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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 8:23
@AlexYoucis Are you invoking faithfully flat descent? :D – user38268 Feb 10 '14 at 15:16
@AlexYoucis Can you please explain what you mean by "flatness preserves normality"? Please see my question:… – user237522 Jun 16 '15 at 13:14
up vote 8 down vote accepted

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.

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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 9 '14 at 20:31
Remarkable answer: you are real strong algebraist! Who are you? :) – evgeniamerkulova Feb 9 '14 at 20:35
@evgeniamerkulova: Thank you, kindly: although I'm often (too much) reminded of how little I know. Thanks for asking a nice question! – zcn Feb 9 '14 at 20:41

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