How to prove that a ring is not flat over $k[t,s]$?

Question

Let $k$ be a field, $A=k[t,s]$, and $C=A[z]/(tz-s)$. How can I prove, using the ideals $tA$ and $sA$, that $C$ is not flat over $A$? (Liu, Algebraic Geometry and Arithmetic Curves, Exercise 2.6(c).)

I know that if $A$ is a Dedekind domain then $A$-module is flat if and only if it is not torsion-free over $A$. But Dedekind domains are new structures for me so I'm not sure if $k[t,s]$ is a Dedekind domain. Or do I have to show that not both of $tA$ and $sA$ can't be maximal or prime over $A_{tA}$ or $A_{sA}$?

Answer here http://math.stackexchange.com/questions/110599/why-isnt-mathbbcx-y-z-xz-y-a-flat-mathbbcx-y-module for $k = \mathbf{C}$, but... — Olórin, Jan 19 '15 at 17:12
@RobertGreen: The answer by Pierre-Yves Gaillard in the link you gave is for any commutative ring, which is much more general than the question asked. — Krish, Jan 19 '15 at 17:17
@Krish Hum yes, the question is for $k=\mathbf{C}$, but... the answer works for any commutative ring $k$ with unit. That's what I meant. — Olórin, Jan 19 '15 at 17:36

user26857 · Accepted Answer · 2015-01-21T15:42:36.303

2

Hint. If $C$ is $A$-flat, then $(sA\cap tA)C=sC\cap tC$. Does this equality hold?

edited Jan 21 '15 at 15:42

answered Jan 19 '15 at 19:16

user26857

52,094

I understand the hint but I don't see why the identity does not hold in this situation. – Jaakko Seppälä Jan 19 '15 at 19:58
Can you see what is $sA\cap tA$? – user26857 Jan 19 '15 at 19:59
Polynomials such that every term is of the form $as^nt^m,n,m\geq 1$? – Jaakko Seppälä Jan 19 '15 at 20:06
For short, it is $stA$, so $(sA\cap tA)C=stC$. Now can you show that $s\in sC\cap tC$ and $s\notin stC$? – user26857 Jan 19 '15 at 20:08
Umm. $s\in sC$ as $s(tz-s)/(tz-s)=s$ and does $s\in tC$ as $A[z]$ is factorial? $s\not\in stC$ as there is no secod degree polynomial $P$ in $A[z]$ such that $P/(tz-s)=st$? – Jaakko Seppälä Jan 19 '15 at 20:23
I'd say that $s\in sC$ since $s=s\cdot 1$. Moreover, in $C$ we have $tz=s$, so $s\in tC$. If $s\in stC$ then $1\in tC$ since $C$ is an integral domain, so $t$ is invertible in $C$ which I don't think is true. – user26857 Jan 19 '15 at 20:34

How to prove that a ring is not flat over $k[t,s]$?

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