Show that a flat morphism \(f: X \to Y\) of finite type of noetherian schemes is open: for every open subset \(U \subseteq X\), the image \(f(U)\) is open in \(Y\).
Hint: show that \(f(U)\) is constructible and stable under generization, using (II, Ex. 3.18) and (II, Ex. 3.19).
Solution
Fix an open subset \(U\subseteq X\).
1. The subset \(f(U)\subseteq Y\) is constructible.
The open subscheme \(U\) is noetherian because \(X\) is noetherian, and the restricted morphism \[ f|_U:U\longrightarrow Y \] is of finite type because finite type is preserved by restriction to an open subscheme. The whole space \(U\) is a constructible subset of itself. Chevalley’s theorem, Hartshorne II, Exercise 3.19, says that the image of a constructible subset under a finite-type morphism of noetherian schemes is constructible. Therefore \[ f(U)=(f|_U)(U) \] is constructible in \(Y\).
2. A flat local homomorphism of local rings is faithfully flat.
Let \[ (A,\mathfrak m)\longrightarrow(B,\mathfrak n) \] be local and flat. For an \(A\)-module \(M\), suppose \[ M\otimes_A B=0. \] If \(M\ne0\), choose \(0\ne u\in M\). Its cyclic submodule is \[ Au\cong A/I \] for a proper ideal \(I\subseteq A\), and because \(A\) is local we have \(I\subseteq\mathfrak m\). Flatness preserves the injection \(Au\hookrightarrow M\), so \[ (A/I)\otimes_A B\hookrightarrow M\otimes_A B. \] But \[ (A/I)\otimes_A B\cong B/IB. \] Because the homomorphism is local, \[ IB\subseteq\mathfrak mB\subseteq\mathfrak n, \] so \(B/IB\ne0\), contradicting \(M\otimes_A B=0\). Thus tensoring with \(B\) detects the zero module, and a flat module with this property is faithfully flat.
3. If \(x\in X\), \(y=f(x)\), and \(y'\) is a generization of \(y\), then there is a generization \(x'\) of \(x\) with\[ f(x')=y'. \]
The morphism on local rings \[ \OO_{Y,y}\longrightarrow\OO_{X,x} \] is flat because \(f\) is flat, and it is a local homomorphism. By step 2 it is faithfully flat. Hence the induced map on spectra \[ \Spec\OO_{X,x}\longrightarrow\Spec\OO_{Y,y} \] is surjective. Indeed, for any prime \(\mathfrak p\subseteq\OO_{Y,y}\), faithful flatness gives \[ \kappa(\mathfrak p)\otimes_{\OO_{Y,y}}\OO_{X,x}\ne0. \] Any nonzero ring has a prime ideal, and a prime of this fibre corresponds to a prime of \(\OO_{X,x}\) contracting to \(\mathfrak p\).
Points of \(\Spec\OO_{Y,y}\) correspond exactly to generizations of \(y\) in \(Y\), while points of \(\Spec\OO_{X,x}\) correspond to generizations of \(x\) in \(X\). The point \(y'\) therefore has a preimage \(x'\) in \(\Spec\OO_{X,x}\), and this \(x'\) is a generization of \(x\) satisfying \(f(x')=y'\).
4. The constructible subset \(f(U)\) is stable under generization.
Let \(y\in f(U)\) and let \(y'\) be a generization of \(y\). Choose \(x\in U\) with \(f(x)=y\). By step 3 there is a generization \(x'\) of \(x\) with \(f(x')=y'\).
Open subsets are stable under generization, so \(x\in U\) implies \(x'\in U\). Therefore \[ y'=f(x')\in f(U). \] Thus \(f(U)\) is stable under generization.
5. The subset \(f(U)\) is open in \(Y\).
Noetherian schemes are Zariski spaces in the sense used in Hartshorne II.3. Exercise II.3.18(c) says that a subset of a Zariski space is open if and only if it is constructible and stable under generization. Step 1 proves constructibility of \(f(U)\), and step 4 proves generization stability. Hence \[ \boxed{f(U)\text{ is open in }Y}. \] Since \(U\subseteq X\) was arbitrary, \(f\) is an open map.
6. Q.E.D.
Step 5 proves the required openness for every open subset of \(X\).