It is a fact that any regular local ring is an integrally closed domain. Thus every variety has a nonempty open subset of normal points.
In this exercise, show directly, without using the regularity fact, that the set of nonnormal points of a variety is a proper closed subset. You will need the finiteness of integral closure.
Solution
Let \(X\) be a variety with function field \(K=k(X)\). Normality is local, so we first work on an affine open \[ U=\Spec A\subseteq X, \] where \(A\) is a finitely generated \(k\)-domain with fraction field \(K\). Let \(\overline A\) denote the integral closure of \(A\) in \(K\).
1. The \(A\)-module\[ M=\overline A/A \]is finite.
The finiteness theorem for normalization says that the integral closure of a finitely generated domain over a field in its finite fraction-field extension is finite as a module over the original ring [@Har10a, Theorem I.3.9A]. Here the extension of fraction fields is the identity \(K/K\), so \(\overline A\) is a finite \(A\)-module. Its quotient \(M\) is therefore finite as well.
2. For a prime \(\mathfrak p\in\Spec A\), the local ring \(A_{\mathfrak p}\) is normal if and only if\[ M_{\mathfrak p}=0. \]
Integral closure commutes with localization: \[ \overline{A_{\mathfrak p}} = (\overline A)_{\mathfrak p} \subseteq K. \] Indeed, localization of an integral extension is integral. Conversely, if \(u\in K\) is integral over \(A_{\mathfrak p}\), choose one denominator \(s\notin\mathfrak p\) containing all denominators in a monic equation for \(u\). For a sufficiently large \(N\), multiplying that equation after setting \(v=s^Nu\) gives a monic equation for \(v\) with coefficients in \(A\). Thus \(v\in\overline A\) and \[ u=v/s^N\in(\overline A)_{\mathfrak p}. \] Thus \((\overline A)_{\mathfrak p}\) is exactly the integral closure of \(A_{\mathfrak p}\) in its fraction field.
Consequently \[ A_{\mathfrak p}\text{ is integrally closed} \quad\Longleftrightarrow\quad A_{\mathfrak p}=(\overline A)_{\mathfrak p}. \] The localized exact sequence \[ 0\longrightarrow A_{\mathfrak p} \longrightarrow(\overline A)_{\mathfrak p} \longrightarrow M_{\mathfrak p} \longrightarrow0 \] shows that this equality is equivalent to \(M_{\mathfrak p}=0\). Since \(A_{\mathfrak p}\) is already a domain, integrally closed is exactly normality here.
3. The nonnormal locus in \(U\) is the closed subset\[ \boxed{\operatorname{NNor}(U) = \operatorname{Supp}_A(M) = V(\Ann_A M).} \]
By step 2, a prime \(\mathfrak p\) is nonnormal exactly when the localization \(M_{\mathfrak p}\) is nonzero. This is the definition of the support of the finite module \(M\). For a finite module over a ring, \[ \operatorname{Supp}_A(M)=V(\Ann_A M). \] Indeed, if generators \(m_1,\ldots,m_s\) all vanish after localization at \(\mathfrak p\), choose denominators outside \(\mathfrak p\) killing them and multiply those denominators to obtain an element of \(\Ann_A M\) outside \(\mathfrak p\). The converse is immediate. Thus the nonnormal locus on \(U\) is closed.
4. This closed subset is proper.
Let \(\eta\) be the generic point of \(U\), corresponding to the zero prime. Its local ring is the fraction field \[ A_{(0)}=K, \] which is integrally closed in itself. By step 2, \[ M_{(0)}=0. \] Thus \(\eta\notin\operatorname{Supp} M\). Since the support is closed and omits the generic point of the irreducible space \(U\), it is a proper closed subset. Equivalently, \(\Ann_A M\) contains a nonzero element, and the principal open defined by that element consists entirely of normal points.
5. The nonnormal locus of \(X\) is a proper closed subset of \(X\).
Choose a finite affine open cover \[ X=U_1\cup\cdots\cup U_N. \] Normality of a local ring is unchanged when computed in an open neighborhood, so the global nonnormal locus \(N\) satisfies \[ N\cap U_i=\operatorname{NNor}(U_i). \] Step 3 makes each intersection closed in \(U_i\). Closedness is local on an open cover, hence \(N\) is closed in \(X\).
The generic point of the irreducible variety \(X\) has local ring \(K\), a field, and is therefore normal. Thus it does not lie in \(N\). Hence \(N\) is proper. Its complement is the nonempty open normal locus.
6. Q.E.D.
Steps <1>1–<1>4 give the finite-module description and proper closedness on every affine chart, and step 5 glues those descriptions on the variety.