Let \(f: X \to Y\) be a morphism of varieties over \(k\). Assume that \(Y\) is regular, that \(X\) is Cohen-Macaulay, and that every fibre of \(f\) has dimension equal to \(\dim X - \dim Y\). Show that \(f\) is flat.
Hint: imitate the proof of (10.4), using (II, 8.21A).
Solution
Put \[ m=\dim X, \qquad n=\dim Y, \qquad d=m-n. \]
1. It is enough to prove that \(f\) is flat at every closed point of \(X\).
Let \(\xi\in X\) be arbitrary. Choose a closed point \[ x\in\overline{\{\xi\}}. \] Then \(f(x)\) is closed, while \(f(\xi)\) is a generization of \(f(x)\). The point \(\xi\) corresponds to a prime of \(\OO_{X,x}\), whose contraction in \(\OO_{Y,f(x)}\) is the prime corresponding to \(f(\xi)\).
Thus, if \[ \OO_{Y,f(x)}\longrightarrow\OO_{X,x} \] is flat, localizing this map at those two primes gives the flat map \[ \OO_{Y,f(\xi)}\longrightarrow\OO_{X,\xi}. \] Hence flatness at all closed points implies flatness everywhere.
2. Fix a closed point \(x\in X\), put \(y=f(x)\), and write\[ A=\OO_{Y,y}, \qquad B=\OO_{X,x}. \]Then \(A\) is a regular local ring of dimension \(n\), while \(B\) is a Cohen–Macaulay local ring of dimension \(m\).
Because \(x\) and \(y\) are closed points of the varieties \(X\) and \(Y\), \[ \dim B=\dim X=m, \qquad \dim A=\dim Y=n. \] The first ring is Cohen–Macaulay because \(X\) is Cohen–Macaulay, and the second is regular because \(Y\) is regular.
Choose a regular system of parameters \[ t_1,\ldots,t_n \] for \(A\). Since \(A\) is regular, these elements generate its maximal ideal \(\mathfrak m_A\) and form an \(A\)-regular sequence.
3. The local ring of the fibre at \(x\) has dimension exactly \(d\):\[ \dim B/\mathfrak m_A B = \dim B/(t_1,\ldots,t_n)B =d. \]
The quotient \[ B/\mathfrak m_A B \] is the local ring \(\OO_{X_y,x}\) of the fibre \(X_y\) at \(x\). By hypothesis, \[ \dim X_y=d, \] so \[ \dim B/\mathfrak m_A B\le d. \]
On the other hand, quotienting a Noetherian local ring by one element can decrease its dimension by at most one. Applying this successively to \(t_1,\ldots,t_n\) gives \[ \dim B/(t_1,\ldots,t_n)B \ge \dim B-n =m-n =d. \] The two inequalities give equality.
4. The images of \(t_1,\ldots,t_n\) in \(B\) form a \(B\)-regular sequence.
By step 2, \(B\) is Cohen–Macaulay of dimension \(m\). Step 3 gives \[ \dim B/(t_1,\ldots,t_n)B =m-n. \] Hartshorne II.8.21A(c) characterizes regular sequences in a Cohen–Macaulay local ring by this dimension drop. Therefore \[ t_1,\ldots,t_n \] is a \(B\)-regular sequence.
5. We have\[ \Tor_1^A\bigl(\kappa(y),B\bigr)=0. \]
Since \(t_1,\ldots,t_n\) is an \(A\)-regular sequence generating \(\mathfrak m_A\), the Koszul complex \[ K_A(t_1,\ldots,t_n) \] is a finite free resolution of \[ \kappa(y)=A/\mathfrak m_A. \] Tensoring this resolution with \(B\) gives \[ K_A(t_1,\ldots,t_n)\tensor_A B \cong K_B(t_1,\ldots,t_n). \] By step 4 the same sequence is \(B\)-regular, so the latter Koszul complex is exact in every positive degree. Its first homology is therefore zero, which is precisely the displayed \(\Tor_1\) group.
6. The local map\[ A\longrightarrow B \]is flat.
Modulo the maximal ideal of \(A\), \[ B/\mathfrak m_A B \] is a module over the field \[ A/\mathfrak m_A=\kappa(y), \] and hence is flat over that field. Step 5 gives \[ \Tor_1^A\bigl(A/\mathfrak m_A,B\bigr)=0. \] The local criterion for flatness now implies that \(B\) is flat over \(A\).
7. Q.E.D.
Step 6 proves flatness at every closed point \(x\in X\). Step 1 localizes these flat local maps to every point of \(X\). Therefore \[ \boxed{f:X\to Y\text{ is flat}.} \]