Posted by Jason Polak on 05. July 2011 · Write a comment · Categories: commutative-algebra

It’s no secret that all sorts of neat theorems and ideas go haywire upon the dropping of their finite generation hypotheses. A paltry but illustrative example is that an injective linear map $ T:V\to V$ on a finite dimensional vector space $ V$ is also surjective. From a counterexample point of view this pretty proposition isn’t terribly interesting.

Let us then make a short journey to the land of modules over arbitrary rings, and consider the trite and trivial observation that if $ M$ is a nontrivial module over some ring, then $ 0M\not= M$. This is hardly worth mention, and any capable mathematician would be wont to want more. Luckily there’s a generalization called Nakayama’s Lemma, which aside from being immensely useful, it has a short and entertaining proof.

Nakayama’s lemma, in one form, states that if $ M$ is a finitely generated left module over a ring $ R$ and $ \mathrm{Jac}(R)M = M$ then $ M = 0$. Note that we have not assumed that $ R$ is commutative. The proof is a fun exercise, and I shall omit it, as most readers have probably already seen it anyway.

The natural thing to do now is give a counterexample to the statement of the theorem when we deprive it of ‘finitely generated’ It’s useless to try $ \mathbb{Z}$ modules, since the Jacobson radical of the integers is zero. So, why not make life simple and take a local ring? The natural choice now is to localize $ \mathbb{Z}$ over some prime ideal $ p$. Now we have our ring $ \mathbb{Z}_{(p)}$ which has a single maximal ideal, the image of $ (p)$, and hence a nontrivial Jacobson radical.

We need only find our module. But this is easy part: the module in $ \mathbb{Q}$ with action by the usual multiplication. For any nonzero subset $ S\subseteq \mathbb{Z}_{(p)}$, such as our maximal ideal, we get $ S\mathbb{Q} = \mathbb{Q}$. Ah, divisibility!

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