If $R/\mathfrak{a}$ is flat, then $\mathfrak{a}=\mathfrak{a}^2$

Question

Problem: Let $R$ be a commutative ring and $\mathfrak{a}$ an ideal such that $R/\mathfrak{a}$ is flat. Then $\mathfrak{a}^2=\mathfrak{a}$.

My attempt: $R/\mathfrak{a}$ is flat iff $\alpha\otimes (R/\mathfrak{a})\hookrightarrow R/\mathfrak{a}$ is an injection induced by the inclusion $\mathfrak{a}\hookrightarrow R$. I know that $\mathfrak{a}\otimes (R/\mathfrak{a})=\mathfrak{a}/\mathfrak{a}\mathfrak{a}=\mathfrak{a}/\mathfrak{a}^2$. Therefore, if I show that the first injection maps to $0$, then $\mathfrak{a}/\mathfrak{a}^2=0$ and so $\mathfrak{a}=\mathfrak{a}^2$.

So, given $x\in\mathfrak{a}$ and $r\in R/\mathfrak{a}$, how do I get to $x\otimes r=0$? Many thanks in advance!

Answer 1

Let $a\otimes \bar r\in\mathfrak a\otimes (R/\mathfrak a)$. This gets mapped to $a\otimes \bar r\in R\otimes(R/\mathfrak a)$. In $R\otimes(R/\mathfrak a)$, we have $$a\otimes \bar r=1\otimes a\bar r=1\otimes 0=0.$$