The Stacks project

1. Suggestion: Write $J \subsetneqq J'$ instead of $J' \neq J, J \subset J'$.
2. We use the fact, that the set ${J' \subset R ideal | J \subset J'}$ has a minimal element. This seems similar to the analogous statement for noetherian rings. Should this be proved?

Dear Jonas, your first suggestion is a style question which I am going to ignore. For your second point: a nonempty ordered set with dcc has a minimal element.

A more general remark is that the Stacks project isn't exactly building everything from scratch. The reader should have some basic knowledge of ring theory, fields, topology, set theory, combinatorics, etc. For example the discussion on Noetherian rings simply states some properties and then starts to do more interesting stuff. This makes the choices we've made in the exposition of earlier stuff in the Stacks project a bit arbitrary at times (the proof here included), but I don't see how to avoid this.

How is nakayama's lemma applied here? It's not clear to me

Add a few words here.

The reference to Lemma 00DV (Nakayama's lemma) in the proof here strikes me as a red herring. I think(!) one can excise it from the proof at zero cost:

Lemma 00J8. Let $R$ be Artinian. The Jacobson radical of $R$ is a nilpotent ideal.

Proof. Let $I \subset R$ be the Jacobson radical. Note that $I \supset I^2 \supset I^3 \supset \ldots$ is a descending sequence. Thus $I^n = I^{n + 1}$ for some $n$. Set $J = \\{ x\in R \mid xI^n = 0\\}$. We have to show $J = R$. Suppose otherwise; by the Artinian property $J$ then has a minimal proper superideal $J'$. In particular, $J'/J$ is simple, so by Lemma 00J2 isomorphic to $R/\mathfrak{m}$ for some maximal $\mathfrak{m} \subset R$. We conclude that $$J \subset J' \subset \\{ x\in R \mid x\mathfrak{m}I^n = 0\\} \subset \\{ x\in R \mid xI^{n+1} = 0\\} = J$$ whence $J=J'$, contradiction. $\Box$

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P.s. (feel free to ignore what follows unless something is wrong!):

I was initially a little baffled what the motivation could be for considering $I^n$'s annihilator. Perhaps this is banal/well-known, but it strikes me that the annihilators of products of maximal ideals of $R$ are precisely what arise in succession as one attempts to iteratively "build-up" $R$ from $(0) \subset R$ via those elements currently minimal above the current ideal with some particular (maximal) denominator $\mathfrak{m}$ (just as one built $J'$ upon $J$ in the above proof).

I.e., having by Lemma 00J7 that $R$ has finitely many maximal ideals $\mathfrak{m}_ 0 , \dots , \mathfrak{m}_ {k-1}$, this view leads us naturally to the observation that $\text{Ann}(\prod_{i=0}^{k-1}\mathfrak{m}_ i^{d_ i}) = R$ (as the former cannot have a minimal proper superideal) where $d_ i$ is the minimal exponent for which $\mathfrak{m}_ i^{d_ i+1}=\mathfrak{m}_ i^{d_ i}$; this is essentially the statement/proof of Lemma 00J8 above.

Moreover the quotients of the obvious (size-$\sum_{i=0}^{k-1} d_ i$) filtration of $\text{Ann}(\prod_{i=0}^{k-1}\mathfrak{m}_ i^{d_ i})$ are each $R/\mathfrak{m}_ i$-modules for some suitable (variable) choice of $i$, and as each must have finite vector space dimension in order to maintain the Artinian property of $R$ we see that $R$ must have finite length over itself; this is essentially (half) the statement/proof of Lemma 00JB.