PROJECTIVE MODULES 11
Then /(A) G ob Cgr with ^ / ( A ) = J(A). Moreover /(A) is projective in CgT. By Frobenius reciprocity
dim
F
Homu(g)(I(\),£gr(v)) = dim
F
Homu(Go)(P(X),£gr(n) \u(G0))
= [£9r(v) \u(G0y. L(\)].
For A = fi the above expression is non-zero. This implies that /(A) has a graded projective
indecomposable summand corresponding to the irreducible Cgr(X). We will denote this graded
projective cover by Vgr(X).
Theore m 1.2.5. For each X G t we have TCgr{X) = £(A) and TVgriX) = V(X).
Proof. For each A G T observe that
$ 3
&mF
Homu{6)(I(X),£gr(ii)) = ^ [ £
5
r O )
\u(G0): L(x)}
net
M€T
E[^)k(G
0
)^(A)]
= ] T d i m
F
# o r a t f
( g )
( / ( A ) , £(//)).
net
The first sum indicates the number of graded projective indecomposable summands in /(A), while
the last sum indicates the number of projective indecomposable summands in /(A). Since FVgr(X)
is a projective U(g) module it follows that the last sum must be greater than or equal to the first.
Hence equality holds and so by the Krull-Schmidt theorem and the fact that V(X) is a projective
summand of ^"P
pr
(A) we have FVgr{X) = V(X). From the computation above for each //, A G T we
also have
[Cgr(n) \u(Go): L(X)] = [£(/*) \u(Goy. 1(A)].
It follows that
d i m
F
£
5 r
( A ) = dim
F
£(A) .
Consequently, JrCgr(X) = C(X) because C(X) is a homomorphic image of TCgr(X).u
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