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I have tried for some time now to prove the following statement from an exercise, and now I wonder if it is even correct:


Let $A$ be a ring and $E$ a left $A$-module. For a left ideal $\mathfrak{a}$ of $A$, $\mathfrak{a}E$ is defined as the submodule $\sum_{x\in E}\mathfrak{a}x$ of $E$. If $\mathfrak{a}$ is a finitely generated left ideal, show that, for every family $(E_\lambda)_{\lambda\in L}$ of modules, $$\mathfrak{a}\cdot\prod_{\lambda\in L}E_\lambda=\prod_{\lambda\in L}\mathfrak{a}E_\lambda.$$


"$\subset$" is true for any ideal. Let $a_1,\ldots,a_n$ generate $\mathfrak{a}$. For "$\supset$", let $x=(x_\lambda)\in\prod_{\lambda\in L}\mathfrak{a}E_\lambda$. For every $\lambda$, there exist $k_\lambda\in\mathbb{N}$ and $x_\lambda^1,\ldots,x_\lambda^{k_\lambda}\in E_\lambda$ such that $$x_\lambda\in\mathfrak{a}x_\lambda^1+\ldots+\mathfrak{a}x_\lambda^{k_\lambda}=\mathfrak{a}x_\lambda^1+\ldots+\mathfrak{a}x_\lambda^{k_\lambda}=(Aa_1+\ldots+Aa_n)x_\lambda^1+\ldots+(Aa_1+\ldots+Aa_n)x_\lambda^{k_\lambda}.$$ But now what? What I would like to see here is $$x_\lambda\in(a_1A+\ldots+a_nA)x_\lambda^1+\ldots+(a_1A+\ldots+a_nA)x_\lambda^{k_\lambda}\subset a_1E_\lambda+\ldots+a_nE_\lambda,$$ which implies $x\in a_1\prod_{\lambda\in L}E_\lambda+\ldots+a_n\prod_{\lambda\in L}E_\lambda\subset\mathfrak{a}\cdot\prod_{\lambda\in L}E_\lambda.$

Hence the proposition is true if $\mathfrak{a}$ is a two-sided ideal that is finitely generated as a right ideal.

I've played around a bit with $2\times 2$-matrix rings, which have left ideals that aren't right ideals. But neither could I find a counterexample, nor did I get any intuition for the general case.


Can someone help me along?

2 Answers 2