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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 $\dim \varnothing =-\infty$.

A standard sufficient hypothesis is that $X$ is both a Noetherian scheme and an integral scheme, and is regular in codimension one; in particular, a Noetherian normal scheme qualifies. A Weil divisor is then a finite sumover integral codimension-one closed subschemes $Z$. The local ring at the generic point of each $Z$ is a discrete valuation ring, so every nonzero rational function $g\in k(X{)}^{\times }$ has a principal divisorThe divisor class group is

A Weil divisor on $X$ is a finite formal sumover integral codimension-one closed subschemes $Z$. Since $X$ is regular in codimension one, the local ring at the generic point of each $Z$ is a discrete valuation ring. A nonzero rational function $f\in k(X{)}^{\times }$ therefore defines the principal divisor $\mathrm{div}(f)={\sum }_Z{v}_Z(f)[Z]$. The divisor class group is

False. For $n\ge 2$, take the affine hypersurfaceIt is an integral scheme, and its only possible singular point is the origin, which has codimension two. Hence it is regular in codimension one; as a hypersurface it satisfies Serre's condition $(S_2)$, so the Serre criterion for normality also makes it a normal scheme. The Divisor class group of an A-type surface singularity isgenerated by $V(x,z)$. Thus a closed affine subscheme satisfying $(\star )$ can have nonzero torsion in its class group.