Another argument I really like. We first present two auxiliary lemmata.
Lemma 1. Let be an injective ring homomorphism such that the associated map
is closed. Then
.
Proof. Since is injective, the map
is dominant. Hence, it is a closed surjective map. For such a map, a closed set
is empty if and only if
is empty. Applying this to
for
, we get
if and only if
, i.e.
if and only if
.
Lemma 2. Let be an injective ring homomorphism such that the associated map
(induced by
) is closed. Then
is integral.
Proof. Let . Consider the ideal
, and let
. Note that
, and
is the image
of the composite map
(first isomorphism theorem). Write
for the inclusion
. Note that the map
induced by is just the restriction of
to the closed subsets
, hence it is a closed map. Since
is injective, Lemma 1 asserts that
But is invertible in
(its inverse is
), and it lies in
since it is the image of
under
. Hence, it is invertible in
, i.e.
. That is, we can write
for certain . Hence,
in
. Hence, some multiple
is
in
, proving that
is integral over
.
Corollary. Let be a ring homomorphism such that the associated map
is proper. Then
is finite.
Proof. Let be the kernel of
; then
lands in
. Then
is proper (Hartshorne, Corollary II.4.8(e)). Moreover, the map
is injective, hence integral by Lemma 2. Since is proper, it is of finite type. Hence,
is integral and of finite type, hence finite. Hence so is
.
A more geometric version is the following:
Theorem. Let be a morphism of schemes that is both affine and proper. Then
is finite.
Proof. Let be an affine open in
. Then
is affine (since
is an affine morphism); say
. Then the restriction
is proper (properness is local on the target), hence
is finite over
by the theorem above. This proves that
is finite.
Remark. The converse is also true, and in fact much easier: a finite morphism is affine by definition, and proper since in the affine case it is in fact projective. The question whether finite morphisms are always projective turns out to be somewhat subtle: it is true in the EGA sense of projective (it is given by the relative Proj of some sheaf of graded algebras, i.e. it embeds into some projective bundle), but not in the sense of Hartshorne (it embeds into for some
, i.e. it embeds into a trivial projective bundle).
I believe that the following is an example of the last statement (i.e. a finite morphism which is not H-projective). Let be a field of infinite degree of imperfection, i.e.
and
is an infinite extension of
. Then Theorem 2 in [B-McL] says that for each
there exists a finite field extension
of
which cannot be generated by fewer than
elements¹. Correspondingly,
cannot be embedded into
for
. Then consider
an infinite disjoint union of
(labelled by
), and
the disjoint union of
for all
. Then
is certainly projective in EGA’s sense (although any projective bundle it embeds into cannot have constant dimension). Yet it cannot be embedded into any
.
References.
[B-McL] M.F. Becker, S. MacLane, The minimum number of generators for inseparable algebraic extensions. Bull. Amer. Math. Soc. Volume 46-2 (1940), p. 182-186.
¹It is probably not very hard to actually come up with an example of such a field with extensions
. I think that
and
should do the trick.
But is there a connected counterexample of the last statement?
I think the following is a counterexample. Let
be an infinite chain of
s over a finite field
, with irreducible components
. Say that the point
in
is attached to the point
in
. For notational reasons, it is easier to remove
from
, so let’s do so. I will define
as another tree of
s based on some local model.
For a subset
, write
for the open subset
. Note that
is not the irreducible component
, but rather the open subset
of
. Moreover,
is the union of the opens
.
For the local model, note that
is isomorphic to
, where the
-axis
corresponds to
and the
-axis
to
. Write
, the union of the three axes in
. The map
given by
induces a finite map
. Note that over
, this is an isomorphism, whereas over
it is two copies of
.
Now to construct
, let
be the disjoint union of
copies of
, with its natural map to
. Above
, this is
copies of
, and above
it is
copies of
. This gives the glueing maps between
and
, and we let
be the union.
Then
is finite since this is a local condition. But it cannot be embedded into
for any
, since
is finite and the fibres of
get arbitrarily many
-points. ![Rendered by QuickLaTeX.com \qedsymbol](https://lovelylittlelemmas.rjprojects.net/wp-content/ql-cache/quicklatex.com-ee0d727242cede2ba48c8d3c4c849723_l3.svg)
That said, in cases of interest this is of course true. For instance, if
is quasi-projective over an affine scheme, then the two notions of projectivity agree; see [Tag 087S].
If
has the resolution property [Tag 0F85], then every quasi-coherent sheaf of finite type is a quotient of a locally free sheaf, so then at least you can always embed
into
for some locally free sheaf
. If
is connected, then
has constant rank.