Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I want to solve an exercise from Liu's book Algebraic Geometry and Arithmetic Curves, namely exercise 4.9 in chapter 2: Let $X$ be a Noetherian scheme. Show that the set of points $x\in X$ such that $\mathcal{O}_{X,x}$ is reduced is open.

I already did the following: Because $X$ is Noetherian, we can write $X=\bigcup_{i=1}^n \operatorname{Spec}(A_i)$ where $A_i$ is a Noetherian ring. If we denote by $Z$ the set of points where $\mathcal{O}_{X,x}$ is reduced, then $Z$ is open if and only if $Z\cap \operatorname{Spec}(A_i)$ is open. So we can suppose $X=\operatorname{Spec}(A)$ with $A$ a Noetherian ring.

Then we also know that $\mathcal{O}_{X,x}=A_P$ with $P$ the prime ideal corresponding to the point $x$. So $\mathcal{O}_{X,x}$ is Noetherian too.

Can someone help me solving this problem?

share|improve this question
    
Hint: for each $x \in Z$, try to find an open neighborhood $U$ such that $\mathcal{O}_{X,y}$ is also reduced for all $y \in U$ (hence $U \subset Z$ and you can cover $Z$ with open sets this way). –  Adeel Dec 2 '12 at 13:24
    
This usually boils down to showing the following. Let $A$ be a local noetherian ring. Then, $A$ is reduced if and only if $A_p$ is reduced for all prime ideals $p$. –  Harry Dec 2 '12 at 15:26
add comment

1 Answer

up vote 4 down vote accepted

Let $N$ be the nilradical of $A$. First observe that $O_{X,x}$ is reduced if and only if $NO_{X,x}=0$. Then show that $NO_{X,x}=0$ implies the same is true in some open neighborhood of $x$.

share|improve this answer
    
Actually, that was exactly my problem: I can't construct an open neighborhood of $x$ where the ring of germs is reduced. –  Alies Dec 3 '12 at 13:14
1  
@Alies: let $f_1,\dots, f_n$ be a system of generators of $N$. Then $NO_{X,x}=0$ if and only if $(f_i)_x=0$ for all $i$. Now, if $f\in A$ and $f_x=0$, can you find an open neighborhood of $x$ in which $f=0$ ? –  user18119 Dec 3 '12 at 21:03
    
thank you very much! I didn't know how to use Noetherianity, but know I do! Thanks again! –  Alies Dec 4 '12 at 16:37
    
Is it true that we didn't use quasi-compactness? So this statement would be true for locally Noetherian schemes as well? –  Alies Dec 4 '12 at 17:00
    
Yes, locally noetherian is enough. –  user18119 Dec 4 '12 at 19:19
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.