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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?

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  • $\begingroup$ what is the vector space structure on the second of the two sets? in fact, what is the second of the two sets? $\endgroup$ Jan 16 '14 at 23:24
  • $\begingroup$ @IttayWeiss Right. I misread the question. Deleted my comment. $\endgroup$ Jan 16 '14 at 23:39
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    $\begingroup$ @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 $\endgroup$
    – user88576
    Jan 16 '14 at 23:40
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The two vector spaces are certainly isomorphic (though I haven't constructed an isomorphism).

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 $I$.)

Since the two vector spaces you wrote both have the cardinality of the continuum, any bases for them must also have the cardinality of the continuum. Thus they are isomorphic.

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  • $\begingroup$ Neither do I see anything wrong with it. Now this brings up the question of constructing an explicit isomorphism, if even possible. The obvious way to embed the localization into the first product is injective but not surjective unfortunately. $\endgroup$
    – anon
    Jan 16 '14 at 23:52
  • $\begingroup$ Is there some explicit equivalent of Cantor-Bernstein for linear mappings of vector spaces? If so, it would be enough to construct a linear injection of the big space into the "little" one. $\endgroup$
    – user121926
    Jan 17 '14 at 0:13
  • $\begingroup$ Not sure what you mean by "explicit," but your argument essentially uses C-B for vector spaces in your claim that we get a linear isomorphism out of a bijection of bases: if two bases inject into each other, then by C-B for sets they're in bijection, and any such bijection extends to a linear isomorphism. $\endgroup$ Jan 17 '14 at 0:21
  • $\begingroup$ 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. $\endgroup$
    – user121926
    Jan 17 '14 at 0:44

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