Are embedded points where the nonreducedness is?

Question

I know that if Spec $A$ is reduced, then there are no embedded points. I was wondering, if I know that $p$ is an embedded point of some Spec $B$, does that imply $B_{p}$ is non-reduced? Thanks!

Answer 1

Yes. If we have $p \in \operatorname{Ass} B$, we have $pB_p \in \operatorname{Ass} B_p$. So $pB_p = Ann(x)$. Clearly $0 \neq x$ is no unit because $p \neq 0$ (since it was an embedded prime, in particular not a minimal prime). We obtain $x \in pB_p$, hence $x^2=0$.