TITLE: Non-flatness of $k[t]$ as a $k[t^2,t^3]$-module
QUESTION [7 upvotes]: 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.

REPLY [3 votes]: $\require{AMScd}$ 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.