An artinian ring is a product of local rings

Question

I am rather confused with the last line of the argument used in 00JA, Stacks Project.

Lemma 00JA. Any ring with finitely many maximal ideals and locally nilpotent Jacobson radical is the product of its localizations at its maximal ideals. Also, all primes are maximal.

Proof. Let $R$ be a ring with finitely many maximal ideals $\mathfrak m_1, \ldots, \mathfrak m_n$. Let $I = \bigcap_{i = 1}^n \mathfrak m_i$ be the Jacobson radical of $R$. Assume $I$ is locally nilpotent. Let $\mathfrak p$ be a prime ideal of $R$. Since every prime contains every nilpotent element of $R$ we see $ \mathfrak p \supset \mathfrak m_1 \cap \ldots \cap \mathfrak m_n$. Since $\mathfrak m_1 \cap \ldots \cap \mathfrak m_n \supset \mathfrak m_1 \ldots \mathfrak m_n$ we conclude $\mathfrak p \supset \mathfrak m_1 \ldots \mathfrak m_n$. Hence $\mathfrak p \supset \mathfrak m_i$ for some $i$, and so $\mathfrak p = \mathfrak m_i$. By the Chinese remainder theorem (Lemma 00DT) we have $R/I \cong \bigoplus R/\mathfrak m_i$ which is a product of fields. Hence by Lemma 00J9 there are idempotents $e_i$, $i = 1, \ldots, n$ with $e_i \bmod \mathfrak m_j = \delta_{ij}$. Hence $R = \prod Re_i$, and each $Re_i$ is a ring with exactly one maximal ideal. $\square$

1. How is the "Hence $R=\prod Re_i$..." deduction made?

2. Why does $Re_i$ have exactly one maximal ideal?

3. Relating back to the statement, how is this a localization at the maximal ideals?

Answer 1

An important thing that is kind of suppressed is that the idempotents form a complete orthogonal family of idempotents, i.e. they satisfy $$e_ie_j=\delta_{ij}e_j \;\;\;\text{ and }\;\;\;e_1+e_2+\dots+e_n=1.$$

To see the orthogonality, note that for $i \neq j$ we in fact have $e_ie_j \in \mathfrak{m}_1 \cap \mathfrak{m}_2 \cap \dots \cap \mathfrak{m}_n=I$. Since $I$ is locally nilpotent, $(e_ie_j)^N=0$ for some $N$. But $(e_ie_j)^N=e_ie_j$ as these are idempotent. So $e_ie_j=0$.

To see the completeness, note that $e:=e_1+e_2+\dots+e_n$ is again an idempotent (thanks to orthogonality above) and we have $e\equiv 1 \pmod{\mathfrak{m}_i}$ for all $i$. So $1-e$ is again an idempotent (one can check) and, by the same argument as above, it is contained in $I$, hence nilpotent, hence $1-e=0$. That is, $e_1+e_2+\dots+e_n=1$, as desired.

Now,

1. Any element $x \in R$ is uniquely written as a sum of elements from $Re_i$'s, $x=x(e_1+e_2+\dots +e_n)=xe_1+xe_2+\dots+xe_n$. The fact that $e_i$ is an idempotent means that $Re_i$ gets a ring structure whose unit element is $e_i$. The correspondence $x \leftrightsquigarrow (xe_1, \dots, xe_n)$ then defines the isomorphism $R \simeq \prod_i Re_i$. (Checking that this correspondence is a ring homomorphism will involve the orthogonality.)

2. Each of the rings $Re_j$ will have at least one maximal ideal, pick one and call it $M_j$. Then clearly $$\mathfrak{M}_j:= Re_1 \times Re_2 \times \dots \times Re_{j-1} \times M_j \times Re_{j+1} \times \dots \times Re_n$$ is a maximal ideal of $\prod_iRe_i$, hence it corresponds to a unique maximal ideal of $R$. But there are only $n$ of those, so by the pigenhole principle, each $Re_j$ has only the one maximal ideal $M_j$.

3. To see that $R$ is the product of its localizations, it is enough to see that the same is true for $\prod_iRe_i$. Well, in this case it is easy to see that the localization of $\prod_iRe_i$ at $\mathfrak{M}_j$ is naturally isomorphic to $Re_j$ (meaning that the localization map is just the projection onto the $j$th component). So the claim is true for $\prod_iRe_i,$ hence for $R$ as well.