关于含幺交换阿廷环中理想幂乘积与乘积幂的包含关系验证问询

Hey there, let's break down this inclusion step by step for commutative Artinian rings with unity—this is a great question about unpacking the nitty-gritty of ideal product and power definitions!

First, let's recap the context from Dummit & Foote's Abstract Algebra (3rd ed., p.752): For a commutative Artinian ring $R$ with unity (whose distinct maximal ideals are $M_1, \dots, M_n$), we have the chain:
$$
\prod_{i=1}^nM_im\subseteq\left(\prod_{i=1}^nM_i\right)m\overset{\text{CRT}}{=}\left(\bigcap_{i=1}^nM_i\right)m=(\text{Jac};R)^m
$$
Your core question is how to verify the left-to-right inclusion using the explicit set representations you've written (labeled $(*)$). Let's unpack this.

First, restate the two ideal definitions clearly #

From your set descriptions:

1. $\prod_{i=1}^nM_im$: This is the product of the $m$-th powers of each maximal ideal $M_i$. By ideal power/product rules, it's the set of finite sums of products where each factor is a finite sum of $m$-fold products from $M_i$.
2. $\left(\prod_{i=1}^nM_i\right)m$: This is the $m$-th power of the product of all maximal ideals. It's the set of finite sums of $m$-fold products, where each factor is a finite sum of products of elements from each $M_i$.

Proving the inclusion $\prod_{i=1}^nM_im \subseteq \left(\prod_{i=1}^nM_i\right)m$ #

We can use the fact that ideals are closed under finite sums, so we only need to show every generating element of the left ideal is in the right ideal, then extend to all sums.

Let's walk through the key steps:

1. Pick a generating element of $\prod_{i=1}^nM_im$:
   Take an element of the form $\prod_{i=1}^n a_i$, where $a_i \in M_i^m$. By definition of $M_i^m$, each $a_i$ is a finite sum of $m$-fold products from $M_i$:
   $$a_i = \sum_{s=1}^{k_i} \prod_{t=1}^m x_{i,t,s} \quad \text{where } x_{i,t,s} \in M_i$$

2. Expand the product:
   Compute $\prod_{i=1}^n a_i = \prod_{i=1}^n \left( \sum_{s=1}^{k_i} \prod_{t=1}^m x_{i,t,s} \right)$. Using the distributive property of ring multiplication, this expands to a finite sum of terms of the form:
   $$\prod_{i=1}^n \left( \prod_{t=1}^m x_{i,t,s_i} \right)$$
   where $s_i$ is an index from the sum for $a_i$.

3. Rearrange the product to match the right ideal's form:
   Rewrite the term above by grouping the $t$-th positions across all $M_i$:
   $$\prod_{t=1}^m \left( \prod_{i=1}^n x_{i,t,s_i} \right)$$
   Notice that $\prod_{i=1}^n x_{i,t,s_i} \in \prod_{i=1}^n M_i$ (since it's a product of one element from each $M_i$). This means the entire term is a $m$-fold product of elements from $\prod_{i=1}^n M_i$—which is exactly a generating element of $\left(\prod_{i=1}^nM_i\right)m$.

4. Extend to all elements of the left ideal:
   Since $\prod_{i=1}^n a_i$ is a finite sum of such generating elements of the right ideal, and the right ideal is closed under finite sums (it's an ideal), $\prod_{i=1}^n a_i \in \left(\prod_{i=1}^nM_i\right)m$.

   Every element of $\prod_{i=1}^nM_im$ is a finite sum of such $\prod_{i=1}^n a_i$ terms, so the entire left ideal is contained in the right ideal.

A concrete small case to make it tangible #

Let's take $n=2$, $m=2$ to see it in action:

• Let $a_1 = x_{1,1}x_{1,2} + y_{1,1}y_{1,2} \in M_1^2$ (all $x,y \in M_1$)
• Let $a_2 = x_{2,1}x_{2,2} + y_{2,1}y_{2,2} \in M_2^2$ (all $x,y \in M_2$)
• Their product:
  $$a_1a_2 = (x_{1,1}x_{1,2})(x_{2,1}x_{2,2}) + (x_{1,1}x_{1,2})(y_{2,1}y_{2,2}) + (y_{1,1}y_{1,2})(x_{2,1}x_{2,2}) + (y_{1,1}y_{1,2})(y_{2,1}y_{2,2})$$
• Rearrange each term:
  • $(x_{1,1}x_{2,1})(x_{1,2}x_{2,2}) \in (M_1M_2)^2$ (since $x_{1,1}x_{2,1} \in M_1M_2$ and $x_{1,2}x_{2,2} \in M_1M_2$)
  • All other terms follow the same pattern, so their sum is in $(M_1M_2)^2$.

Key takeaway #

The inclusion boils down to distributivity of multiplication over addition and the basic closure properties of ideals. We can rearrange the products from the left ideal to fit exactly the form of elements in the right ideal, and since ideals are closed under finite sums, the inclusion holds.

备注：内容来源于stack exchange，提问作者Quay Chern