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The supremum of the lengths of strict chains of nonempty irreducible closed subsets. It is the maximum of the dimensions of the irreducible components. For an affine algebraic set it equals the Krull dimension of its coordinate ring; for a quasi-projective algebraic set it is the supremum of these dimensions over affine open subsets.

For closed irreducible subvarieties $Y,Z$ of a smooth $d$-dimensional algebraic variety $X$, every irreducible component of their intersection has dimension at least $\dim Y+\dim Z-d$. Locally the diagonal of $X\times X$ is cut out by $d$ parameters. Intersecting it with $Y\times Z$ and applying the Krull height theorem gives the bound. The three-dimensional affine quadric cone shows that the same lower bound can fail when $X$ is singular.

Use the usual convention that a variety is an irreducible variety. For a quasi-projective algebraic set $T$, its algebraic dimension is the supremum of the lengths $r$ of strict chainsof nonempty irreducible closed subsets of $T$. It is the maximum of the algebraic dimensions of its irreducible components. On an affine algebraic set, the correspondence between irreducible closed subsets and prime ideals reverses inclusion, so this is the Krull dimension of the coordinate ring. On a quasi-projective algebraic set, it is the supremum of the Krull dimensions of the coordinate rings of its affine open subsets. At a closed point $x$, the Krull dimension of the local ring measures chains through $x$.

Here is a closed-point dimension lemma for affine domains that avoids transcendence degree. Put $A=k[X]$. By Noether normalization, there is a integral extensionThe number of variables is $d$ because integral extensions preserve Krull dimension and $\dim B=d$. For any maximal ideal $\mathfrak{m}$ of $A$, its contraction $\mathfrak{n}$ to $B$ is a maximal ideal. Since $k$ is an algebraically closed field, $\mathfrak{n}=(z_1-c_1,\dots ,z_d-c_d)$ has height of a prime ideal $d$. The going-down theorem applies because $B$ is an integrally closed domain and $A$ is a integral domain. It lifts a length-$d$ chain below $\mathfrak{n}$ to one below $\mathfrak{m}$. HenceThe opposite inequality follows from $\dim A=d$.

For the nonempty open subset $U$, choose a nonempty principal open subset $D(g)\subset U$ and a closed point $x\in D(g)$. Every prime ideal below ${\mathfrak{m}}_x$ avoids $g$, so the preceding chain survives in the localization $A_g$. Consequently $\dim D(g)\ge d$. Conversely, any chain of irreducible closed subsets in $U$ gives a chain of the same length after taking closures in $X$: intersecting those closures with $U$ recovers the original subsets. Therefore

For the principal hypersurface dimension lemma, let $P$ be a minimal prime ideal over $(f)$. Since $f\ne 0$ in the integral domain $A$, $P\ne (0)$. The Krull principal ideal theorem gives $\mathrm{ht}P=1$. Choose a closed point $x$ on $V(P)$ lying on none of the other finitely many irreducible components of $V(f)$. Such a point exists because those other components cut out proper closed subsets of the irreducible variety $V(P)$, and closed points are dense. Set $R=A_{{\mathfrak{m}}_x}$. Then $\dim R=d$ andWrite $s=\dim R/fR$. Choose a system of parameters ${\bar{y}}_1,\dots ,{\bar{y}}_s$ in $R/fR$ and lift it to $R$. The ideal $(f,y_1,\dots ,y_s)$ has radical equal to the maximal ideal of $R$. The Krull height theorem yields $d\le s+1$. On the other hand, any chain of prime ideals containing $f$ can be extended strictly at the bottom by the zero prime ideal of the integral domain $R$, giving $s\le d-1$. Thus $s=d-1$. Since $R/PR$ is a localization of $A/P$, $\dim A/P\ge d-1$. Extending a chain in $A/P$ by $(0)⊊P$ also gives $\dim A/P\le d-1$. Hence every component has the required dimension:This argument uses Noether normalization, going-down theorem, and the Krull height theorem, never the dimension from the function field theorem. Irreducibility is essential: if the word variety were instead allowed to mean an arbitrary reducible affine algebraic set, neither assertion would hold without extra hypotheses. For example, a disjoint union of an affine plane and an affine line has an open component of smaller algebraic dimension; a function equal to $1$ on the plane and a coordinate on the line has a nonempty zero set of algebraic dimension $0$.

A blowup of an algebraic variety supplies a reduced special fibre with all three requested defects. Let $B={\mathbb{A}}^3$ have coordinates $(x,y,z)$, let $p=(0,1,0)$, and let $\pi :X={\mathrm{Bl}}_{p}B\to B$. Take $Y={\mathbb{A}}^1$ and the regular mapFor an explicit construction, $X$ is the closed subvariety of ${\mathbb{A}}^3\times {\mathbb{P}}^2$, with projective coordinates $[u:v:w]$, defined byThese equations say that $(x,y-1,z)$ is proportional to $(u,v,w)$. The three standard projective charts are affine spaces ${\mathbb{A}}^3$, so $X$ is a smooth variety which is an irreducible variety embedded in a quasi-projective algebraic set. Its algebraic exceptional divisor is $E={\pi }^{-1}(p)\cong {\mathbb{P}}^2$.

If $t\ne 0$, the surface $xy=t$ misses $p$, and the blowup of an algebraic variety is an isomorphism over it. HenceIt is a smooth variety, an affine variety, an irreducible variety, and has algebraic dimension $2$.

For $t=0$, the fibre of a morphism has three irreducible components: the strict transform of an algebraic subvariety $\widetilde{V(x)}$, the unchanged plane $V(y)$, and the algebraic exceptional divisor $E\cong {\mathbb{P}}^2$. The centre $p$ lies on $V(x)$ but not on $V(y)$; this is why all three components have multiplicity one. To verify reducedness and the crossing directly near $E$, use the chart $v\ne 0$, puttingHere $x=ab$, $y=1+b$, $z=bc$, and $f=ab(1+b)$. Near $b=0$, the factor $1+b$ is a unit, so the fibre of a morphism is $ab=0$, a reduced union of two crossing planes. The $u$-chart gives $f=x(1+xv)$, whose second factor is a unit along $E$; the $w$-chart gives $f=zu(1+zv)$ and the same reduced crossing. Away from $E$, reducedness follows from $xy=0$ in $B$. Thus the scheme-theoretic fibre of a morphism is itself reduced.

It has singular points of an algebraic variety already along $V(x,y)$, which misses $p$: locally its equation is $xy=0$, with Zariski tangent space of dimension of a vector space $3$ but local dimension of an algebraic variety $2$. Finally, $E$ is a closed subvariety of $X_0$. A closed subvariety of an affine variety is an affine variety, whereas ${\mathbb{P}}^2$ is not: all its global regular functions are constant, which cannot be the coordinate ring of a positive-dimensional affine variety. Consequently $X_0$ is singular, reducible, and nonaffine, as required. Choosing the centre off the intersection of the original planes avoids introducing a nonreduced special fibre of a morphism.

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 $\dim \varnothing =-\infty$.