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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.

An algebraic variety whose irreducible components all have the same dimension of an algebraic set. The components of a generic fibre in the fiber dimension theorem have this property even when that fibre is reducible.

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.

Let $Y$ be the reduced union of the irreducible components of an affine variety $X$ containing $P$. Removing all other components leaves an open neighbourhood of $P$ on which $X$ and $Y$ agree, so their local rings at $P$ agree. In the coordinate ring $B=k[Y]$, every denominator not vanishing at $P$ avoids all minimal primes: each component of $Y$ contains $P$. Such a denominator is a non-zero-divisor. Consequently $B_{{\mathfrak{m}}_P}$ injects into the total quotient ring $Q(B)=\mathrm{Rat}(Y)$. One cannot generally replace $Y$ by a single component through an intersection point.

A nonsingular complex projective plane curve of degree $d\ge 2$, with no common irreducible component with its Hessian curve, has at most $3d(d-2)$ distinct flexes. The Hessian criterion for a flex identifies the flexes with the intersection, and Bézout's theorem counts its total intersection multiplicity. Higher contact can make this bound on distinct points strict.

For a general nonsingular curve of degree $d\ge 3$, all flexes have tangent contact exactly three, making the bound an equality. For a fixed point and line through it, contact at least four imposes four independent linear conditions on the coefficients; the space of such point-line pairs has dimension three. Thus degree-$d$ equations with contact at least four form a proper closed exceptional locus. In local graph coordinates $y=h(x)$, the Hessian determinant along the curve is a nonvanishing factor times $h^{\prime \prime }(x)$, so contact exactly three makes the intersection simple. A nonsingular plane conic has no flexes; a line is excluded because its tangent is a component and its Hessian determinant is identically zero.

Write $C=V(P)$, where $P\in \mathbb{C}[x,y,z]$ is a squarefree polynomial and a homogeneous polynomial of degree $d$. A projective point $p\in C$ is a singular point precisely whenIf $p=[p_0:p_1:p_2]$ is a nonsingular point of an algebraic curve, its projective tangent line isThe coefficients are not all zero, and the Euler theorem for homogeneous functions gives $p_0{P}_x(p)+p_1{P}_y(p)+p_2{P}_z(p)=dP(p)=0$, so the line passes through $p$. Changing the representative of the projective point only rescales the equation.

The strong form of Bézout's theorem says that two projective plane curves $C,D$ of degrees $d,e$, with no common irreducible component, have finitely many intersections andEach intersection multiplicity is a positive integer. At $p\in C\cap D$,This is the criterion for a transverse intersection of plane curves. In particular, an intersection at a singular point of either curve contributes at least two.

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.

Let $d_i=\mathrm{deg}{C}_i$. Under the reduced plane curve convention, the distinct irreducible components have ${\sum }_i{d}_i=d$. Away from intersections with other components, a point of $C_i$ is nonsingular on the union exactly when it is nonsingular on $C_i$. For example, a local equation of the union is the equation of $C_i$ multiplied by a nonvanishing factor. At an intersection of two components, differentiating their product makes every first partial derivative vanish. Therefore

Part (b) bounds the number of singular points of $C_i$ by $d_i(d_i-1)/2$. For each distinct pair of irreducible components, Bézout's theorem bounds the number of distinct intersections by $d_i{d}_j$. The cardinality of a union is at most the sum of the cardinalities, even if some points have been counted several times. Hence

The reduced convention matters. If one instead allows the equation $L^2=0$ as a nonreduced curve and uses its equation to test singularity, every point on its supporting line has vanishing first partial derivatives. A finite bound would then be false.

Choose homogeneous polynomials $F,G$ defining the two projective plane curves. Their absence of a common irreducible component makes them coprime. For a point $O=[1:0:0]$ outside both curves, they have degrees $n,m$ in $X$ and nonzero constant leading coefficients. The resultantis therefore nonzero. Indeed, a vanishing resultant over $k(Y,Z)$ would give a common factor there, and Gauss's lemma would give a common component of the original curves.

The homogeneity of $F,G$ makes $R$ a homogeneous polynomial of degree $nm$: scaling $(Y,Z)$ by $a$ scales their $X$-roots by $a$, and the product of the $nm$ differences of roots scales by $a^{nm}$. Every point in the intersection projects from $O$ to a zero of $R$ in the projective line. Each such projection line contains only finitely many points of either curve, since it passes through $O$ outside them. Thus the intersection is finite already.

We may now choose $O$ also outside the finitely many joining lines of distinct intersection points. Projection is then injective on the intersection. A nonzero degree-$nm$ homogeneous polynomial on the projective line has at most $nm$ distinct zeros, so the resultant bound for intersections of plane curves givesThis derives the required cardinality bound without assuming the full intersection-multiplicity statement of Bézout's theorem.

The dimension of an algebraic set is the supremum of lengths of strict chains of nonempty irreducible closed subsets. It is the maximum of the dimensions of the irreducible components. On an affine variety $X=\mathrm{Spec}A$, it equals the Krull dimension of $A$; if $X$ is irreducible, Noether normalization identifies this with the transcendence degree of its function field over $k$. Nonempty Zariski-open subsets of an irreducible algebraic variety have the same dimension, so the definition is compatible with affine charts and with birational maps. In particular, $\dim {\mathbb{A}}^n=\dim {\mathbb{P}}^n=n$ and $\dim (X\times Y)=\dim X+\dim Y$ for irreducible algebraic varieties.

The fiber dimension theorem says that a dominant morphism of algebraic varieties $f:X\to Y$ between irreducible algebraic varieties has generic fibre dimensionEvery irreducible component of every nonempty fibre over a closed point has dimension at least $r$. There is a nonempty Zariski-open subset of $Y$ on which the fibres are nonempty and are equidimensional algebraic varieties of dimension $r$. Dominance is required; for a general morphism of algebraic varieties, replace $Y$ by the closure of its image. This gives the dimension of image of a morphism formula.

For the generic fibre, the formula is the additivity of transcendence degree in the tower $k\subset k(Y)\subset k(X)$. The lower bound on closed fibres comes from cutting by at most $\dim Y$ local parameters after a finite Noether normalization of an affine target chart: the principal hypersurface dimension lemma drops dimension by at most one per equation, and the finite normalization fibre separates the finitely many target points. For the generic upper bound, take finitely many affine source charts and apply Noether normalization to each coordinate ring over $k(Y)$. Clearing denominators in $k[Y]$ makes each chart finite over ${\mathbb{A}}^r$ relative to a suitable open target chart. Its fibres have dimension at most $r$. Intersecting these finitely many target opens and using the lower bound gives generic equidimensionality of fibres.

The upper semicontinuity of local fibre dimension is a statement on the source: for a finite-type morphism of schemes, the points $x$ satisfying ${\dim }_x{X}_{f(x)}\le d$ form an open set. For a proper morphism, the maximum fibre dimension is upper semicontinuous on the target as well. The properness qualification matters; one should not assert the latter for every finite-type morphism of schemes.

A first application is dimension counting for plane sections and intersections. On an irreducible affine variety, imposing $q$ polynomial equations gives each nonempty component algebraic codimension at most $q$, by repeated principal ideal theorem. On a smooth variety this underlies the intersection dimension bound on a smooth variety. A second application is to images: if a morphism of algebraic varieties has finite nonempty fibres, its image closure has the same dimension as the source. If it is also proper, that image is closed. Thus an isolated fibre point forces dominance in equal dimensions when the irreducible source and target have equal dimension.

For the cubic-surface application, projective lines in ${\mathbb{P}}^3$ form the Grassmannian $G=\mathrm{Gr}(2,4)$ of dimension four. Cubic forms form ${\mathbb{P}}^{19}$, since there are $\left(\genfrac{}{}{0ex}{}{6}{3}\right)=20$ degree-three monomials in four variables. The cubic surface line incidence varietyhas fibre ${\mathbb{P}}^{15}$ over every $L$: restricting a cubic to a projective line gives four independently prescribable coefficients. It is therefore an irreducible projective-space bundle of dimension $4+15=19$. The other projection $\pi :I\to {\mathbb{P}}^{19}$ is a proper morphism, so its image $Z$ is closed and irreducible.

To show $Z={\mathbb{P}}^{19}$, dimension counting alone is not enough; exhibit a fibre with an isolated point. TakeA nearby projective line is the graph $x_2=a{x}_0+b{x}_1$, $x_3=c{x}_0+d{x}_1$. Restriction givesIt vanishes only when $a=b=c=d=0$. Thus $L_0$ is an isolated reduced point of the fibre. Applying the lower bound in the fiber dimension theorem to $I\to Z$ gives $0\ge 19-\dim Z$. Since $Z\subseteq {\mathbb{P}}^{19}$, its dimension is exactly 19, and closedness forces $Z={\mathbb{P}}^{19}$. This is the incidence proof that a cubic surface contains a line. The auxiliary cubic $F_0$ need not be smooth: the argument proves existence for every cubic form, and hence in particular for every smooth cubic surface.

Let $X$ be an affine variety over an algebraically closed field $k$, with reduced coordinate ring $A=k[X]$. On an open subset $U$, define ${\mathcal{O}}_X(U)$ to be the functions $h:U\to k$ which, near every point, have an expression $a/b$ with $a,b\in A$ and $b$ nowhere zero on that neighbourhood. Restriction is ordinary restriction of functions. Compatible local functions glue uniquely and remain locally quotients, so this defines the structure sheaf. On the principal open subset $D(f)$ it gives $A_f$, and at $P$ it gives the local ring $A_{{\mathfrak{m}}_P}$. The principal-open identification can also be seen by identifying $D(f)$ with the affine algebraic set obtained by adjoining a coordinate $t$ and the equation $tf=1$; its coordinate ring is $A[t]/(tf-1)\cong A_f$. This construction permits several irreducible components; no division by a function vanishing identically on a component is assumed.

An abstract algebraic variety is a locally ringed space locally isomorphic to such affine models, with a finite affine cover; we use the usual separated convention. Reducible varieties are allowed here. The separation condition says that the diagonal morphism is a closed immersion. Two varieties are related by a birational map with an inverse when some dense open subset of one is isomorphic as a locally ringed space to a dense open subset of the other.

Define the ring of rational functions on a reduced variety by pairs $(U,h)$, where $U\subseteq X$ is dense open and $h\in {\mathcal{O}}_X(U)$, identifying two pairs when their restrictions agree on a common dense open subset. Finite intersections of dense open subsets are dense, so addition and multiplication on these intersections are well defined. For an irreducible variety this gives the usual function field; for a reducible variety it need not be a field. Restriction to any dense open subset $V$ induces an isomorphismIndeed, intersecting domains with $V$ defines the restriction, and a dense open subset of $V$ is also dense open in $X$, which defines its inverse. An isomorphism between dense opens therefore induces an isomorphism of their rational rings. Thus birational varieties have isomorphic rings of rational functions, also in the reducible case.

For the finite-cover assertion, write the closed complement of $U$ as $V(I)$, where $I=(f_1,\dots ,f_r)$ is an ideal of the Noetherian ring $A$. ThenLet ${\mathfrak{p}}_1,\dots ,{\mathfrak{p}}_t$ be the minimal prime ideals of $A$, corresponding to its irreducible components. Since $U$ is dense, it meets every component, so $I$ is contained in none of these primes. Prime avoidance supplies $s\in I$ outside their union. Consequently $D(s)\subseteq U$ meets a dense open subset of every component and is dense in $X$. Add $D(s)$ to the finite cover if necessary. This proves that a finite principal-open cover can be chosen with a dense member. It does not say that every principal-open cover already has such a member: for $X=V(xy)$, the dense open $X\setminus \{(0,0)\}=D(x)\cup D(y)$ has neither displayed member dense, although adding $D(x+y)$ supplies a dense member.

For clarity, the zero divisors in this reduced Noetherian ring are exactly ${\bigcup }_i{\mathfrak{p}}_i$. If $ab=0$ and $a$ avoids every ${\mathfrak{p}}_i$, reduction in the domain $A/{\mathfrak{p}}_i$ forces $b\in {\mathfrak{p}}_i$ for all $i$, hence $b=0$. Conversely, if $a\in {\mathfrak{p}}_i$, choose for every $j\ne i$ an element $b_j\in {\mathfrak{p}}_j\setminus {\mathfrak{p}}_i$, possible because distinct minimal primes are incomparable. Their product $b$ is not in ${\mathfrak{p}}_i$ but belongs to all the other primes. Then $b\ne 0$ and $ab\in {\bigcap }_j{\mathfrak{p}}_j=0$. For a single minimal prime, reducedness makes that prime zero and this argument uses $b=1$. It follows that

The total quotient ring $S^{-1}A$, where $S$ consists of these non-zero-divisors, maps to $\mathrm{Rat}(X)$ byThe equivalence relation for localization makes this a well-defined $k$-algebra homomorphism. For surjectivity, a rational class represented on $U$ restricts to the dense principal open $D(s)\subseteq U$ constructed above. Its section there belongs to $A_s$, so it is $a/s^r$ and lies in the image. For injectivity, if such a fraction is zero on a dense open subset, its numerator vanishes on that dense subset. The numerator's zero set is closed, so it vanishes everywhere on $X$ and is zero in the reduced coordinate ring. ThusIn particular $A\to S^{-1}A$ is canonically injective: $a/1=0$ implies $sa=0$ for some non-zero-divisor $s$, hence $a=0$.

Finally let $Y$ be the reduced closed union of exactly those irreducible components of $X$ which contain $P$. Removing all the other components gives an open neighbourhood of $P$ on which $X$ and $Y$ coincide, including their structure sheaves. Hence ${\mathcal{O}}_{X,P}\cong {\mathcal{O}}_{Y,P}$. Write $B=k[Y]$ and let ${\mathfrak{n}}_P$ be its maximal ideal at $P$. Every minimal prime of $B$ is contained in ${\mathfrak{n}}_P$, because every component of $Y$ contains $P$. If $b\notin {\mathfrak{n}}_P$, it avoids each such prime and is a non-zero-divisor. Therefore the map of localizationsis defined and injective: if its value is zero, a non-zero-divisor annihilates $a$, so $a=0$. We obtain the desired local ring embeds in rational functions on incident components:At an intersection point one must retain all incident components. For example, the local ring of $V(xy)$ at the origin has two nonzero germs $x,y$ with $xy=0$; it cannot embed into the function field of just one of its axes.

Suppose the map were nonconstant. Since projective space is proper, its image $Z$ is a closed projective subvariety of ${\mathbb{P}}^n$, of positive dimension $r\le n$. Fix $y\in Z$. Regard the map as surjective onto $Z$; the fiber dimension theorem just proved givesChoose a target hyperplane $H$ not containing $y$. It meets $Z$: the affine cone over $Z$ has dimension $r+1\ge 2$, and the hyperplane equation vanishes at its vertex but not identically. By the principal hypersurface dimension lemma its zero set on the cone has dimension $r\ge 1$, so it contains a nonzero cone point and hence a point of $Z\cap H$.

The inverse image $D={\varphi }^{-1}(H)$ is therefore nonempty and proper. Locally the hyperplane equation is a single regular equation, after trivializing the pulled-back line bundle. On the irreducible smooth source it is not identically zero, so the same hypersurface dimension lemma shows that each component of $D$ has dimension $m-1$. Choose one component $D_0$. Its cone is a codimension-one irreducible closed subset of affine $(m+1)$-space, whose prime ideal is principal because the polynomial ring is a unique factorization domain. Thus $D_0$ is a projective hypersurface defined by a homogeneous polynomial $P$ of positive degree.

Choose a positive-dimensional irreducible component $F$ of ${\varphi }^{-1}(y)$. Since $y\notin H$, $F$ is disjoint from $D_0$. On the cone over $F$, the equation $P$ is consequently nonzero everywhere except at the vertex. This contradicts the hypersurface dimension lemma: the cone has dimension at least two, so its nonempty proper zero set for $P$ has dimension at least one and cannot consist just of the vertex. Therefore $\boxed{\varphi \text{ is constant}}$. The argument proves morphisms from projective space to smaller projective space are constant by combining the fibre bound with the forced intersection of a positive-dimensional projective variety and a hypersurface.

Use the usual irreducible algebraic variety convention here, and write $N=\dim X$, $r=\dim Y$. Fix a closed point $y$ and an affine neighborhood $Y_0$ of it. Noether normalization gives a finite surjective map $\pi :Y_0\to {\mathbb{A}}^r$. Put $a=\pi (y)$. Its fibre is a finite set $\{y_1,\dots ,y_s\}$, one of whose members is $y$, because a finite morphism has finite fibres over closed points.

Choose an affine open $U\subseteq {\varphi }^{-1}(Y_0)$ meeting $X_y$. Since $X$ is irreducible, $U$ still has dimension $N$. In $U$, the fibre of $\pi \varphi$ over $a$ is cut out by the $r$ regular functions ${\pi }_j\varphi -a_j$. Apply the supplied principal hypersurface dimension lemma successively on each irreducible component: if a function vanishes identically the dimension does not drop, if its zero set is empty the component disappears, and otherwise each new component loses exactly one dimension. Hence every surviving component after the $r$ cuts has dimension at least $N-r$.

But that composite fibre is the disjoint union of $U\cap {\varphi }^{-1}(y_i)$. Each member is closed, and since there are finitely many members it is also open in the composite fibre. Thus $U\cap X_y$ is a nonempty union of its irreducible components, givingThe same argument around a point on any fibre component gives the componentwise bound. This is the closed-fibre lower bound by finite normalization; it avoids assuming that $Y$ is smooth or that its point ideal has only $r$ generators.

Irreducibility matters if “variety” is generalized to arbitrary reduced reducible spaces. For example $X={\mathbb{A}}^2\bigsqcup {\mathbb{A}}^1\to {\mathbb{A}}^1$ can map the first component constantly to zero and the second by the identity. It is surjective, but nonzero fibres have dimension zero while $\dim X-\dim Y=1$. For reducible sources one uses the individual component dimensions instead; the preceding proof establishes the stated theorem under its standard irreducible-variety convention.

Any isomorphism would permute irreducible components and preserve their intersections. In each configuration, the origin is the unique point lying on all three components, so an isomorphism must take origin to origin.

The coordinate ring of the spatial configuration isThe displayed ideal is indeed the intersection of the three coordinate-axis ideals. Its generators have no linear parts, so at the origin the residue classes of $x,y,z$ form a basis of $\mathfrak{m}/{\mathfrak{m}}^2$. Hence its algebraic cotangent space and Zariski tangent space have dimension three.

For the plane configuration, the coordinate ring is $B=k[s,t]/(st(s-t))$. The three linear factors are distinct in every characteristic, so this principal ideal is radical. It likewise has no linear part, but now $\mathfrak{n}/{\mathfrak{n}}^2$ has basis $s,t$ and dimension two. An isomorphism of local rings induces an isomorphism of these cotangent spaces, which is impossible:Thus the varieties are not isomorphic. This is embedding dimension distinguishes three-branch curves: matching incidence of branches does not determine their local algebra.

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.

If $X$ is an affine scheme, each reduced irreducible component is a closed subscheme, hence an affine scheme. We prove the converse by explicit affine gluing along closed subschemes.

First suppose $X$ is the scheme-theoretic union of two closed subschemes which are affine schemes $Y$ and $Z$: their defining ideal sheaves have zero intersection. Their scheme-theoretic intersection $W$ is a closed subscheme of both, hence an affine scheme. WriteThe two maps $A\to C$ and $B\to C$ are surjective. Form the fiber product of ringsBoth projections are surjective. Let $I=\ker (R\to A)$ and $J=\ker (R\to B)$. ThenFor the last equality, the common image map $R\to C$ is surjective, and its kernel of a ring homomorphism consists exactly of pairs whose two coordinates map to zero. Therefore $\mathrm{Spec}R$ has closed pieces isomorphic to $Y$ and $Z$, with intersection $W$. They cover it because each prime ideal contains $I$ or $J$, as $IJ=0$.

A finite closed cover determines the topology: a subset is closed if and only if its intersections with both closed pieces are closed. The identifications of $Y,Z,W$ therefore give a homeomorphism between $X$ and $\mathrm{Spec}R$. To check the structure sheaf, on $X$ the elementary ideal-intersection sequence givesIt is exact on stalks, by ${\mathcal{I}}_Y\cap {\mathcal{I}}_Z=0$. The same sequence holds on $\mathrm{Spec}R$, using $I\cap J=0$. Under the homeomorphism, the two middle sheaves and their restriction maps identify, so their kernels identify as sheaves of rings. This identifies the structure sheaves, giving an isomorphism of schemes $X\simeq \mathrm{Spec}R$. The intersection $W$ need not be reduced.

Now a Noetherian scheme has finitely many irreducible components $Y_1,\dots ,Y_r$. Induct on $r$. For $r=1$, the reduced scheme is its single reduced irreducible component, so is an affine scheme. For $r>1$, let $Y=Y_1$ and let $Z$ be the reduced union of the remaining irreducible components. Its components are the corresponding $Y_i$, so $Z$ is an affine scheme by induction. The defining ideal sheaves of $Y$ and $Z$ have zero intersection: their intersection is the ideal sheaf of the reduced union of all components, which is zero because $X$ is reduced. The two-piece argument proves that $X$ is an affine scheme. The empty scheme is $\mathrm{Spec}0$ and causes no exception. Hence

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 $\dim \varnothing =-\infty$.