6

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 $m\in \Bbb N$. But, $$\{0_R\}\subseteq M_1^m\dotsb M_n^m=(M_1\dotsb M_n)^m\subseteq \left(\bigcap_{i=1}^{n}M_i\right)^m=\mathrm{Jac}(R)^m=\{0_R\}.$$ 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?

Quay Chern
  • 697
Chris
  • 3,026
  • 1
    For (2): Assuming $R$ is commutative, which it seems like it is, each $R/M_i$ is a field. – Tim May 28 '19 at 18:57
  • @ Thanks, I fixed it. So $R/M_i^m$ is field, so it's a local ring with $\mathfrak{m}={0_{R/M_i^m}}={M_i^m}$? – Chris May 28 '19 at 19:05
  • Sorry I was answering your question (2). $R/M_i$ is a field since $M_i$ is a maximal ideal. Modding out by powers of a maximal ideal is likely not a field. – Tim May 28 '19 at 19:12
  • @Tim, I made a typo on powers and I fixed it. I apologize for that. – Chris May 28 '19 at 19:14
  • See https://math.stackexchange.com/questions/746111/is-quotient-of-a-ring-by-a-power-of-a-maximal-ideal-local – Tim May 28 '19 at 19:22
  • 1
    You have to show that $R/M_i^m$ are Artinian local rings. And if $m>1$ this is certainly not a field. – user26857 May 28 '19 at 20:11
  • @user26857 Thank you very much. Your answers are always to the point. Could you please explain why these rings are Artinian? – Chris May 28 '19 at 20:15
  • 1
    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. – user26857 May 28 '19 at 21:11
  • 1
    @user26857 Splendid! Thank you – Chris May 29 '19 at 11:29
  • 2
    May I know why $M_i^m+M_j^m=R$ so that we could use CRT in 2? – Jun Xu Feb 26 '21 at 06:21
  • @JunXu To show that $M_i^m+M_j^m=R$ for all $i\ne j$, you need the following lemma: let $R$ be a commutative ring with $1$ with comaximal (or called coprime) ideals $I$ and $J$ s.t. $I+J=R$. Then $I^n+J^m=R$ for any $n,m\in\mathbb{Z}_{\geqslant1}$. For the proof, suppose $i+j=1$ where $i\in I$ and $j\in J$. WLOG let $n\leqslant m$. To be continued. – Quay Chern Oct 21 '24 at 17:20
  • @JunXu Continue. Then consider the Binomial Theorem for $(i+j)^{n+m}=1^{n+m}=1$, and divide the expansion of $(i+j)^{n+m}$ into two parts: $\left(\sum_{0\leqslant k<n\atop\Rightarrow\ n-k>0}\mathrm{C}{n+m}^ki^kj^{n-k}\right)j^m=r_1j^m\in r_1J^m\subseteq J^m$ and $\left(\sum{n\leqslant k\leqslant n+m\atop\Rightarrow\ k-n\geqslant0}\mathrm{C}_{n+m}^ki^{k-n}j^{n+m-k}\right)i^n=r_2i^n\in r_2I^n\subseteq I^n$, so that $1=(i+j)^{n+m}\in J^m+I^n$, completing the proof. – Quay Chern Oct 21 '24 at 17:20

1 Answers1

3

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.

rschwieb
  • 160,592
  • Thank you for your answer! Could you please write down some counter examples as an application? And of courseI ll accept the answer. – Chris May 28 '19 at 19:58
  • @Chris There are no counterexamples to a true statement. What kind of counterexample are you looking for? – rschwieb May 28 '19 at 20:06
  • I am sorry, I misunderstood the word counter example. I mean some applications-examples to this theorem. – Chris May 28 '19 at 20:08
  • 1
    @Chris I provided one... i'm not sure what else one should say. – rschwieb May 29 '19 at 13:30
  • Just something like that I had in my mind. What about sth like polynomials in $F[X] $, where $F$ is a field? – Chris May 29 '19 at 14:38
  • 1
    @Chris When $F$ is a field, $F[X]$ is a nonfield domain, hence not Artinian. So your proposition does not apply. – rschwieb May 29 '19 at 14:58
  • The use of CRT requires that the ideals are comaximal, i.e., $M_i^m+M_j^m=R$ for all $i\ne j$. But this has not appeared in the proof, so why you say that his use of CRT is true? – Quay Chern Oct 20 '24 at 18:20
  • @QuayChern It is used implicitly to show the mapping is surjective onto the product. – rschwieb Oct 21 '24 at 00:17
  • Thanks for your replying! But I still have some doubt: how does the questioner obtain that $M_1^mM_2^m\cdots M_n^m\subseteq(M_1M_2\cdots M_n)^m$? It seems not that obvious, and I have posted a question about it: the product of some power of ideals is contained in some power of the product of ideals. Could you help me have a look at this? – Quay Chern Oct 21 '24 at 03:03
  • @QuayChern Consider the products of the generators like this: Think of the $m$ items from $M_i$ that you are multiplying as the $i$th row of an $n\times m$ matrix. If you multiply first elements in the rows, then multiply those results, you would be considering a product in $M_1^mM_2^m\ldots M_n^m$. If instead you multiplied the column elements together, then multiplied those results together, you'd be looking at a product in $(M_1M_2\dots M_n)^m$. – rschwieb Oct 21 '24 at 13:08
  • Thanks again! Following your comment, by considering the generators we may have $\prod_{i=1}^nM_i^m=\left.\left<\ \prod_{i=1}^n\left(\prod_{t=1}^mx_{i,t}\right)\ \right|\ x_{i,t}\in M_i\ \right>$ and $\left(\prod_{i=1}^nM_i\right)^m=\left.\left<\ \prod_{t=1}^m\left(\prod_{i=1}^nx_{i,t}\right)\ \right|\ x_{i,t}\in M_i\ \right>$, hence $\prod_{i=1}^nM_i^m=\left(\prod_{i=1}^nM_i\right)^m$. Is the process above right? ૮(˶ᵔ ᵕ ᵔ˶)ა – Quay Chern Oct 21 '24 at 14:35
  • 2
    @QuayChern I think you are over-concerning yourself with the equalities. Such a thing might be true, but it's actually much simpler just to note all the generators of $\prod M_i^m$ are contained in $(\prod M_i)^m$, and therefore the former is contained in the latter. That's all the OP used. Intersections of coprime ideals are equal to products, but if you want to prove that maybe you would establish the binary case and then do induction. – rschwieb Oct 21 '24 at 15:23