SCHEMES 17
Our goal in this subsection is to show that 9Jtot(X, Y7) is closed in 9Hot(X, Y) if
Yf
is closed in Y and if some additional condition holds. We need some preparation:
Consider an open covering (Xj)jej of X and a closed subfunctor Y' of Y. Let
Pj : dJlot(X, Y) —* DJlox(Xj,Y) denote the obvious restriction map. We claim:
(2) mox(X, Yf) = p | pjifOtotiXj, Y').
jeJ
Of course, one inclusion ("c") is trivial. Consider on the other hand / G 9Jtot(X,
Y)(A) =
MOT(XA,YA)
for some /c-algebra A with pj(A)f G Mor(XJA,YA) for all
j G J, i.e., with XjA C
f~1(YA)
for all j . Now the (XjA)jej are an open covering
of XA and /
_ 1
(1^ ) is a closed subfunctor of XA, SO 1.12(5) yields
f~1(YA)
= XA,
hence / G OTot(X, F'X^)-
(3) Le£ X and Y be k-functors and
Yf
C Y a closed subfunctor. If X admits
an open covering (Xj)jej with affine schemes such that each k[Xj] is free as a
k-module, then 97tot(X, Y') is closed in 0Jlor(X, Y).
(If X is a scheme, then X is called locally free if and only if there is an open covering
as above.)
One sees using (2) that it suffices to prove (3) in the case where X = SpkR
for some ^-algebra R that is free as a /c-module. We now have to show for each
fc-algebra A and each morphism / : SpkA DJlot(X, Y) that
f~19Jlox(X,
Y') is
closed.
We have for each /c-algebra B natural bijections
Mor(SpkB,Wlox{SpkR,Y)) ~ Wlox(SpkR,Y)(B) ~ Mor((SpkR)B,YB)
~ Moi(SpB(R 0 B), YB) ^ Y(R 0 B)
arising from Yoneda's lemma, the definitions, or the universal property of the tensor
product.
Taking B = A, we see that any / as above corresponds to some y G Y(R®A).
One checks then for all B that
f(B) : (SpkA)(B) = Homfe_alg(A, B) - Wlox{SpkR, Y)(B) ~ Y(# 0 B)
maps any /3 G Homfe_aig(A, S) to y(id/j ®/?)(y) G V(ii (g) £) .
On the other hand y G Y(R®A) defines also a morphism f : Spk{R®A) * F
that maps for all B any 7 G Homfc_aig(i? 0 A, B) to F(7)(y). So we get above
f(B)((3) = f'(R®B)(idR®f3).
So
f~lmox{SpkR, Yr){B)
consists of all f3 with /'(i j 0 B)(idfl 0/3) G y ' ( # 0 5) .
Since Y' is closed in Y, there is an ideal
If
in R®A with ( / ' J "
1
^ ' ) = V(J').
We get now for all B
f^MoxiSpkRXXB) = {/3e Homfc_alg(A,£) | V c ker(idR®(3) }.
Now we use for the first time that R is free as a fc-module. It implies that
ker(id# 0/3) = R 0 ker(/?), hence
f^VJloxiSpkR,
Yf)(B)
= { p G Hom^_alg(A, B ) | J ' c f l ® ker(/3) }.
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