# Algebraic Hartog's lemma reduce to the affine case

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.

• Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures. Jun 27, 2019 at 10:51
• There is an affine open containing $X-U$ Jun 27, 2019 at 15:20
• @José Carlos Santos I have edited it. Thank you for your reminding.
– user685167
Jun 28, 2019 at 9:04
• @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 $X$.
– user685167
Jun 28, 2019 at 9:08
• 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 $V$ Jun 28, 2019 at 13:11

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.

• You mean points of codimension 0 are generic points, right? Points of codimension 1 correspond to irreducible hypersurfaces. Sep 25, 2020 at 9:30
• 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?
– user685167
Sep 25, 2020 at 9:32
• @red_trumpet Yes, thank you, fixed.
– Z Wu
Sep 25, 2020 at 14:06
• @Ang Yes, it is based on the glueing property of sheaves.
– Z Wu
Sep 25, 2020 at 14:07
• @Ang My apology, I will add more information. It was not as easy as I thought.
– Z Wu
Sep 25, 2020 at 14:26