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For a surjective morphism of irreducible varieties, choose an affine neighborhood of the target point and a finite Noether normalization to affine space of target dimension $r$. On an affine source neighborhood the composite fibre is cut out by $r$ functions, so repeated principal hypersurface dimension lemma gives dimension at least source dimension minus $r$ on every component. The normalization fibre is finite. The original fibre is therefore an open-and-closed union of components of that composite fibre, proving the bound without assuming target smoothness. For a reducible source the bound must instead be applied using the dimension of each source component.

If $A$ is a finite-type integral domain over an algebraically closed field and $\dim A=d$, every maximal ideal $\mathfrak{m}$ has height of a prime ideal $d$. Choose a Noether normalization $k[z_1,\dots ,z_d]\subset A$. The contraction of $\mathfrak{m}$ is a maximal ideal of the polynomial ring and has height $d$; going-down theorem lifts its full chain since the polynomial ring is integrally closed domain. The upper bound is $\dim A$. A principal open subset containing a closed point consequently has dimension $d$, because the whole chain under that point survives localization.

A complex affine algebraic set is either finite or has the cardinality of $\mathbb{C}$. One proof takes a positive-dimensional irreducible component and applies Noether normalization: the resulting finite dominant map to a positive-dimensional affine space has a point above every closed point.

For a dominant morphism of algebraic varieties between irreducible varieties, there is a nonempty Zariski-open subset of the target over which every fibre is nonempty and every irreducible component has the displayed dimension. Apply Noether normalization to affine source charts over the target function field and clear denominators on the target to get a generic upper bound. Combine it with the closed-fibre lower bound in the fiber dimension theorem.

For a finite integral inclusion $A=k[t_1,\dots ,t_r]\subset B$ of domains, a nonzero $b\in B$ has a monic minimal polynomial over the fraction field with nonzero constant term $a_0\in A$. Its equation gives $a_0\in bB\cap A$. Consequently the image of $V_B(b)$ under the finite morphism to ${\mathbb{A}}^r$ is contained in the proper Zariski-closed set $V_A(a_0)$. Finite morphisms preserve the dimension of a closed subvariety and its image, so this proves dimension drops for proper closed subsets after Noether normalization. A power of $a_0$, up to sign, is the field norm of $b$.

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.

The usual field norm $N_{L/K}$ is defined for a finite field extension. The printed hypothesis “algebraic” alone allows an infinite extension, for which this determinant norm is not defined. We therefore assume $[L:K]<\infty$. For an arbitrary algebraic extension the meaningful elementwise statement is instead $N_{K(b)/K}(b)\in A$, and we prove this as well.

First a unique factorization domain is an integrally closed domain. Indeed, write an integral fraction as $u/v$ with relatively prime numerator and denominator. A monic integral equation, after multiplication by $v^r$, shows that $v$ divides $u^r$. Unique factorization forces $v$ to be a unit.

Let $m_b(T)=T^r+a_{r-1}{T}^{r-1}+\cdots +a_0\in K[T]$ be the monic minimal polynomial of $b$. Because $b$ is an integral element over $A$, it satisfies a monic polynomial $f\in A[T]$, and $m_b$ divides $f$ in $K[T]$. Every root of $m_b$ in an algebraic closure is therefore integral over $A$. This assertion includes repeated roots when the extension is inseparable. The coefficients of $m_b$ are elementary symmetric expressions in these roots. The integral elements form a ring: finitely many integral elements generate a finite $A$-module algebra, and the determinant trick shows every element of that algebra is integral. Thus all $a_i$ are integral over $A$. Since $a_i\in K$ and $A$ is integrally closed, $a_i\in A$.

Multiplication by $b$ on $K(b)$, in the basis $1,b,\dots ,b^{r-1}$, has the companion matrix of $m_b$, whose determinant is $(-1{)}^r{a}_0$. If $e=[L:K(b)]$, take a $K(b)$-basis of $L$. Multiplication by $b$ preserves each summand in the resulting direct sum of $e$ copies of $K(b)$, so its determinant over $K$ isIn particular $N_{K(b)/K}(b)=(-1{)}^r{a}_0\in A$. The argument requires no separability; in fact it holds for every integrally closed domain, not just a unique factorization domain.

For the connection with dimension, let $B$ be the coordinate ring of an irreducible affine variety. By Noether normalization there is a finite integral inclusion $A=k[t_1,\dots ,t_d]\subset B$, with $d=\dim B$. Their fraction fields form a finite extension. For any $0\ne b\in B$, its minimal-polynomial constant term $a_0$ is nonzero and lies in $A$. Its equation givesThus a nonzero function on the normalized variety produces a nonzero function on the base that vanishes wherever $b$ vanishes. Up to sign, the field norm is a power of this $a_0$, so it too belongs to $bB\cap A$.

Under the finite morphism to ${\mathbb{A}}^d$, the image of $V_B(b)$ is therefore contained in the proper Zariski-closed set $V_A(a_0)$ of dimension at most $d-1$. Finite morphisms preserve the dimension of a closed subvariety and its image. Applying this to a nonzero element of the ideal of any proper closed subvariety proves proper closed subvarieties have strictly smaller dimension. This is the useful role of the norm argument: nonzero equations upstairs give nonzero equations downstairs under a finite normalization.

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.

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 $0$.