Let \(f:X\to Y\) be a dominant morphism of affine varieties. Show that for any \(y\in f(X)\), any irreducible component of the fiber \(f^{-1}(y)\) is an affine variety of dimension \(d\geq \dim X-\dim Y\). Show that equality holds on a Zariski-dense open subset of \(Y\).
Solution
Let \(k\) be the algebraically closed ground field, and write \[ A=\mco(Y), \qquad B=\mco(X), \qquad m=\dim Y, \qquad n=\dim X, \qquad e=n-m. \] Affine varieties are irreducible, so \(A\) and \(B\) are finitely generated integral \(k\)-algebras. Dominance of \(f\) says that the comorphism \[ A\hookrightarrow B \] is injective.
1. Let \(y\in f(X)\), let \(\mathfrak p\subseteq A\) be the maximal idealof \(y\), and let \(Z\) be an irreducible component of \(f^{-1}(y)\). Then there isa prime \(\mathfrak q\subseteq B\), minimal over \(\mathfrak pB\), such that\[ Z=V(\mathfrak q)=\Spec(B/\mathfrak q), \]so \(Z\) is an affine variety.
Since \(k\) is algebraically closed, \[ \kappa(y)=k. \] Hence the fibre is \[ f^{-1}(y) = \Spec(B\tensor_A k) = \Spec(B/\mathfrak pB). \] Its irreducible components correspond to the primes of \(B\) minimal over \(\mathfrak pB\). Thus \(Z=V(\mathfrak q)\) for such a prime \(\mathfrak q\), and \[ Z\cong\Spec(B/\mathfrak q). \] Because \(\mathfrak q\) is prime, \(B/\mathfrak q\) is a finitely generated integral \(k\)-algebra. Therefore \(Z\) is an affine variety.
Moreover, \[ \mathfrak p \subseteq \mathfrak q\cap A. \] The contraction \(\mathfrak q\cap A\) is proper and \(\mathfrak p\) is maximal, so \[ \mathfrak q\cap A=\mathfrak p. \]
2. For the prime \(\mathfrak q\) of step 1,\[ \height\mathfrak q\leq m. \]
Localize the injective map \(A\to B\) at \(\mathfrak p\) and \(\mathfrak q\): \[ A_{\mathfrak p}\longrightarrow B_{\mathfrak q}. \] Since \(\mathfrak q\) is minimal over \(\mathfrak pB\), the quotient \[ B_{\mathfrak q}/\mathfrak pB_{\mathfrak q} \] has Krull dimension \(0\).
The standard local dimension inequality for a homomorphism of Noetherian local rings gives \[ \dim B_{\mathfrak q} \leq \dim A_{\mathfrak p} + \dim\bigl(B_{\mathfrak q}/\mathfrak pB_{\mathfrak q}\bigr) = \dim A_{\mathfrak p}. \]
The point \(y\) is closed, so the affine dimension theorem P-AGH2320DIMENSION gives \[ \dim A_{\mathfrak p}=\dim A=m. \] Finally, \[ \dim B_{\mathfrak q}=\height\mathfrak q, \] which proves the claim.
3. Every irreducible component \(Z\) of every nonempty fibre satisfies\[ \boxed{ \dim Z\geq\dim X-\dim Y. } \]
For the prime \(\mathfrak q\) of step 1, the affine dimension formula P-AGH2320DIMENSION gives \[ \dim(B/\mathfrak q)+\height\mathfrak q = \dim B = n. \] By step 2, \[ \height\mathfrak q\leq m. \] Therefore \[ \dim Z = n-\height\mathfrak q \geq n-m = e. \]
4. There is a nonempty Zariski-open subset\[ V\subseteq Y \]such that\[ V\subseteq f(X) \]and\[ f^{-1}(V)\longrightarrow V \]is flat.
Generic flatness T-MORGEN gives a nonempty open subset \[ V_0\subseteq Y \] over which \(f\) is flat.
Since \(f\) is a dominant finite-type morphism, Chevalley’s theorem P-AGH2319CHEVALLEY shows that \(f(X)\) is constructible. Dominance says that \(f(X)\) is dense in the irreducible variety \(Y\). A dense constructible subset of an irreducible Noetherian space contains a nonempty open subset. Hence there is a nonempty open \[ V_1\subseteq Y \] with \[ V_1\subseteq f(X). \] Set \[ V=V_0\cap V_1. \] The intersection is nonempty and open because \(Y\) is irreducible. It has both required properties.
5. If \(y\in V\) and \(Z\) is any irreducible component of \(f^{-1}(y)\),then\[ \boxed{ \dim Z=e=\dim X-\dim Y. } \]
1. Since \(y\in V\), the local map\[ A_{\mathfrak p}\longrightarrow B_{\mathfrak q} \]is flat. Flat ring maps satisfy going-down.
Because \(y\) is a closed point of the \(m\)-dimensional affine variety \(Y\), \[ \height\mathfrak p=m. \] Going-down lifts a chain of primes of length \(m\) ending at \(\mathfrak p\) to a chain of primes of \(B_{\mathfrak q}\) ending at \(\mathfrak q\). Hence \[ \height\mathfrak q\geq m. \] Step 2 gives the reverse inequality, so \[ \height\mathfrak q=m. \] Using the affine dimension formula exactly as in step 3 now gives \[ \dim Z = n-\height\mathfrak q = n-m = e. \]
6. The equality locus contains a Zariski-dense open subset of \(Y\).
The set \(V\) from step 4 is a nonempty open subset of the irreducible variety \(Y\), so it is Zariski dense. Step 5 shows that for every \(y\in V\), every irreducible component of the fibre has dimension \[ \dim X-\dim Y. \]
7. Q.E.D.
Step 1 proves that every fibre component is affine, step 3 proves the dimension lower bound for every \(y\in f(X)\), and steps <1>4–<1>6 prove generic equality on a Zariski-dense open subset of \(Y\).