Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Recall that we define the set of associated points of a locally Noetherian scheme $X$ as $\operatorname{Ass}(\mathcal{O}_X) = \{ x \in X : \mathfrak{m}_x \in \operatorname{Ass}_{\mathcal{O}_{X,x}}(\mathcal{O}_{X,x})\}$. I am having trouble understanding the proof of the following (Liu, 7.1.9):

Let $U$ be an open subset of $X$ and $i : U \to X$ the inclusion. The morphism $\mathcal{O}_X \to i_\ast(\mathcal{O}_U)$ is injective iff $\operatorname{Ass}(\mathcal{O}_X) \subseteq U$.

Since the property is local, he assumes $X = \operatorname{Spec}(A)$ so that the problem reduces to showing $A \to \Gamma(U, \mathcal{O}_X)$ is injective iff $\operatorname{Ass}(A) \subseteq U$. For the reverse direction, he argues as follows:

Let us suppose now that there exists a $\mathfrak{p} = \operatorname{Ann}(a) \in \operatorname{Ass}(A)$ with $\mathfrak{p} \not\in U$. Then for any point $x \in U$, we have $\operatorname{Ann}(a)\mathcal{O}_{X,x} = \mathfrak{p}\mathcal{O}_{X,x} = \mathcal{O}_{X,x}$; hence $a_x = 0$. Consequently, $a|_U = 0$.

The equality $\mathfrak{p}\mathcal{O}_{X,x} = \mathcal{O}_{X,x}$ is what I don't understand. As far as I understand, $\mathfrak{p}\mathcal{O}_{X,x}$ denotes the image of $\mathfrak{p}$ under the homomorphism $A \to \mathcal{O}_{X,x}$, right?

I also tried to prove $a_x = 0$ directly: we want to show that the image of $a$ vanishes under the homomorphism $A \to \mathcal{O}_{X,x} = A_Q$ (where $Q \subset A$ is the prime ideal corresponding to $x$); i.e. $t \cdot a = 0$ for some $t \not\in Q$. This is equivalent to the set $P \setminus Q$ being nonempty for all $Q \in U$. Assuming $U = D(f)$ ($f \in A$), we want $P \setminus Q$ nonempty for all prime ideals $Q$ not containing $f$ where $f \in P$. It is clear that any such $Q$ must contain $a$ (as $af = 0 \in Q$) but I don't know what to do next.

share|cite|improve this question
up vote 4 down vote accepted

The notation $\mathfrak{p}\mathcal{O}_{X,x}$ means the ideal generated in $\mathcal{O}_{X,x}$ by the elements $\frac{p}{1}$, where $p\in\mathfrak{p}$.

One important observation: If $U\subseteq \mbox{Spec}(A)$ and $x\in U$, but $\mathfrak{p}\not\in U$, then it follows that $\mathfrak{p}\not\subseteq x$, considered as ideals of $A$. To see this, take $U=\mbox{Spec}(A)\setminus V(I)$ for $I\subseteq A$ and write out the definitions.

This means that there exists $f\in\mathfrak{p}$, such that $f\not\in x$. Then $f$ is a unit in $\mathcal{O}_{X,x}=A_x$, that is $\frac{1}{f}\in\mathcal{O}_{X,x}$. But $\mathfrak{p}\mathcal{O}_{X,x}$ is an ideal in $\mathcal{O}_{X,x}$, and it contains $1 = f\cdot \frac{1}{f}$, so it must be the whole ring.

share|cite|improve this answer

Your Answer


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.