Endomorphism ring, module, radical

Question

For a module $C$ and over a ring $R$, $E:=\operatorname{End}_R(C)$, if $1-x$ is in $\operatorname{Rad}(E)$,the intersection of all left maximal ideals, then there is $v$ in $E$ such that $\mathit{id}_E=1=vx$.

Why?

Moreover, if $e$ is an idempotent in $E$ and the left annihilator of $f$ in the ring $eEe$ is contained in $\operatorname{Rad}(eEe) = e\operatorname{Rad}(E)e$ then if $g \in eEe$ is such that $g\circ f = e\circ f$, then $g$ is invertible in $eEe$. Why?

That's a fundamental property of the Jacobson radical. Actually, an element $y$ belongs to the Jacobson radical if and only if, for every element $z$ in the ring, $1-yz$ is invertible. — egreg, Feb 03 '14 at 17:25
I changed to some tex notation for easier reading. I translated the composition circles in the last line correctly, right? — rschwieb, Feb 03 '14 at 21:12
Ok, I need help on the question as soon as possible. Thank you. — user122424, Feb 03 '14 at 21:30

score 0 · Answer 1 · answered Feb 03 '14 at 21:35

You should learn the "quasiregular element" definition of the Jacobson radical.

The following are equivalent for a ring $R$:

$x\in J(R)$, the Jacobson radical of $R$.
$1-rx$ is a unit of $R$ for every $r\in R$.

The proof is easy. Hints: what would happen if $R(1-rx)\neq R$? If $x+M$ is nonzero in $R/M$ for some maximal left ideal $M$, there would exist an $R$ such that $rx+M=1+M$... then what?

The second part of the question is more confusingly stated than it has to be.

If $S$ is any ring (such as the ring $eRe$) and $f$ is an element whose left annihilator is contained in $J(S)$, and $gf=1_Sf$ (notice $e$ is the identity of $eRe$) then $g$ is a unit. This follows immediately after rewriting $gf=f$ as $(1_S-g)f=0$ and applying the first part you mentioned.

@OP Some proofs are in this answer. – Bill Dubuque Mar 10 '14 at 20:33 — Bill Dubuque, Mar 10 '14 at 20:33

Endomorphism ring, module, radical

1 Answers1