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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

  1. $R/\mathrm{Jac}(R)\cong (R/M_1)\times \dotsb \times (R/M_n)$.
  2. 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.

  1. Since $R$ is an Artinian ring, we have $\mathrm{Jac}(R)^m=\{0_R\}$, for some $m\in \Bbb Z^+$. 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 $R/M_i^m$ are artinian local rings?

3) Could you please elaborate some examples as an application?

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    $\begingroup$ For (2): Assuming $R$ is commutative, which it seems like it is, each $R/M_i$ is a field. $\endgroup$ – Tim May 28 at 18:57
  • $\begingroup$ @ Thanks, I fixed it. So $R/M_i^m$ is field, so it's a local ring with $\mathfrak{m}=\{0_{R/M_i^m}\}=\{M_i^m\}$? $\endgroup$ – Chris May 28 at 19:05
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    $\begingroup$ You have to show that $R/M_i^m$ are Artinian local rings. And if $m>1$ this is certainly not a field. $\endgroup$ – user26857 May 28 at 20:11
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    $\begingroup$ Every quotient of an Artinian ring is Artinian. Probably you wanted to know why these are local rings. The (prime) ideals of $R/M^k$ are of the form $P/M^k$ with $P\supseteq M^k$. Now if $P$ is prime we get $P\supseteq M$ and since $M$ is maximal we get $P=M$, so there is only one prime (hence maximal) ideal in our ring. $\endgroup$ – user26857 May 28 at 21:11
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    $\begingroup$ @user26857 Splendid! Thank you $\endgroup$ – Chris May 29 at 11:29
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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.

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  • $\begingroup$ Thank you for your answer! Could you please write down some counter examples as an application? And of courseI ll accept the answer. $\endgroup$ – Chris May 28 at 19:58
  • $\begingroup$ @Chris There are no counterexamples to a true statement. What kind of counterexample are you looking for? $\endgroup$ – rschwieb May 28 at 20:06
  • $\begingroup$ I am sorry, I misunderstood the word counter example. I mean some applications-examples to this theorem. $\endgroup$ – Chris May 28 at 20:08
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    $\begingroup$ @Chris I provided one... i'm not sure what else one should say. $\endgroup$ – rschwieb May 29 at 13:30
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    $\begingroup$ @Chris When $F$ is a field, $F[X]$ is a nonfield domain, hence not Artinian. So your proposition does not apply. $\endgroup$ – rschwieb May 29 at 14:58

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