properties of the Jacobson radical
Theorem:
Let R,T be rings and φ:R→T be a surjective homomorphism
. Then φ(J(R))⊆J(T).
Proof:
We shall use the characterization of the Jacobson radical as the set of all a∈R such that for all r∈R, 1-ra is left invertible.
Let a∈J(R),t∈T. We claim that 1-tφ(a) is left invertible:
Since φ is surjective, t=φ(r) for some r∈R. Since a∈J(R), we know 1-ra is left invertible, so there exists u∈R such that u(1-ra)=1. Then we have
φ(u)(φ(1)-φ(r)φ(a))=φ(u)φ(1-ra)=φ(1)=1 |
So φ(a)∈J(T) as required.
Theorem:
Let R,T be rings. Then J(R×T)⊆J(R)×J(T).
Proof:
Let π1:R×T→R be a (surjective) projection.
By the previous theorem, π1(J(R×T))⊆J(R).
Similarly let π2:R×T→T be a (surjective) projection. We see that π2(J(R×T))⊆J(T).
Now take (a,b)∈J(R×T). Note that a=π1(a,b)∈J(R) and b=π2(a,b)∈J(T). Hence (a,b)∈J(R)×J(T) as required.
Title | properties of the Jacobson radical |
---|---|
Canonical name | PropertiesOfTheJacobsonRadical |
Date of creation | 2013-03-22 12:49:43 |
Last modified on | 2013-03-22 12:49:43 |
Owner | yark (2760) |
Last modified by | yark (2760) |
Numerical id | 12 |
Author | yark (2760) |
Entry type | Result |
Classification | msc 16N20 |