Trouble showing flatness

Question

Let $K$ be a field and $\pi: K[x]/(x^2) \to K$ be the ring homomorphism given by the valuation at $0$. I'm stuck in showing that $\pi^*(K)$ (the pullback) is not a flat module (over $K[x]/(x^2)$).

Answer 1

Let $A$ be a commutative ring and $I\subset A$ an ideal. Let us investigate whether $A/I$ is flat over $A$.

Consider the injection $0\to I\to A$ and tensor it with $A/I$.
Using $M\otimes_A A/I=M/IM$ (for any $A$-module $M$), we get the morphism of $A$-modules: $$I\otimes_AA/I\to A\otimes _AA/I \quad\text {identified with} \quad I/I^2\to A/I: \bar i\mapsto \tilde i=\tilde 0 \quad (\bigstar)$$ Hence if $A/I$ is flat, $(\bigstar)$ must be injective so that necessarily $I/I^2=0$, i.e. $I=I^2$ .
Since in our case $I=(\bar x)\neq I^2=(0)\subset A=K[x]/(x^2)=K[\bar x]$ , we conclude that: $$A/I=K \quad \text {is not flat over} \quad A=K[x]/(x^2)$$