Consider $k$ a field and the rings $A=k[X^2,X^3]\subset B=k[X]$. How to prove that $B$ is not flat over $A$ by using only the definition of flatness that it maintains exact sequences after making tensor products?
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I know you wanted to prove this directly, but are you aware of the fact that flatness preserves normality. In your case $A$ is non-normal and $B$ is normal, so $B/A$ can't be flat. – Alex Youcis Feb 10 '14 at 08:23
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@AlexYoucis Are you invoking faithfully flat descent? :D – Feb 10 '14 at 15:16
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@AlexYoucis Can you please explain what you mean by "flatness preserves normality"? Please see my question: http://math.stackexchange.com/questions/1326926/flatness-and-normality/1327174#1327174 – user237522 Jun 16 '15 at 13:14
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Related: https://math.stackexchange.com/questions/587838 – Watson Jan 22 '17 at 21:54
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Let $m = (x^2, x^3)A$, and consider the short exact sequence
$$0 \to m \to A \to A/m \to 0$$
Tensoring with $B$ over $A$ gives
$$m \otimes_A B \xrightarrow{\phi} B \to B/mB \to 0$$
where $\phi(m \otimes b) = mb$. Then $x^2 \otimes_A x - x^3 \otimes_A 1 \in \ker \phi$, but $x^2 \otimes_A x \ne x^3 \otimes_A 1$ in $m \otimes_A B$ (to see this, it suffies to check that they are distinct elements of the $k$-vector space $(m \otimes_A B) \otimes_A k \cong m \otimes_A (k \otimes_k k) \otimes_A B \cong (m/m^2) \otimes_k (B/mB)$. Now $\{\overline{x^2},\overline{x^3}\}$ is a basis of $m/m^2$, and $\{\overline{1},\overline{x}\}$ is a basis of $B/mB$, so $\overline{x^2} \otimes_k \overline{x}$, $\overline{x^3} \otimes_k \overline{1}$ are distinct basis elements of $(m/m^2) \otimes_k (B/mB)$). Thus $\phi$ is not injective.
zcn
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The tensors are over $A$ (except when justifying that element is nonzero and reducing to $A/m$). I've edited to (hopefully) make it clearer – zcn Feb 09 '14 at 20:31
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3Remarkable answer: you are real strong algebraist! Who are you? :) – evgeniamerkulova Feb 09 '14 at 20:35
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@evgeniamerkulova: Thank you, kindly: although I'm often (too much) reminded of how little I know. Thanks for asking a nice question! – zcn Feb 09 '14 at 20:41
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@zcn Sorry to rekindle this question after so long. I'm really not clear why the isomorphism $m \otimes_A (k \otimes_k k) \otimes_A B \cong (m/m^2) \otimes_k (B/mB)$ holds. Could you explain this? Thanks in advance. – smalldog Dec 13 '17 at 13:57
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@Nao. I would write: $(m \otimes_A B) \otimes_A k \cong (m\otimes_A k)\otimes _k (B\otimes k)= (m\otimes_A A/m)\otimes _k (B\otimes A/m) \cong m/m^2\otimes_k B/mB$ – Georges Elencwajg Sep 15 '18 at 19:09