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It is clear that $\text {Hom}_{\mathbb Q} (\mathbb Q, \mathbb Q)\cong \mathbb Q$. How about $\text {Hom}_{\mathbb Q} (\prod \mathbb Q, \mathbb Q)$, where the product runs over a countable index set? To what module is this $\text {Hom}$ isomorphic?

I do know that it is a vector space over the field $\mathbb Q $. What is the dimension thereof?

Thanks to any leading reply.

karparvar
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    What have you tried? Have you considered the case $\operatorname{Hom}(\mathbb{Q}^2,\mathbb{Q})$, just to see what happens when you up the dimension of the domain? If so, can you prove that result without mentioning matrices? – Mastrem Aug 08 '24 at 07:45
  • @Mastrem The group $\text {Hom} (\bigoplus \mathbb Q, \mathbb Q)$ is isomorphic with the direct product of $\mathbb Q$. – karparvar Aug 08 '24 at 08:45
  • Right, and how do you prove that? Does the proof transfer to your case? If not, where do things go wrong? Be as specific as possible. – Mastrem Aug 08 '24 at 08:49
  • @Mastrem It is a straightforward result from a well known theorm which says $\text {Hom}(\bigoplus M_i, N)\cong \prod \text {Hom}(M_i, N)$. – karparvar Aug 08 '24 at 10:58
  • Yes, and how is that theorem proved? – Mastrem Aug 08 '24 at 11:09

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This follows from two applications of the Erdős–Kaplansky theorem. Writing $|X|$ for the cardinality of a set $X$, this theorem asserts that the dimension of the dual space $V^{\ast}$ of an infinite-dimensional vector space $V$ over a field $K$ is

$$\dim V^{\ast} = |K|^{\dim V}.$$

$\mathbb{Q}^{\mathbb{N}}$ is the dual of the countable-dimensional $\mathbb{Q}$-vector space $\bigoplus_{\mathbb{N}} \mathbb{Q}$, so one application of the theorem gives

$$\dim \mathbb{Q}^{\mathbb{N}} = |\mathbb{Q}|^{|\mathbb{N}|} = \aleph_0^{\aleph_0} = \mathfrak{c}$$

which is the cardinality of the continuum $|\mathbb{R}|$. A second application of the theorem then gives

$$\dim (\mathbb{Q}^{\mathbb{N}})^{\ast} = |\mathbb{Q}|^{\mathfrak{c}} = \aleph_0^{\mathfrak{c}} = 2^{\mathfrak{c}}$$

which is strictly larger, the cardinality of the powerset $\mathbb{P}(\mathbb{R})$. This is an absurdly large vector space with no applications whatsoever that I know of, and of course all of this depends on the axiom of choice. In the absence of choice I think it's consistent that $\mathbb{Q}^{\mathbb{N}}$ does not have a basis at all, and that its dual is $\bigoplus_{\mathbb{N}} \mathbb{Q}$.

Qiaochu Yuan
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