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Jacobson radical of a module category
and its power∗
joking†
2013-03-22 3:05:51
Assume that k is a field and A is a k-algebra. The category of (left) Amodules will be denoted by Mod(A) and HomA (X, Y ) will denote the set of all
A-homomorphisms between A-modules X and Y . Of course HomA (X, Y ) is an
A-module itself and EndA (X) = HomA (X, X) is a k-algebra (even A-algebra)
with composition as a multiplication.
Let X and Y be A-modules. Define
radA (X, Y ) = {f ∈ HomA (X, Y ) | ∀g∈HomA (Y,X) 1X −gf is invertible in EndA (X)}.
Definition. The Jacobson radical of a category Mod(A) is defined as a class
[
rad Mod(A) =
radA (X, Y ).
X,Y ∈Mod(A)
Properties. 1) The Jacobson radical is an ideal in Mod(A), i.e. for any
X, Y, Z ∈ Mod(A), for any f ∈ radA (X, Y ), any h ∈ HomA (Y, Z) and any
g ∈ HomA (Z, X) we have hf ∈ radA (X, Z) and f g ∈ radA (Z, X). Additionaly
radA (X, Y ) is an A-submodule of HomA (X, Y ).
2) For any A-module X we have radA (X, X) = rad(EndA (X)), where on the
right side we have the classical Jacobson radical.
3) If X, Y are both indecomposable A-modules such that both EndA (X) and
EndA (Y ) are local algebras (in particular, if X and Y are finite dimensional),
then
radA (X, Y ) = {f ∈ HomA (X, Y ) | f is not an isomorphism}.
In particular, if X and Y are not isomorphic, then radA (X, Y ) = HomA (X, Y ).
Let n ∈ N and let f ∈ HomA (X, Y ). Assume there is a sequence of Amodules X = X0 , X1 , . . . , Xn−1 , Xn = Y and for any 0 ≤ i ≤ n − 1 we have an
∗ hJacobsonRadicalOfAModuleCategoryAndItsPoweri
created: h2013-03-2i by: hjokingi
version: h41916i Privacy setting: h1i hDefinitioni h16N20i
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A-homomorphism fi ∈ radA (Xi , Xi+1 ) such that f = fn−1 fn−2 · · · f1 f0 . Then
we will say that f is n-factorizable through Jacobson radical.
Definition. The n-th power of a Jacobson radical of a category Mod(A) is
defined as a class
[
radn Mod(A) =
radnA (X, Y ),
X,Y ∈Mod(A)
where radnA (X, Y ) is an A-submodule of HomA (X, Y ) generated by all homomorphisms n-factorizable through Jacobson radical. Additionaly define
rad∞
A (X, Y ) =
∞
\
radnA (X, Y ) n=1
Properties. 0) Obviously radA (X, Y ) = rad1A (X, Y ) and for any n ∈ N we
have
radnA (X, Y ) ⊇ rad∞
A (X, Y ).
1) Of course each radnA (X, Y ) is an A-submodule of HomA (X, Y ) and we have
following sequence of inclusions:
HomA (X, Y ) ⊇ rad1A (X, Y ) ⊇ rad2A (X, Y ) ⊇ rad3A (X, Y ) ⊇ · · ·
2) If both X and Y are finite dimensional, then there exists n ∈ N such that
n
rad∞
A (X, Y ) = radA (X, Y ).
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