Let $R$ be any ring with unity (may be non-commutative).

Let $J(R)$ denote the Jacobson radical of $R$: intersection of all maximal left ideals.

Let $M$ be a left $R$-module, and ${\rm Rad}(M)$ the intersection of all maximal submodules of $M$.

Claim: If $M$ is finitely generated, then $J(R)M={\rm Rad}(M)$.

Question: It is easy to prove the claim if $M$ is irreducible (i.e. simple) $R$-module since both terms will be $0$. Then, can we prove the claim from this special case to the general one i.e. when $M$ is finitely generated $R$-module?

Remark: The claim mentioned is given as a Proposition with proof in Representations and Cohomology: Volume 1 by Benson. When I tried to prove the claim myself, I was able to prove it for simple modules; but I was not getting direction for general modules (finitely generated).

  • $\begingroup$ No condition that $R$ be artinian? That seems to be a usual condition in representation theory, but maybe not all the time. $\endgroup$
    – rschwieb
    Commented Aug 26, 2021 at 19:10
  • $\begingroup$ @rschwieb: The statement in Claim is from Representations and Cohomology, Vol. 1 by Benson (Prop. 1.2.5, Page 4); I nowhere saw there that $R$ is Artinian. $\endgroup$ Commented Aug 27, 2021 at 4:05
  • 1
    $\begingroup$ I’ve added to my answer pointing out that 1.2.5 explicitly relies on 1.2.4 which requires the ring to be left Artinian. $\endgroup$
    – rschwieb
    Commented Aug 27, 2021 at 7:28

1 Answer 1


I think you must be missing a hypothesis because of counterexamples like the one I'm about to give. (At the time of writing, the hypotheses are "any ring with identity (including noncommutative), and a finitely generated module over that ring.")

Let $R=\mathbb Z$ and $M=\mathbb Z/4\mathbb Z$, which is a cyclic $\mathbb Z$ module.

Of course $J(R)=\{0\}$ so that $J(R)M=\{0+4\mathbb Z\}$, but $J(M)$ is pretty clearly $\frac{2\mathbb Z}{4\mathbb Z}$ right?

One can easily see why $J(R)M\subseteq J(M)$ for any ring $R$ with module $M$, though.

If additionally $R/J(R)$ is Artinian then

  • $R/J(R)$ and all its modules are semisimple
  • that means $M/J(R)M$ is a semisimple $R/J(R)$ module and hence a semisimple $R$ module
  • $J(M)$ is proper in $M$ since $M$ is f.g.
  • $J(M)$ is then clearly a superfluous $R$ submodule of $M$, which means that the only complement to $J(M)/J(R)M$ could only be $M/J(R)M$, and so $J(M)/J(R)M$ is the zero module
  • that last line implies $J(M)=J(R)M$.

Regarding your assertion that the book does not mention Artinian: the proof given at the page you cite explicitly relies on lemma 1.2.4, which is only about (Jacobson-)semisimple rings satisfying the DCC on left ideals (which just means they are left Artinian rings.)

You’ll note that the proof as given for 1.2.5 does not work for the example I gave above if you interpret it as applying to just any ring $R$. The proof claims $M/J(R)M$ is completely reducible, but it isn’t for the $R$ and $M$ I gave.

If you interpret it as requiring $R/J(R)$ has to be Artinian, then the argument they list there (using the DCC) is basically the same as my proof above. (They should have been more clear why there is a maximal submodule, too.)

It looks like an early failure of exposition clarity. This is not uncommon when the author is dashing through basics in order to “get to the good stuff.”

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    $\begingroup$ The book by Benson is a canonical example of dashing through — through basics and most other things :-) $\endgroup$ Commented Oct 8, 2022 at 6:13

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