In the book algebraic geometry I of Torsten Wedhorn theorem 6.45

Let $X$ be a locally noetherian normal scheme, and let $U\subseteq X$ be an open subset with $\operatorname{codim}_X(X-U)\ge2$. Then the restriction map $\Gamma(X,O_X)\rightarrow\Gamma(U,O_X)$ is an isomorphism. In other words:every function $f\in\Gamma(U,O_X)$ on $U$ extends uniquely to $X$.

Locally noetherian means that noetherian but we do not need quasi-compact .

Question: I do not know why can we reduce to the affine case. Can there exist some affine open subscheme $\operatorname{Spec}A$ such that the intersection of $\operatorname{Spec}A$ and $U$ is empty?($U$ is the open subscheme in the theorem)

Could some one give an answer? Thank you.

  • $\begingroup$ Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures. $\endgroup$ Jun 27, 2019 at 10:51
  • $\begingroup$ There is an affine open containing $X-U$ $\endgroup$
    – reuns
    Jun 27, 2019 at 15:20
  • $\begingroup$ @José Carlos Santos I have edited it. Thank you for your reminding. $\endgroup$
    – user685167
    Jun 28, 2019 at 9:04
  • $\begingroup$ @reuns But $X-U$ is closed and I thought that if there exists an open affine subset which disjoints with $U$, then function $f$ on $U$ can't be extended to $X$. $\endgroup$
    – user685167
    Jun 28, 2019 at 9:08
  • $\begingroup$ There is $V$ open affine containing $X-U$ (a closed set small enough to be of codimension $\ge 2$) then $W=V \cap U$ is open and $f$ extends from $W$ to $V$ $\endgroup$
    – reuns
    Jun 28, 2019 at 13:11

1 Answer 1


If $U$ is an open subset such that the codimension of $X-U$ is greater or equal to 2. Then by defition of codimension, $\min\{\dim \mathcal{O}_{X,x}:x\in X-U\}\geq 2$, i.e. points of $X-U$ are all of codimension $\geq 2$. Thus the points of codimension $0$ and $1$ lie in $U$.

It's easy to see that a point is of codimension 0 if and only if it is a generic point of $X$. So $U$ contains all generic points, its closure must be $X$, so $U$ is open dense.

Now we can safely pick any non-empty open affine subset to consider the restriction of our desired isomorphism.

Pick an affine open covering $\{U_i\}_{i\in I} $ of $X$, assuming we have isormophisms $\Gamma(U_i,\mathcal{O}_X)\to \Gamma(U_i \cap U,\mathcal{O}_X)$, we now try to deduce the isomorphism $\Gamma(X,\mathcal{O}_X)\to \Gamma(U,\mathcal{O}_X)$ from the sheaf properties, the exact sequence of sheaves.

It's easy to see that the isomorphism follows from the isomorphisms $\Gamma(U_i\cap U_j,\mathcal{O}_X)\to \Gamma(U_i \cap U_j \cap U,\mathcal{O}_X)$ for all $i$ and $j$ with a simple diagram chase. But $U_i \cap U_j$ is not necessarily affine.

Nevertheless it can be shown that $U_i \cap U_j$ can be covered by simultaneously distinguished open subsets of $U_i$ and $U_j$, see this question in StackExchange.

Now we just apply the same argument to $U_i \cap U_j$ with this covering, this time the intersection of two simultaneously distinguished open subsets is open affine, then we are done.

  • $\begingroup$ You mean points of codimension 0 are generic points, right? Points of codimension 1 correspond to irreducible hypersurfaces. $\endgroup$ Sep 25, 2020 at 9:30
  • $\begingroup$ You mean that by the density of $U$, we can take an affine covering $U_i$ of $X$, and if all restriction maps $\Gamma(U_i,O_X)\rightarrow\Gamma(U_i\cap U,O_X)$ are isomorphisms, then we deduce $\Gamma(X,O_X)\rightarrow\Gamma(X\cap U,O_X)$ is an isomorphism? $\endgroup$
    – user685167
    Sep 25, 2020 at 9:32
  • $\begingroup$ @red_trumpet Yes, thank you, fixed. $\endgroup$
    – Z Wu
    Sep 25, 2020 at 14:06
  • $\begingroup$ @Ang Yes, it is based on the glueing property of sheaves. $\endgroup$
    – Z Wu
    Sep 25, 2020 at 14:07
  • $\begingroup$ @Ang My apology, I will add more information. It was not as easy as I thought. $\endgroup$
    – Z Wu
    Sep 25, 2020 at 14:26

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