The stalks at embedded points are non-reduced.

Question

I was trying to prove the claim that: Given locally Noetherian scheme $X$, the stalks at embedded points are non-reduced. (where embedded points means those associated point that not coming from generic point of the irreducible component of $X$).

There is a proof goes as follows: given an embedded point $p$, since we try to prove the stalk is a non reduced ring, we can assume we are working with affine scheme $\text{Spec} A$, however the solution in the link also assume that $\text{Spec} A$ is irreducible. That's my question, why we can assume it's irreducible? (does this means scheme $X$ is covered by affine irreducible subset?)

After assuming that, then $\text{Spec} A$ has a unique generic point the nilradical ${\frak{N}}(A)$. Since $p$ is an embedded point its closure is some irreducible component of $\text{supp }a$ for some $a\in A$. then we can prove $a\in {\frak{N}}(A)$ (you can find the detail of this step in the linked question). therefore $A_p$ is non-reduced.

The question is why we can assume the space is irreducible $\text{Spec } A$?

Answer 1

This isn’t what the proof says. It restricts to the case where $A$ is irreducible for the sake of simplicity.

Anyway, if $\mathfrak{p}$ is embedded, then there is some $a \in A$ whose annihilator is exactly $\mathfrak{p}$. So the image $\alpha$ of $a$ in $A_{\mathfrak{p}}$ is nonzero and has exactly $\mathfrak{p}A_{\mathfrak{p}}$ as an annihilator. In particular, it’s not invertible, so that $\alpha^2=0$ (while $\alpha\neq 0$).