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In my version of Noncommutative Algebra, by Benson Farb & R. Keith Dennis, in chapter I, section 2 on the Jacobson radical, it is claimed that

… each maximal left ideal $I$ is the annihilator of some simple $R$-module (namely, $R/I$). This is true because, as stated in Proposition 1.1, simple $R$-modules are precisely $R/I$ for maximal left ideals $I$.

So I believe this is wrong, else every left maximal ideal would be two-sided by the observation far below. Am I missing something?

How else can we see that for the Jacobson radical as defined by $$J(R) := \bigcap_{\text{simple left $R$-modules $M$}} \operatorname{Ann}_R (M)$$ we have $J(R) = \bigcap_{\text{maximal left ideals $I ⊂ R$}} I$, which was the conclusion of this statement?

Here is why I don’t think this is true: Let $R$ be a ring with identity and $M$ a left $R$-module.

Then $\operatorname{Ann}_R(M)$ is a two-sided ideal, because $∀r ∈ R,\, x ∈ \operatorname{Ann}_R (M)\colon$ $$0 ⊂ (xr)M = x(rM) ⊂ xM = 0 \quad \text{and} \quad (rx)M = r(xM) = r0 = 0,$$ and, of course, $\operatorname{Ann}_R(M)$ as additively closed.

And so any maximal left ideal $I ⊂ R$ would be a two-sided ideal, if this was true.

Also, I think that $R = \mathrm{M}_{n×n}(ℤ)$ and $I = \{\big[\begin{smallmatrix} a & 0 \\ b & 0\end{smallmatrix}\big];\; a,b ∈ ℤ\}$ would yield a counterexample, because $\big[\begin{smallmatrix} 1 & 0 \\ 0 & 0\end{smallmatrix}\big]$ doesn’t annihilate $R/I$.

k.stm
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    Every maximal left ideal is the annihilator of an element of a simple module. Your considerations are correct. – egreg Mar 18 '14 at 10:51
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    Wow, if it's printed that way, that's a pretty egregious mistake! (No pun intended, egreg :) your answer addresses that mistake really well!) It's true that every maximal ideal $M$ is the annihilator of a simple module because for any maximal left ideal $L$ containing $M$, $R/L$ is simple with annihilator $M$. – rschwieb Mar 18 '14 at 12:55
  • @rschwieb Huh! I missed this comment, (no @ in it). Yes, every maximal ideal is the annihilator of a simple module, but the annihilator of a simple module need not be a maximal ideal: that's where primitive ideals kick in. – egreg Oct 31 '19 at 22:36
  • @egreg didn’t put an @ because the contents weren’t directed at you (just the joke was) – rschwieb Oct 31 '19 at 23:22

1 Answers1

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The statement that a maximal left ideal is the annihilator of a simple module is incorrect: indeed the annihilator of a module is a two-sided ideal, as you remark, so any maximal left ideal which is not two-sided is a counterexample. Matrix rings provide easy examples of maximal left ideals which are not two-sided.

On the other hand it is true that the intersection of all primitive ideals (that is, annihilators of simple modules) is the same as the intersection of all maximal left ideals.

Indeed, any maximal left ideal $I$ is the annihilator of some element of some simple module (in particular, of $1+I\in R/I$). Therefore any maximal left ideal contains a primitive ideal and so $$ \bigcap\{\text{primitive ideals}\}\subseteq\bigcap\{\text{maximal left ideals}\}. $$ Conversely, any primitive ideal is the intersection of maximal left ideals, so we get the converse inclusion: indeed, if $I$ is the annihilator of a simple module $S$, then $$ I=\bigcap_{x\in S}\operatorname{Ann}_R(x) $$ and $\operatorname{Ann}_R(x)$ is either $R$ or a maximal left ideal.

egreg
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  • Ah, of course. Any primitive ideal is the intersection of maximal left ideals. That’s it. Thanks! – k.stm Mar 18 '14 at 11:15