Basis of dual space involving free Z-module

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.

Re: Basis of dual space involving free Z-module

Quote:

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 (or what you're asking is basically equivalent to this) where is some infinite cardinal and denotes the dual module. I think the easiest way to do this is to recall the way Hom reacts with coproducts so that

So, what you really want to prove is that for an infinite cardinal one has that . But, since is countable this is trivial for a stupid reason-- and aren't equipotent. Indeed, it's trivial from Cantor's theorem that . That said, since is an infinite cardinal--I leave this to you to work out.

Re: Basis of dual space involving free Z-module

That makes perfect sense and yes, that was what I was asking. My apologies for the exposition. Thank you so very much!