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If $d=1$, $C$ is a line and has no singular points. Suppose $d>1$ and take $P$ to be an irreducible polynomial. First, $P_x$ is not the zero polynomial: otherwise $P\in \mathbb{C}[y,z]$, and a homogeneous polynomial in two variables of degree greater than one factors into linear factors over $\mathbb{C}$, contrary to irreducibility.

Let $D=V(P_x)$ be the polar curve in this coordinate direction. Since $\mathrm{deg}{P}_x=d-1<d$, the irreducible polynomial $P$ cannot divide $P_x$. Thus $C,D$ have no common irreducible component, and Bézout's theorem applies. One may count repeated factors of $P_x$ with their multiplicities; alternatively, passing to its reduced equation only decreases $\mathrm{deg}D$.

Every singular point $p$ of $C$ satisfies $P_x(p)=0$ and has $I_p(C,D)\ge 2$ by part (a). Consequently, if $s$ is the number of distinct singular points,This argument also proves that the singular locus is finite.

There is a nonsingular point of an algebraic curve on $C$, since part (b) makes its singular locus finite whereas a projective curve has infinitely many complex points. Apply a projective transformation taking one such point to $p=[1:0:0]$, and use $P$ for the transformed homogeneous polynomial. As in part (b), $P_x$ is nonzero and $D=V(P_x)$ has no common irreducible component with $C$.

The Euler theorem for homogeneous functions and its differentiated identities giveandSince $p$ is nonsingular on $C$, at least one of $P_y(p),P_z(p)$ is nonzero. Because $d>1$, the polar curve $D$ is also nonsingular at $p$, and these identities show that $T_{p}D=T_{p}C$. Hence $I_p(C,D)\ge 2$.

This additional intersection is separate from all the singular points of $C$. Thus Bézout's theorem givesThe improvement comes from making the polar curve tangent at a chosen nonsingular point, thereby reserving at least two of the available intersections.

Choose one projective line from each of two different pairs. They are skew lines by the preceding disjointness result. After a projective linear coordinate change call themFor $A\in L$, $B\in M$, their joining line meets the cubic surface at $A,B$ and a third point. This defines a rational map $L\times M⇢S$ on the Zariski-open set where that third intersection is distinct and the joining line is not contained in $S$.

More explicitly, because $F$ vanishes on both projective lines,The third point is $[VA-UB]$. Both polynomials $U,V$ are nonzero: if either vanished identically, one of the two projective lines would lie in the singular locus. Thus $UV\ne 0$ supplies a nonempty Zariski-open set.

Conversely, a point $P=[x_0:x_1:x_2:x_3]$ outside $L\cup M$ lies on the unique joining line whose endpoints have coordinates $[x_0:x_1]$ on $L$ and $[x_2:x_3]$ on $M$. This is a rational inverse on the same generic locus. The rational parametrization of a cubic surface from two skew lines givesThe latter contains the affine plane as a dense open subset, as does ${\mathbb{P}}^2$. Hence $S$ is birational to ${\mathbb{P}}^2$.

We first prove the local algebra fact behind regularity in codimension one. Let $(A,\mathfrak{m})$ be a one-dimensional normal Noetherian local domain. Choose $0\ne a\in \mathfrak{m}$. Its radical is $\mathfrak{m}$, so some ${\mathfrak{m}}^n$ lies in $(a)$; take minimal $n$. If $n=1$, then $\mathfrak{m}=(a)$. Otherwise choose $b\in {\mathfrak{m}}^{n-1}\setminus (a)$ and put $z=b/a$ in the fraction field. We have $z\notin A$ but $z\mathfrak{m}\subseteq A$.

If $z\mathfrak{m}\subseteq \mathfrak{m}$, express multiplication by $z$ on a finite generating set of $\mathfrak{m}$. The determinant trick gives a monic polynomial in $z$ annihilating $\mathfrak{m}$; as this ideal is nonzero in a domain, the polynomial itself vanishes. Normality would then give $z\in A$, a contradiction. Hence there is $u\in \mathfrak{m}$ with $zu$ a unit of $A$. Every $v\in \mathfrak{m}$ satisfies $v=u(zv)/(zu)$, proving $\mathfrak{m}=(u)$. The maximal ideal therefore has one generator in dimension one, so $A$ is a regular local ring, indeed a discrete valuation ring. This proves one-dimensional normal local rings are discrete valuation rings directly.

Now a codimension-one point of the normal variety has such a local ring, and a codimension-zero local ring is a field. Both are regular. Since the algebraically closed ground field is perfect, regularity and smoothness agree for this finite-type variety, and the Jacobian criterion makes the singular locus closed. It contains no codimension-zero or codimension-one point. Therefore every irreducible component of the singular locus has codimension at least two, givingFor $d\le 1$ this means the singular locus is empty. In particular the singular set itself is the required closed subset, not merely a set contained in some larger one.

Work on affine open subsets. Their coordinate rings are Noetherian rings and integrally closed domains. At a prime ideal of height of a prime ideal one, the local ring is a one-dimensional Noetherian local ring which is an integrally closed domain, hence a discrete valuation ring and therefore a regular local ring. At the generic point the local ring is a field, hence a regular local ring. For a brief justification of the one-dimensional local ring fact, take $0\ne a\in \mathfrak{m}$ in a one-dimensional Noetherian local ring $R$ which is an integrally closed domain. Some ${\mathfrak{m}}^n\subset (a)$; choose the least such $n$, and choose $b\in {\mathfrak{m}}^{n-1}\setminus (a)$. Then $q=b/a\notin R$, but $q\mathfrak{m}\subset R$. If $q\mathfrak{m}\subset \mathfrak{m}$, the determinant trick applied to the finitely generated faithful module $\mathfrak{m}$ would make $q$ integral over $R$, a contradiction. Hence the ideal $q\mathfrak{m}$ contains a unit and equals $R$. Thus $\mathfrak{m}=q^{-1}R$ is principal. A one-dimensional Noetherian local ring with principal maximal ideal is a discrete valuation ring. Therefore a normal variety is regular in codimension one.

The ground field is an algebraically closed field, hence a perfect field, so having a regular local ring is equivalent to being a smooth point of a variety here. Moreover, the smooth locus of a variety is open, making the singular locus closed. If an irreducible component of the singular locus had algebraic codimension zero or one, its generic point would have a regular local ring by the preceding argument, a contradiction. Consequently every such component has algebraic codimension at least two.

One can read the numerical bound directly from chains of prime ideals, without identifying algebraic dimension with transcendence degree. For a prime ideal $P$ of height of a prime ideal at least two in an affine chart, append a chain $(0)⊊P_1⊊P$ below any chain in the quotient by $P$. Its length increases by two. Thus $\dim A/P\le \dim A-2$. Since a nonempty affine open subset of the irreducible variety $X$ has algebraic dimension $d$,For $d=0$ or $1$, this means the singular locus is empty; take $\dim \varnothing =-\infty$.

For a compact Riemann surface of genus $g\ge 1$, the theta divisor in ${\mathrm{Pic}}^{g-1}(X)$ has the displayed multiplicity. A local determinant description of the theta divisor reduces to an $r$ by $r$ matrix vanishing at $L$, where $r=h^0(L)$. Its linear term is cup product with a tangent class in $H^1({\mathcal{O}}_X)$. The invertible cup-product direction for a line bundle on a curve shows that its determinant has order exactly $r$. Consequently the singular locus consists precisely of classes with at least two independent global sections.

A cubic surface with empty singular locus. Over an algebraically closed field, it is irreducible: distinct positive-degree hypersurface components would meet, and every point of their intersection would be singular in their product equation.

Over a perfect field, a point of an algebraic variety is smooth exactly when its local ring is a regular local ring. The smooth locus of a variety is open; its closed complement is the singular locus.