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Let $A,B$ be $\mathbb{C}$-vector space. We can view them as a $\mathbb{Z}$-module. Suppose that there is a $\mathbb{Z}$-module isomorphism $\phi$ between $A$ and $B$. Then can we have a natural $\mathbb{Q}$(or even $\mathbb{R}$, $\mathbb{C}$)-vector space isomorphism between $A$ and $B$ from $\phi$?

I think at least we can have $\mathbb{Q}$-vector space isomorphism.

For example, Let $a\in A$ and denote $\frac{1}{n}a=\alpha\in A$ for some natural number $n$. Let $\phi$ be a $\mathbb{Z}$-module isomorphism from $A$ to $B$. Then, we have

$\phi(a)=\phi(n\alpha)=n\phi(\alpha)=n\phi(\frac{1}{n}a)$.

Thus, $\phi(\frac{1}{n}a)$ has to be $\frac{1}{n}\phi(a)$. Because $\mathbb{Q}$ is the field of fraction of $\mathbb{Z}$, we may have $\phi(\frac{m}{n}a)=\frac{m}{n}\phi(a)$ for all $a\in A$ and $\frac{m}{n}\in\mathbb{Q}$.

Am I right?

Now, I am struggling to extend to $\mathbb{R}$ or $\mathbb{C}$ vector space isomorphism.

PZM
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  • I think your dream is unrealistic (the parts about going beyond $\Bbb{Q}$). After all, $\Bbb{R}$ and $\Bbb{R}^2$ are isomorphic as vector spaces over $\Bbb{Q}$. Both have continuum dimension. The question is interesting (+1), but I also think it has been covered earlier on the site. – Jyrki Lahtonen Aug 30 '23 at 11:33
  • No, use a Hamel basis to construct lots of automorphisms of $\mathbb R / \mathbb Q$. – Eric Aug 30 '23 at 12:32
  • If you add continuity for $\phi$ then you're getting somewhere. (for $\mathbb R$-linearity, anyway) – M W Aug 30 '23 at 14:04
  • If you consider $A$ and $B$ as complex manifolds, assuming they are finite-dimensional, then a $\Bbb Z$-linear map is $\Bbb C$-linear iff it is holomorphic. – Lukas Heger Aug 30 '23 at 15:00
  • For $\Bbb R$-linearity, measurable is enough. – Lukas Heger Aug 30 '23 at 15:01

1 Answers1

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Your argument that $\mathbb{Z}$-linear implies $\mathbb{Q}$-linear is correct but it's not possible to go beyond this. In fact assuming the axiom of choice $\mathbb{C}$ and $\mathbb{C}^2$, for example, are isomorphic as $\mathbb{Q}$-vector spaces (since they both have the same dimension, namely $|\mathbb{R}|$).

On the other hand, it's consistent with ZF that every homomorphism between Polish groups is automatically continuous, and in particular that every homomorphism between finite-dimensional real or complex vector spaces is automatically continuous. If this is true then any $\mathbb{Z}$-linear isomorphism between finite-dimensional real or complex vector spaces is automatically $\mathbb{R}$-linear, so in the real case we're done, and in the complex case they have the same real dimension and hence the same complex dimension, so there is some other $\mathbb{C}$-linear isomorphism between them.

Qiaochu Yuan
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