Dimension of the fibres of a dominant morphism

The equality of contractions says exactly that \(f(\zeta)=\eta'\).

The inverse image of \(Y'\) in \(X_0\) is cut out set-theoretically by \(\mathfrak pA\). Since \(Z\) is an irreducible component of \(f^{-1}(Y')\), its intersection with \(X_0\) is an irreducible component of this inverse image. Thus its generic prime \(\mathfrak q\) is minimal over \(\mathfrak pA\).

Primes of this quotient correspond to primes \[ \mathfrak r\subseteq A \] with \[ \mathfrak pA\subseteq\mathfrak r\subseteq\mathfrak q. \] Minimality of \(\mathfrak q\) over \(\mathfrak pA\) forces \[ \mathfrak r=\mathfrak q. \] Hence the quotient has one prime ideal and dimension \(0\).

Apply the local dimension inequality to \[ B_{\mathfrak p}\longrightarrow A_{\mathfrak q}. \] By 2, \[ \dim A_{\mathfrak q}\le\dim B_{\mathfrak p}. \] Hartshorne II.3.20(c) identifies these local dimensions with the codimensions of the irreducible closed subsets whose generic points are \(\zeta\) and \(\eta'\): \[ \dim A_{\mathfrak q}=\codim(Z,X), \qquad \dim B_{\mathfrak p}=\codim(Y',Y). \] This proves part (a).

Work on affine neighborhoods \[ Y_0=\Spec B\ni y, \qquad X_0=\Spec A \] meeting the generic point of \(W\). Let \(\mathfrak p\subseteq B\) represent \(y\).

By Hartshorne II.3.10, the fibre is \[ \Spec\bigl(A\otimes_B\kappa(\mathfrak p)\bigr). \] An irreducible component \(W\) corresponds to a prime \(\mathfrak q\subseteq A\) minimal among primes containing \(\mathfrak pA\) and disjoint from \(B\setminus\mathfrak p\). Such a prime satisfies \[ \mathfrak q\cap B=\mathfrak p. \]

If a prime \(\mathfrak r\subsetneq\mathfrak q\) contained \(\mathfrak pA\), then \[ \mathfrak p\subseteq\mathfrak r\cap B\subseteq\mathfrak q\cap B=\mathfrak p, \] so \(\mathfrak r\cap B=\mathfrak p\) and \(\mathfrak r\) would define a smaller prime in the fibre, contradicting minimality. Hence \(\mathfrak q\) is minimal over \(\mathfrak pA\).

Thus \(V(\mathfrak q)\) is an irreducible component of the inverse image of \(V(\mathfrak p)=Y'\cap Y_0\). Taking its closure in \(X\) gives the asserted component \(Z\) of \(f^{-1}(Y')\). Its generic point is the same point represented by \(\mathfrak q\), and its image is \(y\).

Let \(w\) be the generic point of \(W\), equivalently of \(Z\). Since \(w\) maps to the generic point \(y\) of \(Y'\), there is an inclusion of function fields \[ K(Y')=\kappa(y)\hookrightarrow\kappa(w)=K(Z). \]

The fibre component \(W\) is an integral scheme of finite type over the field \(\kappa(y)\), with function field \(\kappa(w)\). Hartshorne II.3.20(b) gives \[ \dim W = \operatorname{trdeg}_{\kappa(y)}\kappa(w). \] By additivity of transcendence degree, \[ \operatorname{trdeg}_{\kappa(y)}\kappa(w) = \operatorname{trdeg}_k\kappa(w) - \operatorname{trdeg}_k\kappa(y). \] Applying II.3.20(b) to \(Z\) and \(Y'\) gives the displayed equality.

Choose an affine open neighborhood of the generic point of \(Y\), and then an affine open neighborhood of the generic point of \(X\) contained in its inverse image. Both replacements are nonempty open subsets of integral finite-type schemes, so Hartshorne II.3.20(e) preserves their dimensions. Their function fields are unchanged by II.3.6, so the relative dimension \(e\) is unchanged as well.

By II.3.20(b), \[ e = \operatorname{trdeg}_{K(Y)}K(X). \] Since \(K(X)\) is generated as a field over \(K(Y)\) by finitely many elements of the finitely generated \(B\)-algebra \(A\), a maximal algebraically independent subset of algebra generators may be chosen inside \(A\); call it \(t_1,\ldots,t_e\).

Their algebraic independence over \(K(Y)\) implies that \[ B[t_1,\ldots,t_e] \] is a polynomial ring over \(B\) and injects into \(A\). Hence \(g\) is dominant.

The field extension \[ K(X)/K(Y)(t_1,\ldots,t_e) \] is algebraic and finitely generated as a field extension, hence finite. Thus the generic fibre of \(g\) is finite, so \(g\) is generically finite.

Hartshorne II.3.7 applied to the generically finite morphism \(g\) gives a dense open \(W\) such that \(U\to W\) is finite.

The restriction remains dominant because \(U\) and \(W\) contain the generic points. A finite morphism is closed, so its image is both closed and dense in the irreducible space \(W\). Hence its image is all of \(W\).

The morphism \(X_1\to Y\) is affine \(e\)-space. Its fibre at \(y\) is \[ (X_1)_y\cong\mathbb A^e_{\kappa(y)}. \] The open set \(W_y\) is nonempty because \(y\in f(U)\) and \(U\to W\) is surjective. Hence \[ \dim W_y=e \] by Hartshorne II.3.20(e), applied over the field \(\kappa(y)\).

Base changing the finite surjective map \(U\to W\) to \(\Spec\kappa(y)\) gives a finite surjective morphism \[ U_y\longrightarrow W_y. \] Integral finite extensions preserve Krull dimension, so \[ \dim U_y=\dim W_y=e. \] This proves part (c).

Given \(x\in X\), put \(y=f(x)\) and choose an irreducible component of \(X_y\) containing \(x\). By part (b), every such component has dimension at least \(e\). Hence \(x\in E_e\).

Let \(U\subseteq X\) be the dense open subset from part (c). We claim \[ E_h\cap U=\varnothing. \]

Let \(x\in U\) and \(y=f(x)\). If \(Z\) is an irreducible component of \(X_y\) containing \(x\), then \[ Z\cap U_y \] is a nonempty open subset of \(Z\). Therefore \[ \dim Z = \dim(Z\cap U_y) \le \dim U_y=e. \] Part (b) gives the reverse inequality, so in fact every fibre component through a point of \(U\) has dimension exactly \(e\). Thus no point of \(U\) lies in \(E_h\) for \(h>e\).

Since \(U\) is nonempty open, \(E_h\) cannot be dense.

If \(\dim X=0\), every nonempty fibre component has dimension \(0\), so each \(E_h\) is either \(X\) or \(\varnothing\).

Assume the assertion for dominant morphisms whose integral source has dimension \(<\dim X\).

For \(h\le e\), part (b) gives \[ E_h=X, \] which is closed. Now suppose \(h>e\). Let \(U\) be the dense open from part (c). By 12, \[ E_h\subseteq F:=X\setminus U, \] a proper closed subset.

Let \[ F_1,\ldots,F_r \] be the irreducible components of \(F\), with their reduced induced structures, and let \[ Y_i=\overline{f(F_i)} \] with reduced induced structures. Each \[ f_i:F_i\longrightarrow Y_i \] is a dominant finite type morphism of integral schemes. Indeed, the composite \(F_i\to Y\) is finite type, while \(F_i\to Y_i\) is quasi-compact because \(F_i\) is noetherian; Hartshorne II.3.13(f) therefore makes the factor map finite type. Since \(F_i\) is a proper closed irreducible subset of the integral finite-type scheme \(X\), II.3.20(d) gives \[ \dim F_i<\dim X. \] Thus the induction hypothesis applies to every \(f_i\).

Let \(x\in E_h(f)\). Choose an irreducible component \[ Z\subseteq X_{f(x)} \] containing \(x\) with \(\dim Z\ge h\).

Every point of \(Z\) belongs to \(E_h(f)\) because the same component \(Z\) witnesses the condition. Hence \[ Z\subseteq E_h(f)\subseteq F. \] As \(Z\) is irreducible and \(F\) is the finite union of the \(F_i\), it lies in some \(F_i\). Then \(Z\) is also an irreducible component of the fibre \((F_i)_{f(x)}\): any larger irreducible subset of that fibre would be a larger irreducible subset of \(X_{f(x)}\), contradicting maximality of \(Z\). Thus \[ x\in E_h(f_i). \]

Conversely, if \(x\in E_h(f_i)\), an irreducible component \(W\) of \((F_i)_{f(x)}\) through \(x\) has dimension at least \(h\). It lies in some irreducible component \(Z\) of the full fibre \(X_{f(x)}\), and \[ \dim Z\ge\dim W\ge h. \] Hence \(x\in E_h(f)\).

This completes the induction and proves the closedness assertion in part (d), together with <1>11–<1>12.

If \(y\in D_h\), some irreducible component of \(X_y\) has dimension at least \(h\); any point on that component belongs to \(E_h\) and maps to \(y\).

Conversely, if \(y=f(x)\) for some \(x\in E_h\), the component witnessing \(x\in E_h\) has dimension at least \(h\), so \(\dim X_y\ge h\).

By part (d), \(E_h\) is closed in the noetherian finite-type scheme \(X\). Give it the reduced induced closed subscheme structure. The restricted morphism \[ E_h\longrightarrow Y \] is finite type. Chevalley’s theorem, Hartshorne II.3.19, therefore says that its image \[ f(E_h)=D_h \] is constructible.

A nonempty fibre has dimension exactly \(h\) if and only if its dimension is at least \(h\) but not at least \(h+1\). For \(h\ge0\), an empty fibre belongs to neither side. Thus the displayed equality holds.

Constructible subsets form a Boolean algebra, so the difference of the constructible sets \(D_h\) and \(D_{h+1}\) is constructible.

With the standard convention \(\dim\varnothing=-1\), \[ C_{-1}=Y\setminus f(X), \] which is constructible because \(f(X)\) is constructible by Hartshorne II.3.19. For \(h<-1\), \(C_h=\varnothing\). Hence \(C_h\) is constructible for every integer \(h\).

Work on affine neighborhoods \[ Y_0=\Spec B, \qquad X_0=\Spec A \] of the generic points, with \(B\hookrightarrow A\) an inclusion of domains. The generic fibre has coordinate ring \[ S^{-1}A, \qquad S=B\setminus\{0\}, \] which is a domain. Thus \(X_\eta\) is integral.

Its function field is \(K(X)\) and its ground field is \[ \kappa(\eta)=K(Y). \] Hartshorne II.3.20(b) gives \[ \dim X_\eta = \operatorname{trdeg}_{K(Y)}K(X). \] Additivity of transcendence degree and II.3.20(b) for \(X\) and \(Y\) give \[ \operatorname{trdeg}_{K(Y)}K(X) = \dim X-\dim Y=e. \]

By 19, the generic point \(\eta\) of the irreducible space \(Y\) belongs to \(C_e\). Step 18 shows that \(C_e\) is constructible.

Hartshorne II.3.18(b) says that a constructible subset of an irreducible Zariski space which contains the generic point is dense and contains a nonempty open subset. Thus \(C_e\) contains an open dense subset of \(Y\).

Steps <1>1–<1>3 prove part (a), <1>4–<1>6 prove part (b), <1>7–<1>10 prove part (c), <1>11–<1>15 prove part (d), and <1>16–<1>20 prove part (e).