# Nonexistence of a vector space isomorphism

I feel that the $\mathbf{Q}$ vector spaces $\prod_{n=0}^\infty \mathbf{Q}$ and $(\mathbf{Z}-0)^{-1}\prod_{n=0}^\infty\mathbf{Z}$ are not isomorphic, what is the quickest way to demonstrate it? By a cardinality of basis argument?

• what is the vector space structure on the second of the two sets? in fact, what is the second of the two sets? Jan 16 '14 at 23:24
• @IttayWeiss Right. I misread the question. Deleted my comment. Jan 16 '14 at 23:39
• @IttayWeiss: it is the localization at $\mathbf{Z}-0$, which is a $\mathbf{Q}=(\mathbf{Z}-0)^{-1}(\mathbf{Z})$ module in a natural manner; the other is just the product vector space
– user88576
Jan 16 '14 at 23:40

If a $\mathbf{Q}$-vector space $V$ has an infinite basis $I$, then $V$ and $I$ have same cardinality. ($V$ consists, more or less, of all sequences of rationals indexed by a finite subset of $I$.)
• That's not what I mean. I used Cantor-Bernstein for sets only. What I mean is the following conjecture: If we have an explicit linear embedding $f$ of $V$ into $W$, and an explicit linear embedding $g$ of $W$ into $V$, then we can define an explicit isomorphism $h$ between $V$ and $W$. Are you saying that this conjecture is true? The proof I know uses the fact that every vector space has a basis, and wouldn't result in an explicit description of $h$. I mean, for example, I don't know of any particular basis for the vector spaces above, so the proof I gave doesn't yield an explicit isomorphism. Jan 17 '14 at 0:44