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Let $A \subseteq B=A[b]$ be two commutative $k$-algebras, where $k$ is a field of characteristic zero and $b \in B-A$.

Let $b_1,b_2 \in B-A$ be two elements of $B$ such that $\frac{B}{\langle b_i \rangle}=\frac{A[b]}{\langle b_i \rangle}$ is flat over $A$, $1 \leq i \leq 2$.

In that case, is it true that $\frac{B}{\langle b_1,b_2 \rangle}=\frac{A[b]}{\langle b_1,b_2 \rangle}$ is flat over $A$? If not, it would be nice to see a counterexample.

I know that there are surjections $\frac{B}{\langle b_i \rangle} \to \frac{B}{\langle b_1,b_2 \rangle}$, $1 \leq i \leq 2$, but do not see the connection to flatness (if such connection exists). Am I missing something trivial here (probably yes)?

Any hints and comments are welcome!

user237522
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1 Answers1

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Take $A=k[X]$, $B=k[X,Y]$, $b_1=X^2+Y$, $b_2=Y$. Then $B/\langle b_i\rangle$ is free over $A$ of rank $1$ (hence flat). However, $B/\langle b_1,b_2\rangle=A/(X^2)$ is certainly not flat over $A$.

Aphelli
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  • Thank you very much! Could you take $b_1,b_2$ having an invertible Jacobian? In your example $\operatorname{Jac}(b_1,b_2)=\operatorname{Jac}(x^2+y,y)=\operatorname{Jac}(x^2,y)=2x \notin k^{\times}$. If I am not wrong, the following $b_1=x^2+y$, $b_2=x$ (having an invertible Jacobian) is not a counterexample, since a field is flat. – user237522 Feb 09 '20 at 12:46
  • $k[x,y]/\langle x^2+y,x \rangle= k[x,y]/ \langle x,y \rangle = k$. – user237522 Feb 09 '20 at 12:57
  • Yes, and $k$ isn’t flat over $k[X]$, as it isn’t torsion-free. I’m not sure but I think that if the Jacobian is invertible, the quotient ring has dimension $0$ and is an integral domain (see smoothness implying regularity implying integral domain). So it’s finite over $k$, so not torsion-free over $A$, so not flat. – Aphelli Feb 09 '20 at 17:06
  • Yes, and $k$ is not flat over $k[x]$ by the flatness criterion presented in the first answer to this question https://math.stackexchange.com/questions/1259551/how-to-show-fraction-field-is-flat-without-localization since the kernel contains the principal ideal $(x)$ (the action of $k[x]$ on $k$ is $x\lambda=0$ for every $\lambda \in k$ and scalar-the usual multiplication in $k$). Please see also another question of mine: https://math.stackexchange.com/questions/3536084/flatness-of-a-fx-1-dots-x-n-f-1-dots-f-n?noredirect=1&lq=1 – user237522 Feb 09 '20 at 17:07
  • In the special case $B=k[x,y]=k[x][y]=k[y][x]$, $f_1=p, f_2=q$: What if the following four conditions are satisfied: (1) $\frac{k[x,y]}{\langle p \rangle}$ is $k[x]$-flat. (2) $\frac{k[x,y]}{\langle p \rangle}$ is $k[y]$-flat. (3) $\frac{k[x,y]}{\langle q \rangle}$ is $k[x]$-flat. (4) $\frac{k[x,y]}{\langle q \rangle}$ is $k[y]$-flat. Is it true that now $\frac{k[x,y]}{\langle p,q \rangle}$ is $k[x]$-flat and $k[y]$-flat? (probably no?) – user237522 Feb 09 '20 at 17:23
  • Your example satisfies (1) and (3), but not (2) and (4) ($k[x]$ is not flat over $k[y]$, etc.). – user237522 Feb 09 '20 at 17:34
  • Perhaps $p=x^2y, q=xy^2$ satisfies all four conditions, but $\frac{k[x,y]}{\langle x^2y,xy^2 \rangle}$ is not $k[x]$-flat and not $k[y]$-flat. I should check this more carefully. – user237522 Feb 09 '20 at 18:08
  • This seems dubious: you get torsion in the partial quotients! – Aphelli Feb 09 '20 at 18:10
  • So please what do you think about my claim concerning the four conditions? – user237522 Feb 09 '20 at 18:12
  • I suspect that in that case necessarily $\langle p,q \rangle =\langle r \rangle$, for some $r \in k[x,y]$. Or perhaps I am missing something. I am now trying to post this as a question in MO. – user237522 Feb 09 '20 at 18:14
  • How about (in characteristic not $2$), $p=x+y$, $q=x-y$? – Aphelli Feb 09 '20 at 20:17
  • Thank you! This is exactly the counterexample I have received in MO, so I added a fifth condition: https://mathoverflow.net/questions/352299/flatness-of-certain-quotient-rings – user237522 Feb 09 '20 at 20:35