How to prove this ideal is not locally free?

Question

Let $I$ be an infinite set, $k$ a field, let $A=k^I$ (the product over the index set $I$), and let $\mathfrak a=k^{(I)}$ (the direct sum over the index set $I$). Then how can we prove $\mathfrak a$ is not a locally free module over $A$?

Since if it is locally free, then in some $D(f)$ we can prove $\mathfrak a_f=0$ or $\mathfrak a_f=A_f$. I want to prove $A_f \ncong a_f$ for any $f\in A$. But then I have no idea...

Answer 1

Suppose the contrary. Then $\mathfrak a$ is finite locally free since $\mathfrak a_f\subseteq A_f$. But finite locally free implies finite (see here). It is easily shown that $\mathfrak a$ is not finite.