# Removing codimension $\ge 2$ subscheme does't change the module of global regular function

Suppose $$(U, \mathcal{O}_U)$$ be an affine scheme isomorphic to the spectrum of an integral, integrally closed, Noetherian domain, $$A$$.

Let $$Y$$ be a closed subscheme of $$A$$ with codimension $$\ge 2$$.

Then, is it true $$\mathcal{O}_U(U) \cong \mathcal{O}_{U-Y}(U-Y)?$$

I guess it's geometrically intuitive because removing codimension $$\ge 2$$ from $$U$$ is just open dense subset of $$U$$ and so we can extend the regular function on $$U-Y$$ to the regular function on $$U$$.

But I don't know how to prove it with rigour.

• A list of near-duplicates, some abstract: 1, 2, 3, any question mentioning "algebraic Hartogs". Commented Apr 24, 2020 at 17:30

The result you mentioned is called algebraic Hartog's lemma. Its proof is long and involves many heavy tools from commutative algebra. Here I give a short outline in which I assume you are familiar with the notion of primary decomposition. For detailed proofs, you can find in books that I cite in below. It will take some efforts to finish all of them but believe me they will do you a favor in future.

The theorem is a consequence of the following theorem

Theorem $$1$$. Let $$A$$ be a Noetherian normal domain. Then we do have $$A = \bigcap_{\mathrm{ht}\mathfrak{p}=1} A_{\mathfrak{p}}$$ in which the intersection is taken in the field of fraction $$K(A)$$.

Its proof is based on series of fundamental theorems. From now on, let assume all the rings are commutative ring with unity.

Lemma $$2$$. (determinant trick) Let $$M$$ be a finitely-generated $$A$$-module and $$\alpha$$ an ideal of $$A$$. Let $$\phi$$ be an $$A$$-endormorphism of $$A$$ such that $$\phi(M) \subset \alpha M$$. Then $$\phi$$ satisfies an equation of form $$\phi^n + a_1 \phi^{n-1} + ... + a_{n-1} = 0, a_i \in \alpha$$ Proof: [M.Atiyah & I.G.McDonald, Introduction to Commutative Algebra, chapter $$2$$, p.$$21$$]

Theorem $$3$$. Let $$\alpha$$ be a decomposable ideal and $$\alpha = \bigcap_{i=1}^n \mathfrak{q_i}$$ be a minimal primary decomposition of $$\alpha$$. Then $$\sqrt{\mathfrak{q_i}}$$ are precisely prime ideals which occur in the set of ideal $$\sqrt{(\alpha:x)} (x \in A)$$.

Proof. [M.Atiyah & I.G.McDonald, Introduction to Commutative Algebra, chapter $$4$$, p.$$52$$]

About notation, $$(\alpha:x)=\left \{y \in A \mid xy \in \alpha \right \}$$.

Prime ideals $$\sqrt{q_i}$$'s are called the prime ideals belong to $$\alpha$$ or prime divisors of $$\alpha$$. In case the ring $$A$$ is Noetherian, we have a stronger result.

Proposition $$4$$. Let $$\alpha \neq (1)$$ be an ideal in a Noetherian ring $$A$$. The the prime ideals which belong a $$\alpha$$ are precisely the prime ideals which occur in the set of ideals $$(\alpha:x), (x \in A)$$.

Proof. [M.Atiyah & I.G.McDonald, Introduction to Commutative Algebra, chapter $$7$$, p.$$83$$]

The above results are implicitly used in theorems below.

Now let assume $$A$$ is integral and $$K(A)$$ its field of fraction. A $$A$$-submodule $$I$$ of $$K(A)$$ is said to be a fractional ideal if $$I \neq 0$$ and there exists $$a \in A$$ such that $$aI \subset A$$. Denote the set $$\left \{a \in A \mid aI \subset A \right \}$$ then we say $$I$$ is invertible if $$II^{-1}=R$$.

The following proposition gives us another way to look at invertible ideals. Most of proofs below can be found in [Matsumura, Commutative Ring Theory] but I think it is worth writing down everything here to help you figure out what is going on.

Proposition $$5$$. Let $$A$$ be an integral domain and $$I$$ a fractional ideal of $$A$$. Then the following conditions are equivalent:

1. $$I$$ is invertible.

2. $$I$$ is finitely generated and for every prime ideal $$\mathfrak{p}$$ of $$A$$, the fractional ideal $$I_{\mathfrak{p}} = IA_{\mathfrak{p}}$$ of $$A_{\mathfrak{p}}$$ is principal.

Proof. $$(1) \Rightarrow (2)$$ If $$II^{-1}=A$$ then there exists $$(a_i,b_i) \in I \times I^{-1}$$ such that $$\sum a_i b_i = 1$$. Then $$(a_i)$$'s generate $$I$$ since for any $$x \in I$$ we do have $$\sum (xb_i)a_i = x$$ and $$xb_i \in A$$. Moreover, at least one of $$a_ib_i$$ is invertible in $$A_{\mathfrak{p}}$$ and hence $$I_{\mathfrak{p}}=a_i A_{\mathfrak{p}}$$.

$$(2) \Rightarrow (1)$$ We always have $$(I^{-1})_{\mathfrak{p}} \subset (I_\mathfrak{p})^{-1}$$. If $$I$$ is finitely generated we shall prove that the equality holds. Let $$I = \sum Aa_i$$. Let $$x \in (I_{\mathfrak{p}})^{-1}$$ so $$xa_i \in A_{\mathfrak{p}}$$ and this implies $$xa_i c_i \in A$$ for some $$c_i \in A - \mathfrak{p}$$. Consequently, $$cx_ia_i \in A \ \forall i$$ for $$c = \prod c_i$$ and particularly we do have $$cx \in I^{-1}$$ or $$x \in (I^{-1})_{\mathfrak{p}}$$. By the hypothesis, $$I_{\mathfrak{p}}$$ is principal so $$I_{\mathfrak{p}}(I_{\mathfrak{p}})^{-1} = A_{\mathfrak{p}}$$ $$(\bullet)$$. Now if $$II^{-1} \neq A$$ then $$II^{-1}$$ is contained in a maximal ideal $$\mathfrak{m}$$, and then $$I_{\mathfrak{m}}(I_{\mathfrak{m}})^{-1}=I_{\mathfrak{m}}(I^{-1})_{\mathfrak{m}} \subset \mathfrak{m}A_{\mathfrak{m}}$$ which contradicts to $$(\bullet)$$.

Corollary $$6$$. Let $$A$$ be a Noetherian domain and $$\mathfrak{p}$$ a prime ideal. If $$\mathfrak{p}$$ is invertible then $$\mathfrak{ht}\mathfrak{p}=1$$ and $$A_{\mathfrak{p}}$$ is a discrete valuation ring (DVR). In particular, $$A_{\mathfrak{p}}$$ is normal because a DVR is a one-dimensional normal Noetherian local ring.

Proof. If $$\mathfrak{p}$$ is invertible then by $$2^{th}$$ condition in lemma $$5$$ we see that $$\mathfrak{p}A_{\mathfrak{p}}$$ is a principal ideal of $$A_{\mathfrak{p}}$$. Furthermore, $$A_{\mathfrak{p}}$$ is Noetherian local ring so $$A_{\mathfrak{p}}$$ is a DVR. Thus, $$\dim A_{\mathfrak{p}}=\mathrm{ht}\mathfrak{p}=1$$.

Corollary $$7$$. Let $$A$$ be a normal Noetherian local ring then every prime divisor of a principal ideal has height $$1$$.

Proof. Suppose $$a \in A, a \neq 0$$ and $$\mathfrak{p}$$ is a prime divisor of $$aA$$. By proposition $$4$$, there exists $$b \in A$$ such that $$(aR:b)=\mathfrak{p}$$. Denote $$\mathfrak{p}A_{\mathfrak{p}}=\mathfrak{m}$$, the unique maximal ideal of $$A_{\mathfrak{p}}$$ then $$(aR_{\mathfrak{p}}:b)=\mathfrak{m}$$ so by definition $$b/a \in \mathfrak{m}^{-1}$$ and $$b/a \notin A_{\mathfrak{p}}$$. If $$(b/a)\mathfrak{m} \subset \mathfrak{m}$$ then by using determinant trick we see that $$b/a$$ is integral over $$A_{\mathfrak{p}}$$ which is a contradiction with the normality of $$A_{\mathfrak{p}}$$. Consequently, $$(b/a)\mathfrak{m} = A_{\mathfrak{p}}$$ and $$\mathfrak{m}^{-1}\mathfrak{m}=A$$. By previous theorem, $$\mathrm{ht}A_{\mathfrak{p}}=\mathfrak{ht}\mathfrak{m}=1$$.

Now is the main theorem.

Proof of theorem $$1$$. We always have $$A \subset \cap_{\mathrm{ht}\mathfrak{p}=1}A_{\mathfrak{p}}$$. Let take $$b/a \in K(A)$$ such that $$a \neq 0$$ and $$b \in aA_{\mathfrak{p}}$$ for every prime ideal $$\mathfrak{p}$$ of height $$1$$. We are going to show that $$b/a \in A$$ by showing that $$J=(aA:b)=A$$. It is eay to see that $$JA_{\mathfrak{p}} = A_{\mathfrak{p}}$$ for each prime ideal $$\mathfrak{p}$$ of height $$1$$ so $$J$$ is not contained in any prime ideal of height $$1$$. $$(\bullet)$$

Let take a primary decomposition of $$aR$$ $$aR = \mathfrak{q}_1 \cap \mathfrak{q}_2 \cap ... \cap \mathfrak{q}_n$$ By corollary $$7$$, each $$\mathfrak{p}_i = \sqrt{\mathfrak{q}_i}$$ has height $$1$$ but by proposition $$4$$, the set of prime ideals belonging to $$J$$ is contained in the set of prime ideal belonging to $$aR$$ - but every such a ideal has height $$1$$ which contradicts to $$(\bullet)$$ so $$J = A$$ or equivalently $$b/a \in A$$.

The following lemma is much important because it transfers theorem $$1$$ to a much more geometric theorem, say, your original desired result.

Lemma $$8$$. Let $$X$$ be a topological space and $$Y \subset X$$ is a closed irreducible subset. Let $$U \subset X$$ be an open set such that $$U \cap Y \neq \varnothing$$. Then $$\mathrm{codim}(Y,X) = \mathrm{codim}(Y\cap U,U)$$ Proof. See StackProject.

And finally,

Theorem $$9$$. (Algebraic Hartog's lemma) Let $$X$$ be a locally Noetherian normal scheme, and let $$U \subset X$$ be an open subset with $$\mathrm{codim}(X - U) \geq 2$$. Then the restriction map $$\Gamma(X,\mathcal{O}_X) \to \Gamma(U,\mathcal{O}_X)$$ is an isomorphism. In other words: every function $$f \in \Gamma(U,\mathcal{O}_X)$$ extends uniquely to $$X$$.

Proof. [U.Gortz and T.Wedhorn, Algebraic Geometry, p.$$164$$]