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Let $A \subseteq B$ be commutative (finitely generated) $k$-algebras, $k$ is a field of characteristic zero. Let $J$ be an ideal of $B$ and assume that its contraction $I:=J \cap A$ is a maximal ideal of $A$.

(1) Is it true that $J$ is a maximal ideal of $B$ or at least a prime ideal of $B$? Probably not?

(2) What additional conditions are required in order to guarantee that $J$ is a prime ideal in $B$? For example, integrality of $A \subseteq B$.

(3) What if we assume that $J$ is a radical ideal?

Relevant questions: i and ii.

Any hints and coments are welcome!

user237522
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1 Answers1

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Hint: Consider $k \subset k[t]$ and some proper ideal $J \subset k[t]$. Then $J$ cannot contain any non-zero element of $k$ so that the intersection with $k$ is the zero ideal and hence a maximal ideal.

Edit: For an example that is not a field, consider $k[t] \subset k[t,s]$ and the (non-prime) ideal $J = (t,s^2) \subset k[t,s]$. Then $J \cap k[t] = (t) \subset k[t]$ is maximal.

Con
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  • Thank you very much. You meant, for example, $J=(t^2)$ or more generally $J=(f)$, where $f$ is reducible. Please, what if $A$ is not a field? – user237522 Feb 27 '20 at 00:03
  • Added another example where $A$ is not a field. – Con Feb 27 '20 at 00:07
  • Nice, thank you! What if $A \subseteq B$ is an algebraic ring extension? – user237522 Feb 27 '20 at 00:08
  • Perhaps $A=k[x^2,y]$, $B=k[x,y]$, $J=Bx^2+By$ is also a counterexample, since $I=Ax^2+Ay$ is a maximal ideal of $A$, but $J$ is not a prime ideal of $B$ ($J$ is a radical ideal of $B$). – user237522 Feb 27 '20 at 00:12
  • Perhaps the correct claim is: If $I$ is a maximal ideal of $A$, then $J$ is a radical ideal of $B$? – user237522 Feb 27 '20 at 00:18
  • What do you mean by an algebraic ring extension? An integral extension? The latter claim is wrong as can be seen by my second example ($J$ contains $s^2$ but not $s$). Its radical is given by $(t,s)$. – Con Feb 27 '20 at 00:23
  • I do not know how to prove (if true) that: $Ax^2+Ay^2 \supseteq (Bx^2+By) \cap A$ (the other inclusion is trivially true). – user237522 Feb 27 '20 at 00:25
  • By an algebraic extension $A \subseteq B$ I mean that for every element $b \in B$, there exist $a_0,a_1,\ldots,a_n \in A$ such that $a_nb^n+\cdots+a_1b+a_0=0$. By an integral extension I mean a special case of an algebraic extension, namely, an algebraic extension with $a_n=1$ (for every $b \in B$). – user237522 Feb 27 '20 at 00:28
  • Thank you for mentioning that my latter claim is wrong. I understand your counterexample. Again, your counterexample is for non-algebraic $A \subseteq B$. In my latter claim I meant to further assume that $A \subseteq B$ is algebraic. – user237522 Feb 27 '20 at 00:36
  • Actually, I guess that my above example of $A=k[x^2,y]$, $B=k[x,y]$, $J=Bx^2+By$ also shows that my latter claim is wrong: $(x^2,y)$ is not radical in $B$. But I am not sure if $Ax^2+Ay=(Bx^2+By) \cap A$. – user237522 Feb 27 '20 at 00:42
  • Have only checked quickly on the bus now, but seems true to me. By intersecting with $A$ you only take the polynomials whose degree in $x$ is even which would also exactly be the polynomials that you are allowed to use over $A$. – Con Feb 27 '20 at 16:56
  • Thanks you. I wonder if my question can be slightly changed to have a more interesting result. What about requiring separability of $A \subseteq B$? – user237522 Feb 27 '20 at 18:34
  • Please, what if $\dim(A)=\dim(B)$, with $\dim()$ being the Krull dimension, for example. In your counterexample, $\dim(A)=1, \dim(B)=2$. Thank you! – user237522 May 31 '20 at 01:57