Let $A=\mathbb{C}[t]$ and $B=\mathbb{C}[t^2,t^3]$. We can naturally regard $A$ as a $B$-module. I want to show that $A$ is not flat $B$-module.
Let $f:A\otimes_B A \to B\otimes_B A,\ x\otimes y \mapsto t^2x\otimes y$.
$f(1\otimes t)=f(t\otimes 1)$, so it is enough to show that $1\otimes t \neq t\otimes 1$ in $A\otimes_B A$. But I can't show it. How to prove it?
Thanks!
Here is one of many ways to show this. Let $D=A/(t^2)$. It suffices to show $1\otimes t\neq t\otimes 1$ as elements of $D\otimes_B D$. But notice that the $B$-algebra structure homomorphism $B\to D$ factors through the quotient $B\to \mathbb{C}$ that sends $t^2$ and $t^3$ to $0$. It follows that the natural map $D\otimes_\mathbb{C} D\to D\otimes_B D$ is an isomorphism (every element of $B$ just acts on $D$ via its constant term). Finally, it is clear that $1\otimes t\neq t\otimes 1$ in $D\otimes_\mathbb{C} D$ (for instance, because $\{1,t\}$ is a basis for $D$ over $\mathbb{C}$).
