I found a example about the module is not a flat module: "Let $R=\mathbb{C}[t]$ be a ring of polynomials variable $t$ with coefficients in $\mathbb{C}$. Consider $R$-module $N=R[x]/\langle tx-t\rangle$. Then $N$ is not a flat $R$-module".
They solved as follows:
Let $M=\langle t\rangle$ be a $R$-module generated by $t\in R$ and $f:M\to R$ is a homomorphism. We proof $g:N\otimes_R M \to N\otimes R$ is not a homomorphism. Indeed, let $I=\langle tx-t\rangle\subseteq R$. Consider element $(x-1+I)\otimes_Rt\in N\otimes_R M$. Remark that $1\notin M$, then we not have $(x-1+I)\otimes_Rt=(tx-t+I)\otimes_R1=0$ in $N\otimes_R M$. However, when we map $g:N\otimes_R M \to N\otimes R$, we have $1\in R$. So $(x-1+I)\otimes_Rt=(tx-t+I)\otimes_R1=0$ in $R$. Hence $g$ is not a homomorphism so $N$ is not a flat $R$-module.
I have a question:" Why we must let $R=\mathbb{C}[t]$? Why is not $\mathbb{R}[t]$?" Thankyou