Direct proof of non-flatness

Question

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?

Answer 1

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.