank you very much!

Edit: One direction of the proof of Lemma 3 says: "If $p$ is maximal then $A/p$ is a field and every element of $B/q$ is algebraic over $A/p$. Hence $B/q$ is a field" (so $q$ is maximal).

Therefore, if I am not wrong (but I may be wrong), the following claim is true: If $A \subseteq B$ is algebraic, then: If $p$ is maximal in $A$, then $q$ is maximal in $B$.

Reason: If I am not worng, exactly the same proof holds; the critical line is: and every element of $B/q$ is algebraic over $A/p$.

Am I right?

strong text [1]: http://math.stanford.edu/~conrad/210BPage/handouts/math210b-going-up.pdf [2]: Extensions and contractions of prime ideals under integral extensions

user237522
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  • Is there anything saying that $B$ is a finitely generated $A$-algebra? If so $B/q$ is algebraic over $A/p$ by Zariski lemma. If $B/p$ is local and $B/A$ is flat then $p$ is maximal, as for $a\in A-A/p^\times$, $A/(p,a)$ injects into $B/(p,a)$ so $a\in q$ thus $\in p$. – reuns Mar 07 '21 at 14:04
  • Thank you very much! Please: (1) Could you make your comment an answer (with slightly more elaboration)? (2) Is my claim in the edit true? – user237522 Mar 07 '21 at 14:20
  • https://en.wikipedia.org/wiki/Zariski%27s_lemma – user237522 Mar 07 '21 at 14:41
  • I think I was wrong and should have assumed that $p,q$ are prime ideals, since $B/q$ should be an integral domain for my claim to hold. – user237522 Mar 07 '21 at 15:00
  • What does algebraic mean for you? If you're not restricting attention to domains this notion can get hairy. – Badam Baplan Mar 10 '21 at 21:51
  • @BadamBaplan, thank you for your comment. By $A \subseteq B$ algebraic I mean that for every $b \in B$ there exist $a_n,\ldots,a_1,a_0 \in A$ such that $a_nb^n+\cdots+a_1b+a_0=0$. I do not mind to assume that $B$ (hence $A$) is a domain. However, I do not want to assume that $p,q$ are prime ideals, so $B/q, A/p$ may not be domains. – user237522 Mar 10 '21 at 21:55

1 Answers1

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For $(1), (2), (4)$, take $k \subseteq k[x]$ for any field. This is a free (a fortiori faithfully flat) extension of UFDs in which the non-maximal ideal $0$ of $k[x]$ contracts to the maximal ideal $0$ of $k$.

I highly doubt that separability is even remotely related to the property in question, and will politely decline to think about it until you provide some motivation.

As for algebraic..... no, even if we restrict attention to $A, B$ domains this doesn't work out. The key property of integral extensions in your "lemma 3" is that they are stable under quotients, but such is not the case for algebraic extensions.

For example $\mathbb{C}[x,xy] \subseteq \mathbb{C}[x,y]$ is an algebraic extension but the height $1$ (non-maximal) prime $(x)\mathbb{C}[x,y]$ contracts to the maximal ideal $(x,xy) \mathbb{C}[x,xy]$.

Badam Baplan
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