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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.

user26857
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user
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2 Answers2

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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.

Rene Schipperus
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    Just one question ... why is $t \otimes t$ non-zero in $k[t]/(t^2) \otimes_{k[t^2,t^3]} k[t]$ ? – user Nov 22 '17 at 13:56
  • Its a good exercise in tensor products. Map to $k[t]/(t^2)\otimes k[t]/(t^2)$ this is a $k[t^2,t^3]$ module but multiplication by $t^2$ and $t^3$ is always zero, so it has no relations that are not present an a $k$ module. So now the question is reduced to tensor products of vector spaces over a field. A much simpler and well understood case. – Rene Schipperus Nov 22 '17 at 14:06
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$\require{AMScd}$ 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.

Mariano Suárez-Álvarez
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    Alternatively, as $R$ is noetherian and $M$ finitely generated, $M$ is finitely presented. It follows then from Lazard's theorem on flat modules that $M$ is projective if it is flat. But then the above exact sequence splits, and it also splits if we tensor it with $N$: but it doesn't. – Mariano Suárez-Álvarez Nov 21 '17 at 22:46