Let $R = \mathbb{C}[t]$ be a ring of polynomials in variable $t$ with coefficients in the field of complex numbers $\mathbb{C}$ and let $$N = R[x]/(tx-t).$$
I claim that $N$ is not a flat $R$- module. If we consider the exact sequence $$0 \rightarrow(t) \rightarrow R$$ such that the ideal $(t)$ is viewed as an $R$-module.
We know that the above sequence exact iff the map $(t) \rightarrow R$ is injective. Let us assume that N is flat then this means that $N \otimes (t) \rightarrow N \otimes R$ has to be injective. How do I go from here?
Thanks.
Best Answer
First of all, you should be more precise when denoting tensor products. Actually, in your framework, $\otimes$ may mean $\otimes_{\mathbb{C}}$ or $\otimes_R$.
If $\otimes=\otimes_{\mathbb{C}}$ then you are thinking at $R$ and $N$ as $\mathbb{C}$-vector spaces. Thus they are free and hence flat.
If $\otimes=\otimes_R$ then proceed as follows. Set $I:=\langle tx-t\rangle\subseteq R$. Consider the element $\left(x-1+I\right)\otimes_R t\in N\otimes_R \langle t\rangle$. Notice that since $1\notin \langle t\rangle$ we do not have $\left(x-1+I\right)\otimes_R t=\left(tx-t+I\right)\otimes_R 1=0$ in $N\otimes_R \langle t\rangle$. However, when you map this element to $N\otimes_R R$ via $N\otimes_R \langle t\rangle \to N\otimes_R R$ then now we have $1\in R$ and so $\left(x-1+I\right)\otimes_R t=\left(tx-t+I\right)\otimes_R 1=0$ in $R$.
Conclusion: the map is not injective and hence $N$ cannot be flat.