Let \(X\) be a projective variety over an algebraically closed field \(k\) of characteristic zero. Show that \[ \operatorname{Sing}(X) \] is a proper Zariski-closed subset of \(X\). If \(X\) is normal, show that every irreducible component \(Z\) of \(\operatorname{Sing}(X)\) satisfies \[ \codim(Z,X)\geq2. \]
Solution
Choose an embedding \[ X\subseteq\PP^n_k \] and let \[ U_i=\{x_i\ne0\}\subseteq\PP^n_k, \qquad X_i=X\intersect U_i. \] The nonempty \(X_i\) are affine varieties and cover \(X\).
1. For every \(i\),\[ \operatorname{Sing}(X)\intersect X_i = \operatorname{Sing}(X_i). \]
Smoothness and regularity are local properties. The open immersion \[ X_i\hookrightarrow X \] identifies the local ring of \(X_i\) at any point \(p\in X_i\) with the local ring of \(X\) at \(p\). Hence \(p\) is smooth in \(X_i\) exactly when it is smooth in \(X\), giving the displayed identity.
2. For every nonempty affine chart \(X_i\), the subset\[ \operatorname{Sing}(X_i) \]is a proper Zariski-closed subset of \(X_i\).
For an affine variety over an algebraically closed field, the regular locus is a nonempty Zariski-open subset; equivalently the singular locus is proper and closed [@Har10a, Theorem I.5.3]; the proof uses the Jacobian criterion PR-MORJAC.
Applying that theorem to the affine variety \(X_i\) gives the assertion.
3. The singular locus\[ \operatorname{Sing}(X) \]is Zariski closed in \(X\).
By step 1 and step 2, for every chart \(X_i\) the intersection \[ \operatorname{Sing}(X)\intersect X_i \] is closed in \(X_i\). Therefore its complement has open intersection with every member of the open cover \(\{X_i\}\). Hence \[ X\sm\operatorname{Sing}(X) \] is open in \(X\), so \(\operatorname{Sing}(X)\) is closed.
4. The singular locus is a proper subset of \(X\).
Since \(X\) is nonempty, at least one chart \(X_i\) is nonempty. By step 2, that chart contains a smooth point. Step 1 says that this point is smooth in \(X\) as well. Hence \[ \operatorname{Sing}(X)\ne X. \] Together with step 3, this proves that the singular locus is proper Zariski closed.
5. Assume now that \(X\) is normal. Every codimension-one point of \(X\) issmooth.
Let \(\eta\) be the generic point of an irreducible closed subset of codimension one. Then \[ \dim\mco_{X,\eta}=1. \] Normality of \(X\) makes \(\mco_{X,\eta}\) a normal domain. Since \(X\) is a variety, this local ring is Noetherian.
A one-dimensional Noetherian normal local domain is a discrete valuation ring by D-QJ5M9. Hence \(\mco_{X,\eta}\) is regular. The field \(k\) is perfect because it is algebraically closed, so regularity is equivalent to smoothness for varieties over \(k\) by T-MORSMREG. Therefore \(\eta\) is a smooth point of \(X\).
6. If \(X\) is normal, every irreducible component \(Z\) of\(\operatorname{Sing}(X)\) has\[ \boxed{\codim(Z,X)\geq2.} \]
By step 4, the singular locus is a proper closed subset, so every irreducible component has positive codimension.
Suppose some component \(Z\) had codimension one, and let \(\eta_Z\) be its generic point. Since \(Z\subseteq\operatorname{Sing}(X)\) and the singular locus is closed, \[ \eta_Z\in\operatorname{Sing}(X). \] But step 5 says every codimension-one point is smooth, a contradiction. Thus codimension one cannot occur, and every component has codimension at least two.
7. Q.E.D.
Steps <1>1–<1>4 prove that the singular locus is proper Zariski closed. Steps <1>5–<1>6 prove the codimension bound under normality.