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Noncommutative Algebra / Farb & Dennis defines the Jacobson Radical $J(R)$ of a ring $R$ as the intersection of all annihilators of simple $R$-modules (page 58).

It is then claimed that $J(R)$ is also the intersection of all maximal left ideals in $R$.

The given explanation is that "simple $R$-modules are precisely $R/I$ for maximal left ideals $I$".

I've been thinking about it and I'm confused. I do believe that a simple $R$-module is isomorphic to $R/I$ for some maximal ideal $I$, but it doesn't mean that it actually is $R/I$.

Is there really a problem here? If so, how can it be fixed?

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Yes, but you'll notice that isomorphic $R$-modules have equal annihilators. – Miha Habič Feb 8 '12 at 11:04
up vote 3 down vote accepted

Corrected with thanks to Cihan for the pertinent comments to this question.

It should have said "the annihilators of simple $R$ modules are precisely the annihilators of modules $R/I$ for maximal left ideals $I$, and these annihilators are intersections of the annihilators of their individual nonzero elements; the latter annihilators are all maximal left ideals, among which $I$ itself as the annihilator of the element $1+I\in R/I$". The first part is because every simple module is isomorphic to some $R/I$, and isomorphic modules have identical annihilators.

I would also like to note that the definition "the intersection of all annihilators of simple $R$-modules" is unnecessarily referring to the class of all simple $R$-modules, which is too large to be called a set. Strictly speaking this is forming a class of sets, whose intersection is taken. One can indeed define the intersection of any non-empty class of sets, but in this context this seems set-theoretic overkill to me: for one thing the annihilators being intersected actually form a set, because of the above equivalence and also because they are all subsets of $R$ (the set of annihilators is a subset of the powerset of $R$). Saying "the intersection of all maximal left ideals in $R$" would be a much clearer as definition to me (and is heavy enough definitional artillery as it stands).

However as Cihan noted, this description does not make it obvious that the radical is a two-sided ideal, whereas the annihilator description does make this obvious (to those who, unlike I did when I first wrote this answer, know enough about such matters).

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"any maximal left ideal $I$ is the annihilator of the particular simple $R$ module $R/I$": That's not true, annihilators of modules are always two sided ideals. (Actually given a left ideal $I$ the annihilator of the left $R$-module $R/I$ is the largest ideal that's contained in $I$) That's why the "intersection of annihilators" definition is useful, the fact that $J(R)$ is an ideal (not only a left ideal) is immediate from there. – Cihan Apr 17 '12 at 23:38
@Cihan: You are right, the annihilator of a module is a two-sided ideal. And I'm somewhat embarassed because I have difficulty correcting the answer. Given a maximal left ideal $I$, its annihilator of $R/I$ is the largest two-sided ideal that is contained in $I$. I'll admit that that is a maximal two-sided ideal, but how do I see it is an intersection of maximal left-ideals? I'd like to say the intersection of all conjugates of $I$, but that does not makes sense, I think. And without it, the "given explanation" doesn't explain all that much. – Marc van Leeuwen Apr 18 '12 at 17:11
$ann(R/I)$ may not be a maximal two-sided ideal even if $I$ is a maximal left ideal. There exist left primitive rings (rings which has a faithful simple left module) which are not simple. That is, there is a non-simple ring $R$ with a maximal left ideal $I$ such that $ann(R/I) = 0$. To see the annihilator of a simple left module as an intersection of maximal left ideals, observe that for any left $R$-module $M$ we have $$ ann(M) = \bigcap_{x \in M - \{ 0 \} }ann(x) $$ Now if $M$ is simple $ann(x)$'s above are maximal left ideals. – Cihan Apr 18 '12 at 21:21

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