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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?

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

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