Jacobson Radical $J(R)$ is a proper ideal

Question

I found a remark on my notes:

Jacobson Radical $J(R)$ is a proper ideal. Hint: Zorn's Lemma

I know $J(R)$ is the intersection of all the maximal left ideals of ring $R$. I know the maximal ideals are proper by definition. However, in the remark i guess it must be a two sided ideal.
But, we have showed that $J(R)$ is two sided by the claim that it is the intersection of all annihilators of all simple $R$-modules which are two sided ideals.

Back to my remark, how come Zorn's Lemma is on the table? Could you please enlighten me?

Answer 1

Since $J(R)$ is the intersection of maximal left ideals, $J(R)$ will be a proper subset of $R$ as long as there exists a maximal left ideal to contain it.

But in rings with identity $1$($\neq 0$), maximal left ideals (and maximal right ideals) exist by a Zorn's lemma argument (which is probably the link you're looking for.)

In case you haven't seen it before, one argues that the poset of left ideals which don't contain the identity satisfies the Zorn's lemma hypothesis. The partial ordering is containment, of course.

For rings without identity, it can happen that there aren't any maximal left or maximal right ideals at all, and indeed in these cases $J(R)$ is defined to be all of $R$.

Answer 2

This also may help .

If M is an R-module, then Jacobson radical J(M)= $J_{R}$(M) of R-module M is the intersection of all maximal submodules of M. (Maximal submodules mean maximal proper submodules ).

So if M is finitely generated, then every submodule $N$ of M is contained in a maximal submodules, by Zorn's Lemma. (Otherwise, If the union of a chain of proper submodules is equal to M, then the union contains all the generators, hence some member of the chain contains all the generators, contradiction.) Now, take $N=0$, so we get J(M) is a proper submodule of M. Since R is finitely generated, J(R) is always proper left ideal.