Originally Posted by

**empyreandance** Hello friends,

I am truly stuck on this problem. Any guidance would be supremely appreciated.

Let F=Sum over x in X of **Z**x be a free **Z**-module with an infinite basis X. Then {f_x : x in X} does not form a basis of F*.

Relevant notes:

f_x is a map F -> Z given by f_x(y) = d_xy, where d is the Kronecker delta

Hints given: F* is isomorphic to the direct product over x in X **Z**x, but under this isomorphism, f_y -> {d_xy * x}, which is in the direct product over x in X of **Z**x.

Hey man, it's a little hard to decipher precisely what's going on here, but I'm guessing you're supposed to show that $\displaystyle \left(\mathbb{Z}^{\oplus\lambda}\right)^\vee \not\cong \mathbb{Z}^{\oplus\lambda}$ (or what you're asking is basically equivalent to this) where $\displaystyle \lambda$ is some infinite cardinal and $\displaystyle ^\vee$ denotes the dual module. I think the easiest way to do this is to recall the way Hom reacts with coproducts so that

$\displaystyle \left(\mathbb{Z}^{\oplus\lambda}\right)^\vee=\text {Hom}_\mathbb{Z}(\mathbb{Z}^{\oplus\lambda}, \mathbb{Z})\cong \left(\text{Hom}_\mathbb{Z}(\mathbb{Z},\mathbb{Z}) )^\lambda\cong \mathbb{Z}^{\lambda}$

So, what you really want to prove is that for an infinite cardinal one has that $\displaystyle \mathbb{Z}^{\oplus\lambda} \not\cong \mathbb{Z}^\lambda$. But, since $\displaystyle \mathbb{Z}$ is countable this is trivial for a stupid reason--$\displaystyle \mathbb{Z}^\lambda$ and $\displaystyle \mathbb{Z}^{\oplus\lambda}$ aren't equipotent. Indeed, it's trivial from Cantor's theorem that $\displaystyle |\mathbb{Z}^\lambda|\geqslant |2^\lambda|>\lambda$. That said, $\displaystyle |\mathbb{Z}^{\oplus\lambda}|=\lambda$ since $\displaystyle \lambda$ is an infinite cardinal--I leave this to you to work out.