Why is the following homomorphism not flat?

Question

Let k be a field of characteristic 2. Consider the map $k[x,y]/(y^2+x^3) \rightarrow k[x,y]/(y^2+x^3)$ given by $x \rightarrow x^2$ and $y \rightarrow y^2$. Why is it that this map is not flat? I have tried to look at the dimension of the fibers, and the dimension criterion but nothing comes up here to my mind.

Answer 1

Since flatness is stable under base change, we may mod out $x$ and it suffices to prove that $R=k[y]/(y^2) \to k[y]/(y'^2)=R'$, $y \mapsto 0$ is not flat. For this we prove that $(y) \otimes_R R' \to R'$ is not injective. In fact this homomorphism is zero, so we only have to show $(y) \otimes_R R' \neq 0$. We have $\mathrm{Ann}_R(y)=(y)$, hence $R/(y) \cong (y)$, $[1] \mapsto y$. It follows $(y) \otimes_R R' \cong R/(y) \otimes_R R' \cong R'/(0) = R'$.

Answer 2

Let us put $R=R'=k[x,y]/(y^2+x^3)$ and let $f: R \to R'$ be your map. I use two different names for the same ring so that we don't get confused. In fact let us write $x'$ and $y'$ for the variables in $R'$, to distinguish them from $x$ and $y$ in $R$. So the map is $x \mapsto x'^2$, $y \mapsto y'^2$. Let $I=\left<x,y\right> \subseteq R$ and consider the inclusion $I \hookrightarrow R$. I claim that $I \otimes_R R' \to R'$ is no longer an inclusion. To see this, remark that the tensor $y \otimes 1 + x \otimes x'$ is mapped to $y'^2+x'^3=0$, so it suffices to show that it is not zero in $I \otimes_R R'$. Can you do this?