# Proving that the Jacobson radical of a (not necessarily unital) ring is contained in the intersection of all left modular ideals.

I am following Lams book "A first course in non-commutative rings". I am attempting to prove that the Jacobson radical of a ring is precisely the intersection of all maximal modular left ideals of said ring (following the exercises on page 63 -64).

I am almost done, the only thing I need to do is prove that the Jacobson radical is contained in every maximal modular left ideal. On this front I am stuck.

How does one prove this?

• What is the definition of "jacobson radical" you are using? Is it in terms of quasiregular elements? – rschwieb Dec 4 '18 at 13:40
• Yes, the sum of all quasiregular left ideals – user2628206 Dec 4 '18 at 13:47

Suppose $$A$$ is a quasregular left ideal and that it is not contained in a maximal modular left ideal $$L$$.

Then $$A+L=R$$. Let $$e$$ be the right identity afforded by $$L$$ (that is, $$x-xe\in L$$ for all $$x\in R$$). Then $$x+\lambda_1=e$$ for some $$x\in A$$ and $$\lambda_1\in L$$.

By quasiregularity of $$x$$ there is a $$y$$ such that $$yx=x+y$$, and premultiplying the equality we just had by $$y$$:

$$yx+y\lambda_1=ye$$

and rewriting

$$y\lambda_1=ye-yx=ye-y-x=\lambda_2-x$$ for some $$\lambda_2\in L$$.

Rewriting again, $$x=\lambda_2-y\lambda_1\in L$$. But this implies $$e=x+\lambda_1\in L$$ too.

But now for any $$z\in R$$ at all, we have that $$z=(z-ze)+ze\in L$$, which is absurd. Therefore the initial assumption that $$A\nsubseteq L$$ is incorrect.