$R^{\infty}$ does not have a finite spanning set

Question

Let $R^{\infty}$ be an $R$-module as the set of infinite sequences of elements of $R$, where $R$ is just a commutative ring.

I'm trying to show that $R$ does not have a finite spanning set, but the easy proofs I want rely on $R$ being a field.

I think I'm missing something easy, but I can't find a sequence that can't be written as a linear combination of a given finite subset of $R^{\infty}$.

Answer 1

Let $M$ be a maximal ideal of $R$. Then $V=R^{\infty}/M^{\infty} \cong (R/M)^{\infty}$, which is an infinite dimensional vector space over the field $R/M$. If $R^{\infty}$ was finitely generated, then so would $V$.