# The stalks at embedded points are non-reduced.

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 $$\text{Spec } A$$?

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 $$\alpha\neq 0$$).
• Thank you Aphelli , can you provide a bit hint why there is some $a\in A$ with annihilator exactly $\frak{p}$? Commented Jan 26, 2023 at 15:25
• Isn’t it what it means for $\mathfrak{p}$ to be an associated prime? Commented Jan 26, 2023 at 15:27
• I learned that associated prime is the generic point of some irreducible component of $\text{supp }a$? Commented Jan 26, 2023 at 15:28