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I want to show that $\mathbb{C}[s, t, u] \to \mathbb{C}[x, y] : s \mapsto x^2, t \mapsto xy, u \mapsto y^2$ is not flat.

If $s \otimes y \neq t \otimes x$ in $(x^2, xy, y^2) \otimes_{\mathbb{C}[s, t, u]} \mathbb{C}[x, y]$, then we can show it.

But I have no idea how I can show inequality of elements of a tensor.

EDIT

I'm sorry, I have misunderstood.

I want to show that the inclusion $\mathbb{C}[x^2, xy, y^2] \to \mathbb{C}[x, y]$ is not flat.
In this case, it seems that the arguments of 2 comments do not work. So I think that I must show $x^2 \otimes y \neq xy \otimes x$ in $(x^2, xy, y^2) \otimes_{\mathbb{C}[x^2, xy, y^2]} \mathbb{C}[x, y]$.

I have seen this. But I can't find nice $\mathbb{C}[x^2, xy, y^2]$-linear map $(x^2, xy, y^2) \times \mathbb{C}[x, y] \to M$.

user26857
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zom
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  • Hi. I am not certain if I follow your argument, but you can use that the image of $s,t$ is not a regular sequence to show the non-flatness. – Youngsu Jun 20 '20 at 20:21
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    For a ring homomorphism of your kind, $A\to B$, where $\dim A>\dim B$, the map can not be flat. One way of seeing this, write $0\to I\to A\to B\to 0$, with $I\neq 0$. Then, tensor over $A$ by $B$ and notice that $I\otimes B\neq 0$ and the tensored sequence is thus no longer exact. – Mohan Jun 20 '20 at 20:26
  • @Youngsu Thank you for your comments. But I have misunderstood my text. – zom Jun 21 '20 at 06:41
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    @zom - You can use the $\mathbb C[x^2,xy,y^2]$-bilinear map $(x^2,xy,y^2)\times\mathbb C[x,y)\to\mathbb C\simeq\mathbb C[x,y]/(x,y)$, $$(f,g)\mapsto\frac{\partial^2f}{\partial x^2}(0,0)\ \frac{\partial g}{\partial y}(0,0).$$ – Pierre-Yves Gaillard Jun 21 '20 at 12:24
  • Sorry to bother. Why $x^2 \otimes y \neq xy \otimes x$ in$(x^2, xy, y^2) \otimes_{\mathbb{C}[x^2, xy, y^2]} \mathbb{C}[x, y]$ means the map is not flat? – NEMO Jul 21 '20 at 05:46

1 Answers1

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One of the standard way such things are done is as follows.

I state a standard theorem whose proof can be found in most text books which touches upon flatness and homological algebra.

Let $A\to B$ be a flat and finite map of Noetherian rings. Then $B$ is $A$-projective.

Assuming this, we get $B=\mathbb{C}[x,y]$ is a projective module over $A=\mathbb{C}[x^2,xy,y^2]$ if you assume flatness. Easy to check that rank of $B$ as an $A$-module is $2$ and then $B/MB$ is a $\mathbb{C}$ -vector space of dimension two for any maximal ideal $M\subset A$. But, $B/MB$ has dimension three, where $M=(x^2,xy,y^2)$, leading to a contradiction.

If you knew that a Noetherian local ring is regular if and only if its global dimension is finite, you could also argue that if $A\to B$ is finite and flat with $B$ regular, then so is $A$, giving you another argument for what you want.

Mohan
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  • Thank you very much! Pierre-Yves Gaillard's comment is so direct, and your answer is very cool. I like both! – zom Jun 25 '20 at 12:53