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:
- If
is a scheme that is proper over a noetherian base
, then there exists a projective
-scheme
and a surjective
-morphism
that induces an isomorphism
for some dense open ![{\displaystyle U\subseteq X.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a8327ae44ec07e23dfb0d8d7a5e36177d16d7286)
Proof
The proof here is a standard one.
Reduction to the case of
irreducible
We can first reduce to the case where
is irreducible. To start,
is noetherian since it is of finite type over a noetherian base. Therefore it has finitely many irreducible components
, and we claim that for each
there is an irreducible proper
-scheme
so that
has set-theoretic image
and is an isomorphism on the open dense subset
of
. To see this, define
to be the scheme-theoretic image of the open immersion
![{\displaystyle X\setminus \cup _{j\neq i}X_{j}\to X.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cd981fd9cc5d34d726d1246e4e5eaffc4c368698)
Since
is set-theoretically noetherian for each
, the map
is quasi-compact and we may compute this scheme-theoretic image affine-locally on
, immediately proving the two claims. If we can produce for each
a projective
-scheme
as in the statement of the theorem, then we can take
to be the disjoint union
and
to be the composition
: this map is projective, and an isomorphism over a dense open set of
, while
is a projective
-scheme since it is a finite union of projective
-schemes. Since each
is proper over
, we've completed the reduction to the case
irreducible.
can be covered by finitely many quasi-projective
-schemes
Next, we will show that
can be covered by a finite number of open subsets
so that each
is quasi-projective over
. To do this, we may by quasi-compactness first cover
by finitely many affine opens
, and then cover the preimage of each
in
by finitely many affine opens
each with a closed immersion in to
since
is of finite type and therefore quasi-compact. Composing this map with the open immersions
and
, we see that each
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
is quasi-projective over
.
Construction of
and
Now suppose
is a finite open cover of
by quasi-projective
-schemes, with
an open immersion in to a projective
-scheme. Set
, which is nonempty as
is irreducible. The restrictions of the
to
define a morphism
![{\displaystyle \phi :U\to P=P_{1}\times _{S}\cdots \times _{S}P_{n}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8b9f591de35abe4587061da24a5205bd2113f006)
so that
, where
is the canonical injection and
is the projection. Letting
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
is separated) followed by the open immersion
; as
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
be the scheme-theoretic image of
, and factor
as
![{\displaystyle \psi :U{\stackrel {\psi '}{\to }}X'{\stackrel {h}{\to }}X\times _{S}P}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6a9ba7efc297b4fe42af5b51f1ede44b9af94b13)
where
is an open immersion and
is a closed immersion. Let
and
be the canonical projections. Set
![{\displaystyle f:X'{\stackrel {h}{\to }}X\times _{S}P{\stackrel {q_{1}}{\to }}X,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fab85c152d79164410c6c19792e3845ea4a1bcee)
![{\displaystyle g:X'{\stackrel {h}{\to }}X\times _{S}P{\stackrel {q_{2}}{\to }}P.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0ca30bf32fed9350f94206581949718d4abbd88b)
We will show that
and
satisfy the conclusion of the theorem.
Verification of the claimed properties of
and
To show
is surjective, we first note that it is proper and therefore closed. As its image contains the dense open set
, we see that
must be surjective. It is also straightforward to see that
induces an isomorphism on
: we may just combine the facts that
and
is an isomorphism on to its image, as
factors as the composition of a closed immersion followed by an open immersion
. It remains to show that
is projective over
.
We will do this by showing that
is an immersion. We define the following four families of open subschemes:
![{\displaystyle V_{i}=\phi _{i}(U_{i})\subset P_{i}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/aafc308ac0357c1e00b86bc4628c64074beef3f8)
![{\displaystyle W_{i}=p_{i}^{-1}(V_{i})\subset P}](https://wikimedia.org/api/rest_v1/media/math/render/svg/afce4945fa68995c0947be30e2b1b34301671360)
![{\displaystyle U_{i}'=f^{-1}(U_{i})\subset X'}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c71026735dcd55f3a4591858ed98ad89e78b8c9c)
![{\displaystyle U_{i}''=g^{-1}(W_{i})\subset X'.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ec72fbde4c878979d2aaacc8cba175f6cc83769d)
As the
cover
, the
cover
, and we wish to show that the
also cover
. We will do this by showing that
for all
. It suffices to show that
is equal to
as a map of topological spaces. Replacing
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
is topologically dense in
. Therefore
for all
and the claim is proven.
The upshot is that the
cover
, and we can check that
is an immersion by checking that
is an immersion for all
. For this, consider the morphism
![{\displaystyle u_{i}:W_{i}{\stackrel {p_{i}}{\to }}V_{i}{\stackrel {\phi _{i}^{-1}}{\to }}U_{i}\to X.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b1f0a1ee34f7500f4757e6f6c7a0239dfc9e47d5)
Since
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
via our observation that
is an isomorphism over
from earlier), then the map from
must also factor through this graph by construction of the scheme-theoretic image. Since the restriction of
to
is an isomorphism onto
, the restriction of
to
will be an immersion into
, and our claim will be proven. Let
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
and
, that
. But
and
, so the desired conclusion follows from the definition of
and
is an immersion. Since
is proper, any
-morphism out of
is closed, and thus
is a closed immersion, so
is projective.
Additional statements
In the statement of Chow's lemma, if
is reduced, irreducible, or integral, we can assume that the same holds for
. If both
and
are irreducible, then
is a birational morphism.
References
Bibliography