Non-flatness of $k[t]$ as a $k[t^2,t^3]$-module

Question

Let $k$ be a field. How to show that $k[t]$ is not flat as a module over $k[t^2,t^3]$ ?

Since the ring extension $k[t^2,t^3]\subseteq k[t]$ is integral, it is clear that $k[t]$ is a finitely generated $k[t^2,t^3]$-module , and also torsion free. I am unable to proceed further.

Please help. Thanks in advance.

Answer 1

Here is a proof direct from the definition. Consider the sequence of $k[t^2,t^3]$ modules $$0\rightarrow k[t]/(t^2)\rightarrow k[t]/(t^3)$$ Where the mapping is

$$f(t)+(t^2)\mapsto tf(t)+(t^3)$$ this is well defined and injective. However

$$k[t]/(t^2)\otimes k[t]\rightarrow k[t]/(t^3)\otimes k[t]$$ is not injective as $$t\otimes t\mapsto t^2\otimes t=1\otimes t^3=t^3\otimes 1=0$$

Since tensor product is right exact this suffices.

Answer 2

$\require{AMScd}$ Let $R=k[t^2,t^3]$. There is a short exact sequence of the form

\begin{CD} 0 @>>> R @>{\left(\begin{smallmatrix}t^3\\-t^2\end{smallmatrix}\right)}>> R\oplus R @>{\left(\begin{smallmatrix}1&t\end{smallmatrix}\right)}>> M @>>> 0 \end{CD}

Applying the functor $N\otimes(-)$, with $N=R/(t^2,t^3)$ results in a non-exact sequence: the map $N\otimes R\to N\otimes(R\oplus R)$ is zero and does not have a zero kernel. It follows that $\operatorname{Tor}^R_1(N,M)\neq0$ and, then, that $M$ is not flat.