Let \[ X\subseteq\AA^n_k \] be an affine variety of dimension \(d\) over an algebraically closed field \(k\) of characteristic zero. Show that the singular points of \(X\) form a proper Zariski-closed subset. Deduce that the smooth points form a dense Zariski-open subset.
Solution
Choose generators \[ I(X)=(f_1,\ldots,f_m) \] and put \[ c=n-d. \] Let \[ J = \left( \frac{\partial f_i}{\partial x_j} \right)_{i,j} \] be the Jacobian matrix.
1. The singular locus is the common zero set on \(X\) of all\(c\times c\) minors of \(J\).
If \(c=0\), the Jacobian criterion says that every point is smooth, so the singular locus is empty and the assertion is immediate.
Assume \(c>0\). By the Jacobian criterion PR-MORJAC, a point \[ p\in X \] is smooth exactly when \[ \rank J(p)=c. \] For points of the pure \(d\)-dimensional variety \(X\), one has \[ \rank J(p)\leq c. \] Indeed, \[ \dim T_pX = n-\rank J(p) \geq \dim_pX = d. \] Hence \(p\) is singular exactly when \[ \rank J(p)<c, \] which is equivalent to the vanishing at \(p\) of every \(c\times c\) minor of \(J\).
Therefore \[ \operatorname{Sing}(X) \] is the intersection of \(X\) with the zero loci of those minors, hence is Zariski closed.
2. Over the function field\[ K=k(X), \]the Jacobian matrix has rank exactly\[ c=n-d. \]
Let \[ A=k[X] = k[x_1,\ldots,x_n]/I(X). \] Since \(X\) is irreducible, \(A\) is a domain and \[ K=\Frac A. \]
The conormal sequence for \[ A=k[x_1,\ldots,x_n]/I(X) \] becomes, after tensoring with \(K\), \[ (I/I^2)\tensor_A K \longrightarrow K^n \longrightarrow \Omega_{K/k} \longrightarrow 0. \] The first map sends a defining equation \(f\) to \[ \left( \frac{\partial f}{\partial x_1}, \ldots, \frac{\partial f}{\partial x_n} \right), \] so its image is the row space of \(J\) over \(K\).
The extension \(K/k\) is finitely generated of transcendence degree \[ \trdeg_kK=d. \] Because \(\characteristic k=0\), it is separably generated, and therefore \[ \dim_K\Omega_{K/k}=d. \] Exactness gives \[ \rank_KJ = n-\dim_K\Omega_{K/k} = n-d = c. \]
3. At least one \(c\times c\) minor of \(J\) is a nonzero regular functionon \(X\).
By step 2, the rank of \(J\) over \(K\) is \(c\). Hence some \(c\times c\) minor \(\Delta\) has nonzero determinant in \(K\).
The entries of \(J\) lie in \(A\), so \(\Delta\in A\). Its image in the fraction field is nonzero, hence \[ \Delta\ne0 \] in the domain \(A\).
4. The smooth locus contains the nonempty distinguished open\[ D_X(\Delta). \]
Since \(\Delta\ne0\) in the domain \(A\), the distinguished open \[ D_X(\Delta) \] is nonempty.
For every \[ p\in D_X(\Delta), \] the chosen \(c\times c\) minor is nonzero at \(p\), so \[ \rank J(p)\geq c. \] Step 1 gives the opposite inequality \[ \rank J(p)\leq c. \] Thus \[ \rank J(p)=c, \] and the Jacobian criterion makes \(p\) smooth.
5. The singular locus is a proper Zariski-closed subset of \(X\).
Step 1 proves closedness. Step 4 produces a nonempty open subset consisting entirely of smooth points, so \[ \operatorname{Sing}(X)\ne X. \] Hence the singular locus is proper closed.
6. The smooth locus\[ X_{\reg} = X\sm\operatorname{Sing}(X) \]is dense Zariski open.
By step 5, the complement of the singular locus is a nonempty open subset of the irreducible variety \(X\). Every nonempty open subset of an irreducible space is dense. Therefore \[ \boxed{X_{\reg}\text{ is dense and Zariski open}.} \]
7. Q.E.D.
Steps <1>1–<1>5 prove that the singular locus is proper Zariski closed, and step 6 gives the dense-open smooth locus.