# Restriction maps in an integral scheme are injective

Suppose $X$ is an integral scheme. I would like to show that the restriction maps $res_{U,V} : O_X(U) \rightarrow O_X(V)$ is an inclusion so long as $V$ is not empty. I was wondering if someone could give me some assistance with this. This is exercise 5.2 I on Ravi Vakil's notes. Thank you!

• You may want to look at my comment below the answer you accepted, since there is a subtlety in the question/solution which that answer doesn't make explicit, and which you may not be aware of. Commented Dec 20, 2014 at 15:02

Let $X$ be an integral scheme. Let $\xi$ be its generic point. If we show that the canonical map $\mathcal O_X(U) \to \mathcal O_\xi$ is injective, we're done. Since $U$ can be covered by affine open subsets, we can assume that $U = \operatorname{Spec} A$ is affine. Now the map $\mathcal O_X(U) \to \mathcal O_\xi$ corresponds to the canonical map $A \to \operatorname{Frac}(A)$, where $\operatorname{Frac}(A)$ is the quotient field of $A$. This map is clearly injective as desired.
In other words, we want to show that a section $s\in O_X(U)$ that vanishes on $V$ is necessarily the trivial section. Since $X$ is integral, $U$ is irreducible, thus $V$ is dense in $U$, and we're done.
• You could probably give more details, since this depends on integrality, not just irreducibility. E.g. if $U =$ Spec $k[x,y]/(xy,y^2)$, and $V$ is the complement of the origin (i.e. that complement of the closed point corresponding to the maximal ideal $(x,y)$), then $U$ is irreducible, $V$ is dense in $U$, and the section $y$ of $\mathcal O_U$ vanishes when restricted to $V$. Commented Dec 20, 2014 at 15:02
• @tracing Thank you for pointing it out. I filled in the details myself, and I did use the fact that $X$ is reduced. By the way is there an easy way to see that $U$ is irreducible? Commented Dec 21, 2014 at 17:17
• @user201886: Irreducible is a topological property, and so depends only on the underlying reduced scheme (or ring, in the affine scheme setting). In this case the underlying reduced ring is just $k[x]$ (since $y$ gets sent to zero), which is a domain, hence has irreducible Spec. Commented Dec 22, 2014 at 15:43