Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

2 Answers 2

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.

share|improve this answer

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.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.