# Hartshorne Exercise I.5.13: subset of nonnormal points is proper and closed

I'm trying to solve the following exercise, Exercise I.5.13 from Hartshorne's Algebraic Geometry:

Here's my attempt:

I've been able to show that the set of nonnormal points is a subset of a proper closed subset as follows. By intersecting with an affine open, we may assume that the variety in consideration, say $$Y$$, is an affine variety. Let $$A(Y)$$ coordinate ring of $$Y$$, $$\overline{A(Y)}$$ its closure in $$K(Y)$$, the field of fractions of $$A(Y)$$. By Finiteness of Integral Closure, we obtain that $$\overline{A(Y)}$$ is generated by some finite number of $$f_i/g_i\in K(Y)$$ (for $$i\in \{1,\ldots, n\}$$), over $$A(Y)$$ as an $$A(Y)$$-module. Since localization preserves integral closure, $$\overline{A(Y)}_{g_1\ldots g_n}$$ is integrally closed. But it's easy to see that $$\overline{A(Y)}_{g_1\ldots g_n}\cong A(Y)_{g_1\ldots g_n}$$, as subrings of $$K(Y)$$. Thus, the distinguished open set $$D(g_1\ldots g_n)\cap Y$$ with coordinate ring isomorphic to $$A(Y)_{g_1\ldots g_n}$$ is normal, by Exercise I.3.17 (d), which states that an affine variety is normal if and only if its coordinate ring is integrally closed.

However, I seem to be unable to prove that the set of nonnormal points is closed too. I would be really grateful if someone could provide a hint on how to finish the proof.

Thank you.

• This is Bourbaki, Commutative Algebra, Ch. V, §1, no. 5, Corollary 5. Commented Oct 20, 2023 at 17:46

Let $$\mathfrak{m}\subset A(Y)$$ be the maximal ideal corresponding to a point $$y\in Y$$, and let $$\mathfrak{n}\subset\overline{A(Y)}$$ be a maximal ideal containing $$\mathfrak{m}\overline{A(Y)}$$ corresponding to some point $$y'\in \overline{Y}$$ mapping to $$y$$. We get a series of inclusions of rings $$\mathcal{O}_{Y,y} = A(Y)_{\mathfrak{m}} \subset \overline{A(Y)}_{\mathfrak{m}} \subset \overline{A(Y)}_{\mathfrak{n}} = \mathcal{O}_{\overline{Y},y'}.$$ Since integral closure commutes with localization, if $$A(Y)_\mathfrak{m}$$ is integrally closed, then $$\overline{A(Y)}_{\mathfrak{m}} = \overline{A(Y)_{\mathfrak{m}}}$$ is already integrally closed and $$\mathfrak{m}=\mathfrak{n}$$, so all of the inclusions are actually equalities. If the first inclusion is strict, then $$A(Y)_\mathfrak{m}$$ is not integrally closed and $$y$$ is a non-normal point, so $$y$$ is a normal point exactly when $$A(Y)_\mathfrak{m}=\overline{A(Y)}_\mathfrak{m}$$. Since $$\overline{A(Y)}=A[f_i/g_i]$$, this inclusion is an equality iff all the $$f_i/g_i$$ are in $$\mathcal{O}_{Y,y}$$.
Now our goal is to show that the set of $$y$$ so that any of the $$f_i/g_i$$ are not in $$\mathcal{O}_{Y,y}$$ is closed. Since there are finitely many $$f_i/g_i$$, it suffices to show that for a rational function $$h\in K$$ the set of $$y\in Y$$ so that $$h\notin \mathcal{O}_{Y,y}$$ is closed. Let $$I(h)$$ be the ideal of $$A$$ given by $$\{a\in A(Y) \mid ah\in A(Y)\}$$: this is known as the ideal of denominators of $$h$$. I claim that the vanishing locus of $$I(h)$$ is exactly the set of points where $$h\notin \mathcal{O}_{Y,y}$$.
To show this, suppose $$\mathfrak{m}\subset A(Y)$$ is a maximal ideal corresponding to $$y\in Y$$. Then the statement $$h\in \mathcal{O}_{Y,y}=A(Y)_\mathfrak{m}$$ is equivalent to $$h=\frac{b}{a}$$ for some $$a\in A(Y)\setminus \mathfrak{m}$$, so $$I(h)\subset\mathfrak{m}$$ iff $$h\notin \mathcal{O}_{Y,y}$$. This finishes the problem, since it tells us that the closed subset $$V(I(h))$$ is exactly the locus of points $$y\in Y$$ where $$h\notin \mathcal{O}_{Y,y}$$.