Proposition. Let $R$ be commutative ring with $1_R$. We assume that $R$ is an Artinian ring and $M_1,\dots,M_n$ its maximal ideals. Then
- $R/\mathrm{Jac}(R)\cong (R/M_1)\times \dotsb \times (R/M_n)$.
- The ring $R$ is isomorphic to the direct product of a finite number of Artinian local rings.
Proof. 1. It's obvious that $M_i+M_j=R,\ \forall 1\leq i \neq j \leq n$. So, from Chinese Remainder Theorem, we have
$$R/\bigcap_{i=1}^{n}M_i = R/\mathrm{Jac}(R) \cong (R/M_1)\times \dotsb \times (R/M_n),$$
as we wanted.
- Since $R$ is an Artinian ring, we have $\mathrm{Jac}(R)^m=\{0_R\}$, for some $m\in \Bbb N$. But,
$$\{0_R\}\subseteq M_1^m\dotsb M_n^m=(M_1\dotsb M_n)^m\subseteq \bigcap_{i=1}^{n}M_i=\mathrm{Jac}(R)^m=\{0_R\}.$$
So, if we apply CRT we will take
$$R\cong R/\{0_R\}\cong R/M_1^m\dotsb M_n^m\cong (R/M_1^m)\times \dots \times (R/M_n^m).$$
Questions.
1) Are these thoughts complete and correct?
2) Why are $R/M_i^m$ artinian local rings?
3) Could you please elaborate on some examples as an application?
Best Answer
Yes, your use of the CRT is OK.
It is not hard to show that $R/M^k$ is a local ring for any natural number $k$ and maximal ideal $M$.
The pieces of the ring are Artinian because they are all homomorphic images of the original ring (you just factor out the ideal that is the complement of the factor you're interested in.)
One application of this theorem is that if $n=\prod_{i\in I} p_i^{e_i}$ where $I$ is a finite index set, $p_i$ are distinct primes, and $e_i$ are positive integral exponents, then $\mathbb Z/n\mathbb Z\cong \prod_{i\in I}\mathbb Z/p_i^{e_i}\mathbb Z$.
Another practical corollary is that there are finitely many maximal ideals in such a ring (one for each local summand.)
I'm not sure what other applications you are looking for... It is already quite nice that the ring decomposes into "nicer" ones.