Chow's lemma, named after Wei-Liang Chow, is one of the foundational results in algebraic geometry. It roughly says that a proper morphism is fairly close to being a projective morphism. More precisely, a version of it states the following:

Proof

The proof here is a standard one.

Reduction to the case of X irreducible

We can first reduce to the case where X is irreducible. To start, X is noetherian since it is of finite type over a noetherian base. Therefore it has finitely many irreducible components X_i, and we claim that for each X_i there is an irreducible proper S-scheme Y_i so that Y_i\to X has set-theoretic image X_i and is an isomorphism on the open dense subset of X_i. To see this, define Y_i to be the scheme-theoretic image of the open immersion Since is set-theoretically noetherian for each i, the map is quasi-compact and we may compute this scheme-theoretic image affine-locally on X, immediately proving the two claims. If we can produce for each Y_i a projective S-scheme Y_i' as in the statement of the theorem, then we can take X' to be the disjoint union and f to be the composition : this map is projective, and an isomorphism over a dense open set of X, while is a projective S-scheme since it is a finite union of projective S-schemes. Since each Y_i is proper over S, we've completed the reduction to the case X irreducible.

X can be covered by finitely many quasi-projective S -schemes

Next, we will show that X can be covered by a finite number of open subsets U_i so that each U_i is quasi-projective over S. To do this, we may by quasi-compactness first cover S by finitely many affine opens S_j, and then cover the preimage of each S_j in X by finitely many affine opens X_{jk} each with a closed immersion in to since X\to S is of finite type and therefore quasi-compact. Composing this map with the open immersions and, we see that each X_{ij} is a closed subscheme of an open subscheme of. As is noetherian, every closed subscheme of an open subscheme is also an open subscheme of a closed subscheme, and therefore each X_{ij} is quasi-projective over S.

Construction of X' and f:X'\to X

Now suppose {U_i} is a finite open cover of X by quasi-projective S-schemes, with an open immersion in to a projective S-scheme. Set, which is nonempty as X is irreducible. The restrictions of the \phi_i to U define a morphism so that, where U\to U_i is the canonical injection and is the projection. Letting j:U\to X denote the canonical open immersion, we define, which we claim is an immersion. To see this, note that this morphism can be factored as the graph morphism (which is a closed immersion as P\to S is separated) followed by the open immersion ; as X\times_S P is noetherian, we can apply the same logic as before to see that we can swap the order of the open and closed immersions. Now let X' be the scheme-theoretic image of \psi, and factor \psi as where \psi' is an open immersion and h is a closed immersion. Let and be the canonical projections. Set We will show that X' and f satisfy the conclusion of the theorem.

Verification of the claimed properties of X' and f

To show f is surjective, we first note that it is proper and therefore closed. As its image contains the dense open set U\subset X, we see that f must be surjective. It is also straightforward to see that f induces an isomorphism on U: we may just combine the facts that and \psi is an isomorphism on to its image, as \psi factors as the composition of a closed immersion followed by an open immersion. It remains to show that X' is projective over S. We will do this by showing that g:X'\to P is an immersion. We define the following four families of open subschemes: As the U_i cover X, the U_i' cover X', and we wish to show that the U_i'' also cover X'. We will do this by showing that for all i. It suffices to show that is equal to as a map of topological spaces. Replacing U_i' by its reduction, which has the same underlying topological space, we have that the two morphisms are both extensions of the underlying map of topological space, so by the reduced-to-separated lemma they must be equal as U is topologically dense in U_i. Therefore for all i and the claim is proven. The upshot is that the W_i cover g(X'), and we can check that g is an immersion by checking that is an immersion for all i. For this, consider the morphism Since X\to S is separated, the graph morphism is a closed immersion and the graph is a closed subscheme of ; if we show that factors through this graph (where we consider U\subset X' via our observation that f is an isomorphism over f^{-1}(U) from earlier), then the map from U_i'' must also factor through this graph by construction of the scheme-theoretic image. Since the restriction of q_2 to T_i is an isomorphism onto W_i, the restriction of g to U_i'' will be an immersion into W_i, and our claim will be proven. Let v_i be the canonical injection ; we have to show that there is a morphism so that. By the definition of the fiber product, it suffices to prove that, or by identifying U\subset X and U\subset X', that. But and, so the desired conclusion follows from the definition of \phi:U\to P and g is an immersion. Since X'\to S is proper, any S-morphism out of X' is closed, and thus g:X'\to P is a closed immersion, so X' is projective.

Additional statements

In the statement of Chow's lemma, if X is reduced, irreducible, or integral, we can assume that the same holds for X'. If both X and X' are irreducible, then f: X' \to X is a birational morphism.