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For an integral normal variety that is Noetherian, removing a closed subset of codimension at least two does not create new regular functions. Affine-locally, a normal Noetherian domain is the intersection of its localizations at height-one primes inside its fraction field. The complement still contains every height-one point, so a regular function on it lies in each such localization and therefore in the original ring. On projective space this also follows directly by excluding irreducible polynomial factor of the denominators on each polynomial affine chart.

For a normal variety with affine coordinate ring $A$ and a Weil divisor $D$, the fractional ideal of functions whose principal Weil divisor is at least $D$ isFor an effective prime divisor this is its height-one prime ideal. If the divisor is Cartier locally, this ideal is locally principal. A nonprincipal height-one prime ideal in a normal local ring can therefore detect a non-Cartier divisor.

An algebraic variety whose local rings are integrally closed domains. In particular an irreducible variety that is a normal variety has normal affine coordinate rings, enabling codimension-two extension of regular functions on a normal variety.

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.

Write $a=t_1$, $b=t_2$, $c=t_3$, and $A=\mathbb{C}[a,b,c]/(ac-b^2)$. It is the quadratic cone invariant ringwhere the involution negates both $u$ and $v$. The invariant monomials have even total degree and are generated by $u^2,uv,v^2$; reduction by $ac=b^2$ proves the asserted presentation.

To verify the normal variety property, suppose $q\in \mathrm{Frac}(A)$ is integral over $A$. The same monic equation makes $q$ integral over $\mathbb{C}[u,v]$. This polynomial ring is an integrally closed domain, so $q\in \mathbb{C}[u,v]$. Since $q$ belongs to the fraction field of $A$, it is invariant under the involution, and hence belongs to $A$. Thus $X$ is normal. The gradient of its equation is $(c,-2b,a)$, so the Jacobian criterion shows that the origin $o$ is its only singular point of an algebraic variety. In particular $X$ is not smooth.

Take the prime Weil divisor $D=V_X(a,b)$ and $x=o$. We show that $D$ is not a Cartier divisor at $o$. In the local ringits prime ideal is $I=(a,b)R$. The quotient $I/\mathfrak{m}I$ has dimension of a vector space $2$ over $\mathbb{C}$: $a,b$ generate it, and they are independent since $\mathfrak{m}I\subset {\mathfrak{m}}^2$ while the relation $ac-b^2$ has no linear term. By Nakayama lemma, $I$ cannot be generated by one element.

On a normal variety, the ideal of an effective prime Weil divisor is the divisorial ideal of functions with order at least one along that divisor and order at least zero along every other prime Weil divisor. If $D$ were a Cartier divisor at $o$, a local defining rational function $h$ would identify this ideal with $hR$, because an integrally closed domain is the intersection of its height-one localizations at a prime ideal inside its fraction field. That would make $I$ principal, contradicting the calculation. Therefore $D$ has a nonzero class in the local divisor class group.

Now let $D^{\prime }=D+\mathrm{div}(q)$ be any linearly equivalent Weil divisor. If $o$ were outside its support of a Weil divisor, there would be an open subset containing $o$ on which $D^{\prime }$ is zero. On that neighbourhood $D=-\mathrm{div}(q)$ would be a principal Weil divisor, hence a Cartier divisor, which is impossible. Therefore this point is unavoidable in every representative:The argument does not assume that $D^{\prime }$ is effective. For comparison, $\mathrm{div}(a)=2D$ because on the chart $c\ne 0$, $a=b^2/c$; the obstruction has order two, as in the Divisor class group of an A-type surface singularity.

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 $\dim \varnothing =-\infty$.

The closed subset of points where an algebraic variety is not smooth. Over a perfect field it is equivalently the locus whose local rings are not regular local rings. A normal variety has no singular points in codimension zero or one.

If a Weil divisor $D$ on a normal variety is not Cartier at $x$, every linearly equivalent divisor has $x$ in its support of a Weil divisor. Otherwise that representative would vanish near $x$, expressing $D$ as a principal Weil divisor there. On $V(ac-b^2)$ the prime divisor $V(a,b)$ is not Cartier at the origin: its ideal has two independent generators modulo the maximal ideal times that ideal. This gives an unavoidable support point even for representatives with negative coefficients.