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Let $k$ be a field. Let $I,J$ be two sets such that $k^I\cong k^J$ as $k$-vector spaces. I have read that this isomorphism implies that $I=J$. In the finite case, this is obvious, but how does it follow if both $I$ and $J$ are infinite?

Let $\phi:k^I\rightarrow k^J$ be the isomorphism in question. My idea is to show that this isomorphism implies $k^{\oplus I}\cong k^{\oplus J}$ (these are the subspaces of eventually zero elements), which would imply that $I\cong J$. So let $\alpha\in k^I$. We have to show that $$\{j\in J\ |\ \phi(\alpha)(j)\ne0\}$$ is finite. By definition, the set $\{i\in I|\alpha_i\ne0\}$ is finite. I am not clear on how to proceed. Any hints?

Edit: Sorry I don't understand how the linked question answers the question. Why should a basis of $k^I$ be indexed by $I$? It's clear that there exists a generating system of $k^I$ indexed by $I$ but not a basis.

Edit: The duplicate only shows that if we have two bases $(v_i)_{i\in I}$ and $(w_j)_{j\in J}$ then $I\cong J$. This is trivial. But here we don't know that $k^I$ has a basis indexed by $I$ or $J$.. If $B$ is a basis of $k^I$, we can only conclude that $\text{card}(B)\leq\text{card}(I)$.

user997942
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  • The note I read states that "$I$ is equal to $J$". But I think that it should be "$I$ is isomorphic to $J$", hence the symbol $\cong$. Regarding $k^I$: it is the product of $k$ with itself $I$ times, i.e. $\prod_{i\in I}k$. – user997942 Jan 09 '23 at 19:21
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    Bijective--isn't that what isomorphism means in category of sets? – user997942 Jan 09 '23 at 19:23
  • thanks, I will have a look – user997942 Jan 09 '23 at 19:26
  • The bases of $k^I$ and $k^J$ have cardinality $2^{|I|}$ and $2^{|J|}$, respectively. But from that to get $|I|=|J|$ we might need more axioms than those in ZFC. – plop Jan 09 '23 at 19:29
  • That presumptive duplicate doesn't answer the question. – plop Jan 09 '23 at 19:29
  • Sorry I don't understand how the linked question answers the question. Why should a basis of $k^I$ be indexed by $I$? – user997942 Jan 09 '23 at 19:35
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    The new duplicate doesn't work. The duplicate only shows that if we have two bases $(v_i){i\in I}$ and $(w_j){j\in J}$ then $I\cong J$. But here we don't know that $k^I$ has a basis indexed by $I$ or $J$.. – user997942 Jan 09 '23 at 19:45
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    See here for the question of whether $2^{|I|}=2^{|J|}$ implies $|I|=|J|$. – plop Jan 09 '23 at 19:49
  • @owl thank you! In the paper that I am reading, the author merely says "...along with the trivial observation that $k^I\cong k^J$ implies $I=J$," in a proof! But it seems that this isn't necessarily true..The author doesn't mention anything about necessity of axioms beyond ZFC. – user997942 Jan 09 '23 at 19:53
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    The author may have assumed some more before, e.g., at the beginning of that paper. Do you have a link? And the proof here is not as trivial as you say. – Dietrich Burde Jan 09 '23 at 19:56
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    Probably either the author is assuming $I, J$ are finite or taking the direct sum. I guess the isomorphism could also be an isomorphism of $k$-algebras. – Qiaochu Yuan Jan 09 '23 at 19:58

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Your idea does not work because (assuming choice) there are automorphisms of $k^I$ that do not fix the subspace $k^{\bigoplus I}$. You can see this by choosing a basis containing at least one vector in this subspace, then permuting it so that this vector is exchanged with another vector which isn't.

As far as I know, the answer to this question is undecidable in ZFC. We need two results:

  • If $k$ is any field and $I$ is any infinite set, then the dimension of $k^I$ is the cardinality $|k|^{|I|}$ of $k^I$. See this MO thread.
  • In ZFC, it's undecidable whether $2^I \cong 2^J$ implies $I \cong J$. See this math.SE thread, linked in the comments.

Now we can take $k = \mathbb{F}_2$. (This isn't quite a complete argument because we need to know the analogue of this result for other fields as well, but it means the result is either undecidable or false, and in any case it is not provable in ZFC.)

Qiaochu Yuan
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  • On the other hand one can make the case that if $I$ is infinite it does not really make sense to consider $k^I$ as a discrete vector space and we ought to actually give it the product topology. I don't know what happens if you ask for an isomorphism $k^I \cong k^J$ of topological vector spaces. I think it might be the case that the continuous dual of $k^I$ is $k^{\bigoplus I}$; if so the answer would turn out to be positive. – Qiaochu Yuan Jan 09 '23 at 21:00