# When does multiplication with an ideal commute with the product of modules?

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?

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It seemed fishy but I could not completely verify that my idea for a counterexample worked. I'll put it here and see if it helps you.

Let $R$ be a simple domain with a nontrivial cyclic left ideal $\mathfrak{a}=Ra$. I'm proposing we use $E=\prod_{i=1}^\infty R$ as our module.

Since $R$ is simple $\mathfrak{a}R=R$, and so $\prod_{i=1}^\infty\mathfrak{a}R=E$.

I am really having trouble believing $\mathfrak{a}\prod_{i=1}^\infty R$ is also equal to $E$, but I suppose it's possible. Subjectively it just seems like you cannot get the variety of coefficients on the left of elements of $E$ that you need.

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Here is a counterexample: Let $K$ be a field and let $A=K\langle X,Y\rangle$ be the free algebra in two variables over $K$. Let $\mathfrak{a}$ be the left ideal generated by $Y$, then $(Y,XY,X^2Y,X^3Y,\ldots) \in \prod_{i=0}^{\infty}\mathfrak{a}A$, but it's not contained in $\mathfrak{a}\prod_{i=0}^{\infty}A$ as one easily checks.

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