Let \(f: X \to Y\) be a morphism of finite type of noetherian schemes. Then the image of any constructible subset of \(X\) is a constructible subset of \(Y\). In particular \(f(X)\), which need not be either open or closed, is a constructible subset of \(Y\).
Prove this theorem in the following steps.
Reduce to showing that \(f(X)\) itself is constructible, in the case where \(X\) and \(Y\) are affine, integral, noetherian schemes and \(f\) is a dominant morphism.
In that case, show that \(f(X)\) contains a nonempty open subset of \(Y\) using the following result from commutative algebra. Let \(A \subseteq B\) be an inclusion of noetherian integral domains such that \(B\) is a finitely generated \(A\)-algebra. Then given a nonzero element \(b \in B\), there is a nonzero element \(a \in A\) with the following property: if \(\phi: A \to K\) is any homomorphism of \(A\) into an algebraically closed field \(K\) with \(\phi(a) \neq 0\), then \(\phi\) extends to a homomorphism \(\phi'\) of \(B\) into \(K\) with \(\phi'(b) \neq 0\).
Now use noetherian induction on \(Y\) to complete the proof.
Give some examples of morphisms \(f: X \to Y\) of varieties over an algebraically closed field \(k\) showing that \(f(X)\) need not be either open or closed.
Prove the algebraic result of (b) by induction on the number of generators of \(B\) over \(A\); for one generator prove it directly. In the application, take \(b = 1\). This statement is Chevalley’s theorem; see Cartan and Chevalley, exposé 7, and Matsumura.
Solution
1. It is enough to prove that the image of the whole source is constructible for finite type morphisms of noetherian schemes.
Let \(E\subseteq X\) be constructible. By Hartshorne II.3.18(a), write \[ E=E_1\amalg\cdots\amalg E_r \] with each \(E_i\) locally closed.
Give each \(E_i\) its reduced induced locally closed subscheme structure. Its inclusion \[ E_i\hookrightarrow X \] is a composition of a closed immersion and an open immersion. Since \(X\) is noetherian, every open subset is quasi-compact, so Hartshorne II.3.13(a),(b),(c) shows that this inclusion is of finite type. Hence \[ E_i\longrightarrow Y \] is of finite type.
If images of whole sources under finite type morphisms are constructible, every \[ f(E_i) \] is constructible, and then \[ f(E)=\bigcup_{i=1}^r f(E_i) \] is constructible as a finite union.
2. To prove constructibility of \(f(X)\), it is enough to treat the case in which \(X\) and \(Y\) are affine.
The noetherian scheme \(Y\) is quasi-compact, so choose a finite affine cover \[ Y=V_1\cup\cdots\cup V_m. \] Then \[ f(X)\cap V_j = f\bigl(f^{-1}(V_j)\bigr). \] Since \(f\) is finite type, each \(f^{-1}(V_j)\) is quasi-compact and has a finite affine cover \[ U_{j1},\ldots,U_{jr_j}. \] Therefore \[ f(X)\cap V_j = \bigcup_\ell f(U_{j\ell}). \] If the affine-source/affine-target case is known, each \(f(U_{j\ell})\) is constructible in \(V_j\), hence locally a finite union of locally closed subsets of \(Y\). Thus \(f(X)\cap V_j\) is constructible in \(Y\). Taking the finite union over \(j\) proves constructibility of \(f(X)\).
3. In the affine case it is enough to treat \(X\) and \(Y\) integral and \(f\) dominant.
Write \[ X=\Spec B, \qquad Y=\Spec A. \] Because \(X\) is noetherian, it has finitely many irreducible components \[ X_1,\ldots,X_n. \] Give each \(X_i\) its reduced induced structure. By Hartshorne II.3.11, each \(X_i\) is a closed affine integral subscheme of \(X\), and \[ f(X)=\bigcup_i f(X_i). \]
For each \(i\), let \[ Y_i=\overline{f(X_i)} \] with its reduced induced structure. Then \(Y_i\) is an integral closed affine subscheme of \(Y\), and the morphism \[ X_i\longrightarrow Y_i \] is dominant. It is finite type: its composite with the closed immersion \(Y_i\hookrightarrow Y\) is finite type, while \(X_i\to Y_i\) is quasi-compact because \(X_i\) is affine, so Hartshorne II.3.13(f) applies.
If the theorem is known for these affine integral dominant morphisms, each \(f(X_i)\) is constructible in \(Y_i\), hence constructible in \(Y\) because \(Y_i\) is closed. Their finite union is \(f(X)\).
4. This proves the reduction required in part (a).
Step 1 reduces from an arbitrary constructible subset to the image of the whole source, 2 reduces to affine source and target, and 3 reduces to the affine integral dominant case.
5. We prove the commutative-algebra lemma in part (b) first when\[ B=A[x] \]is generated by one element over \(A\).
Let \(0\ne b\in B\). There are two cases.
Case 1: \(x\) is transcendental over \(K(A)\). Then \[ B\cong A[T] \] and \(b\) corresponds to a nonzero polynomial \[ q(T)=q_0+q_1T+\cdots+q_mT^m\in A[T]. \] Choose one nonzero coefficient \[ a=q_j\in A. \] Let \[ \phi:A\to K \] map into an algebraically closed field with \(\phi(a)\ne0\). Then the polynomial \[ \phi(q)(T)\in K[T] \] is nonzero. Since an algebraically closed field is infinite, choose \(\lambda\in K\) with \[ \phi(q)(\lambda)\ne0. \] Sending \(x\mapsto\lambda\) extends \(\phi\) to \[ \phi':B\to K \] and gives \(\phi'(b)\ne0\).
Case 2: \(x\) is algebraic over \(K(A)\). Let \[ F(T)\in K(A)[T] \] be the monic minimal polynomial of \(x\). Write \[ b=q(x) \] for \(q(T)\in A[T]\). Since \(b\ne0\), \(F\) does not divide \(q\) in \(K(A)[T]\), so \[ \gcd(F,q)=1. \] Hence there are polynomials \(P,Q\in K(A)[T]\) with \[ PF+Qq=1. \]
Clear all denominators occurring in \(F,P,Q\). There exists \(0\ne a\in A\) such that
- \(F\in A_a[T]\) is still monic, and
- after multiplying the Bézout identity by a unit of \(A_a\), it has the form \[ P_0F+Q_0q=c \] with \(P_0,Q_0\in A_a[T]\) and \(c\in A\) nonzero and invertible in \(A_a\).
Replace \(a\) by \(ac\). If \(\phi:A\to K\) satisfies \(\phi(a)\ne0\), it extends to \(A_a\). The monic polynomial \(\phi(F)\) has a root \(\lambda\in K\). Evaluating the Bézout identity at \(\lambda\) gives \[ \phi(Q_0)(\lambda)\phi(q)(\lambda)=\phi(c)\ne0. \] Thus \(\phi(q)(\lambda)\ne0\). Sending \(x\mapsto\lambda\) defines the desired extension \(B\to K\) with \(b\) nonzero.
Indeed, this assignment respects every relation defining \(A[x]\): if \(r(T)\in A[T]\) satisfies \(r(x)=0\), then the minimal polynomial \(F\) divides \(r\) in \(K(A)[T]\). After working in the localization \(A_a\) chosen above, division by the monic polynomial \(F\) shows that \(r=FH\) for some \(H\in A_a[T]\). Hence \[ \phi(r)(\lambda)=\phi(F)(\lambda)\phi(H)(\lambda)=0. \]
6. The algebraic lemma holds for every finitely generated \(A\)-algebra domain\[ B=A[x_1,\ldots,x_n]. \]
Induct on \(n\). The case \(n=1\) is 5.
Put \[ C=A[x_1,\ldots,x_{n-1}], \qquad B=C[x_n]. \] Apply the one-generator case over the domain \(C\) to the nonzero element \(b\in B\). It gives a nonzero element \[ c\in C \] such that every homomorphism \(C\to K\) to an algebraically closed field which does not kill \(c\) extends to \(B\to K\) without killing \(b\).
Apply the induction hypothesis to \[ A\subseteq C \] and the nonzero element \(c\). There is \(0\ne a\in A\) such that every \(\phi:A\to K\) with \(\phi(a)\ne0\) extends to a map \(C\to K\) with \(c\) nonzero. Applying the property of \(c\) then extends further to \(B\) with \(b\) nonzero. This proves the lemma.
7. Let\[ f:\Spec B\longrightarrow\Spec A \]be dominant, with \(A\subseteq B\) noetherian integral domains and \(B\) finitely generated over \(A\). Then \(f(\Spec B)\) contains a nonempty principal open subset of \(\Spec A\).
Apply 6 with \[ b=1. \] Obtain \(0\ne a\in A\) with the extension property.
We claim \[ D(a)\subseteq f(\Spec B). \] Let \(\mathfrak p\in D(a)\). Compose \[ A\longrightarrow\kappa(\mathfrak p) \hookrightarrow \overline{\kappa(\mathfrak p)}. \] The image of \(a\) is nonzero. By the algebra lemma, this map extends to \[ B\longrightarrow\overline{\kappa(\mathfrak p)}. \] Let \(\mathfrak q\) be its kernel. Since the extension agrees with the original map on \(A\), \[ \mathfrak q\cap A=\mathfrak p. \] Thus \(f(\mathfrak q)=\mathfrak p\).
Because \(a\ne0\) and \(A\) is a domain, \(D(a)\) contains the generic point and is nonempty. Hence the image contains a nonempty open subset.
8. This proves part (b).
Steps <1>5–<1>6 prove the algebraic lemma requested in the exercise, and 7 applies it with \(b=1\) to the affine integral dominant morphism.
9. We now prove constructibility of \(f(X)\) for every finite type morphism of noetherian schemes by noetherian induction on the target.
For a closed subset \(Y'\subseteq Y\), let \(\mathcal P(Y')\) be the assertion:
for every finite type morphism \(h:W\to Y'\) of noetherian schemes, the image \(h(W)\) is constructible in \(Y'\).
Assume inductively that \(\mathcal P\) holds for every proper closed subset of \(Y\). We prove \(\mathcal P(Y)\).
10. In proving \(\mathcal P(Y)\), we may first assume that \(Y\) is irreducible. Under that assumption, if no irreducible component of \(X\) dominates \(Y\), then \(f(X)\) is constructible by the induction hypothesis.
If \(Y\) is reducible, write it as the finite union of its irreducible components \[ Y=Y_1\cup\cdots\cup Y_m. \] Each \(Y_j\) is a proper closed subset of \(Y\). The morphism \[ f^{-1}(Y_j)\to Y_j \] is finite type by base change, so the induction hypothesis makes \[ f\bigl(f^{-1}(Y_j)\bigr) \] constructible in \(Y_j\), hence in \(Y\). Their finite union is \(f(X)\). Thus only irreducible \(Y\) remains.
Now give the finitely many irreducible components \(X_i\) of \(X\) their reduced induced structures, and let \[ Y_i=\overline{f(X_i)}. \] If no \(X_i\) dominates the now irreducible \(Y\), then every \(Y_i\) is a proper closed subset of \(Y\).
3. The induction hypothesis gives \(f(X_i)\) constructible in \(Y_i\), hence in \(Y\). Their finite union is \(f(X)\).
11. Suppose an irreducible component \(X_0\subseteq X\) has dense image in an irreducible component \(Y_0\subseteq Y\). Then \(f(X)\) contains a nonempty open subset of \(Y_0\).
Replace \(X_0,Y_0\) by their reduced induced structures. The map \[ X_0\to Y_0 \] is dominant and finite type.
Let \(\eta\in Y_0\) be the generic point. Choose an affine neighborhood \[ V\subseteq Y_0 \] of \(\eta\). Since the map is dominant, the generic point of \(X_0\) lies over \(\eta\). Choose an affine neighborhood \(U\) of that generic point with \[ U\subseteq f^{-1}(V). \] Then \[ U\to V \] is an affine integral dominant finite type morphism. By 7, its image contains a nonempty open subset \[ O\subseteq V. \] Thus \[ O\subseteq f(X_0)\subseteq f(X). \] Since \(O\) contains the generic point of \(Y_0\), it is dense in \(Y_0\).
12. The noetherian induction in 9 proves that \(f(X)\) is constructible.
By 10, the reducible-target case and the case in which no component of \(X\) dominates the irreducible target are already handled.
Otherwise 11 gives a nonempty open \[ O\subseteq Y \] with \[ O\subseteq f(X). \] Put \[ F=Y\setminus O, \] a proper closed subset. Then \[ f(X) = O\cup f\bigl(f^{-1}(F)\bigr). \] The base change \[ f^{-1}(F)\to F \] is finite type, so the induction hypothesis gives its image constructible in \(F\), hence in \(Y\). The open set \(O\) is constructible, so their union is constructible.
Thus \(\mathcal P(Y)\) follows from \(\mathcal P\) for proper closed subsets. Hartshorne II.3.16 gives \(\mathcal P(Y)\) for every noetherian target \(Y\).
13. Therefore the image of every constructible subset under a finite type morphism of noetherian schemes is constructible.
Step 12 proves constructibility for the image of the whole source. Step 1 reduces the general constructible-subset statement to that case.
14. A finite type morphism can have image which is open but not closed.
The projection \[ \pi:V(xy-1)\subseteq\mathbb A^2_k\longrightarrow\mathbb A^1_k, \qquad (x,y)\longmapsto x, \] has image \[ \mathbb A^1_k\setminus\{0\}. \] Indeed, the equation \(xy=1\) has a solution \(y=x^{-1}\) exactly when \(x\ne0\). This image is open and not closed.
15. A finite type morphism can have image which is closed but not open.
The closed immersion \[ \{0\}\hookrightarrow\mathbb A^1_k \] is finite type and has image the closed point \(\{0\}\), which is not open.
16. A finite type morphism can have image which is neither open nor closed.
Consider \[ F:\mathbb A^2_k\longrightarrow\mathbb A^2_k, \qquad (x,y)\longmapsto(x,xy). \] Its image is \[ D(x)\cup\{(0,0)\}. \] For \(x\ne0\), any second coordinate \(v\) occurs by taking \(y=v/x\). For \(x=0\), the second coordinate is forced to be \(0\).
This subset is not closed because it contains the dense open \(D(x)\) but is not all of \(\mathbb A^2\). It is not open because every Zariski-open neighborhood of \((0,0)\) meets the line \(x=0\) away from the origin, while those points are not in the image. Equivalently, its complement \[ \{(0,v):v\ne0\} \] is not closed: its closure is the whole line \(V(x)\), which also contains \((0,0)\).
17. Q.E.D.
Step 4 proves part (a), 8 proves part (b), <1>12–<1>13 prove part (c) and the theorem, and <1>14–<1>16 give the examples requested in part (d).