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On page 752 in Dummit&Foote's Abstract Algebra, 3rd-edition, it states that $$ \prod_{i=1}^nM_i^m\subseteq\left(\prod_{i=1}^nM_i\right)^m\overset{\text{CRT}}{=}\left(\bigcap_{i=1}^nM_i\right)^m=(\text{Jac}\;R)^m $$ where $M_1,\dots,M_n$ are all the distinct maximal ideals of the Artinian ring $R$ (commutative with $1$).

Now we have $$ (*)\begin{cases} \displaystyle\prod_{i=1}^nM_i^m=\left.\left\{\ \sum^{<\infty}_j\left(\prod_{i=1}^n\left(\sum_{s}^{<\infty}\left(\prod_{t=1}^mx_{i,t,s,j}\right)\right)\right)\ \right|\ x_{i,t,s,j}\in M_i\ \right\}. \\\displaystyle\left(\prod_{i=1}^nM_i\right)^m=\left.\left\{\ \sum^{<\infty}_s\left(\prod_{t=1}^m\left(\sum_{j}^{<\infty}\left(\prod_{i=1}^nx_{i,j,t,s}\right)\right)\right)\ \right|\ x_{i,j,t,s}\in M_i\ \right\}. \end{cases} $$ But how can we prove the containment?

Update: correct the expansion of $\prod_{i=1}^nM_i^m$ and $\left(\prod_{i=1}^nM_i\right)^m$.

Update: thank rschwieb for reminding me to consider the generators of the two ideals above:

by expanding the sums and products in $(*)$ we will obtain that $$ \begin{cases} \displaystyle\prod_{i=1}^nM_i^m=\left.\left<\ \prod_{i=1}^n\left(\prod_{t=1}^mx_{i,t}\right)\ \right|\ x_{i,t}\in M_i\ \right>. \\\displaystyle\left(\prod_{i=1}^nM_i\right)^m=\left.\left<\ \prod_{t=1}^m\left(\prod_{i=1}^nx_{i,t}\right)\ \right|\ x_{i,t}\in M_i\ \right>. \end{cases} $$ Hence $\prod_{i=1}^nM_i^m=\left(\prod_{i=1}^nM_i\right)^m$.

Update: Moreover, in order to show that $R\cong R/0=R/\left(\prod_{i\in[1,n]}^{\text{ideals}}M_i^m\right)\underset{\text{CRT}}{\cong}\prod_{i\in[1,n]}^{\times}\left(R/M_i^m\right)$ by the Chinese Remainder Theorem, we need to show that $M_i^m+M_j^m=R$ for all $i\ne j$, by 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 $n\leqslant m$. 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
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    Both equalities you give are incorrect: no power of $m$ should appear on the right-hand side. For instance $I^2$ is not the sum of elements of the form $x^2$ with $x\in I$, but rather the sum of elements of the form $xy$ with $x,y\in I$. – Captain Lama Oct 20 '24 at 16:06
  • @CaptainLama Thanks for pointing out the mistake I had made, and I have corrected it. – Quay Chern Oct 20 '24 at 17:42
  • The Lemma on powers being coprime is already proved here in many places, e.g. here. Please clarify what remains for the question after all the updates. – Bill Dubuque Oct 21 '24 at 18:17
  • @BillDubuque thanks, I had completed the proof of this question after all the updates. – Quay Chern Oct 22 '24 at 03:23

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