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Since every map of vector spaces is a map of abelian groups, I was wondering if the converse also holds:

Given an additive map $\phi: V \to W$ between two vector spaces, does it follow that $\phi$ is also $\mathbf{k}-$linear? I'm interested in the case of $\mathbf{k}$ having characteristic zero, specially if $\mathbf{k}$ is a famous field like rational or real or complex numbers.

I'm guessing it's false, but I tried to come up with counter-examples for $\mathbf{k} = \mathbf{Q}, \mathbf{R}$ and couldn't find any. Finding a counter-example in characteristic $p>0$ may not be that hard, for instance since taking $p-$th power is additive. However that's not the case I care most about. Appreciate any help!

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2 Answers 2

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This is true only for $k\cong\mathbb{Q}$. In the case $k=\mathbb{Q}$, suppose $\phi:V\to W$ is a homomorphism of abelian groups between two vector spaces, $v\in V$ and $\frac{a}{b}\in\mathbb{Q}$ (with $a,b\in\mathbb{Z}$). Note that then $$a\phi(v)=\phi(av)=\phi\left(b\cdot\frac{a}{b}v\right)=b\phi\left(\frac{a}{b}v\right),$$ so multiplying by $\frac{1}{b}$ we find that $\phi(\frac{a}{b}v)=\frac{a}{b}\phi(v)$ so $\phi$ is linear.

On the other hand, if $k$ is not isomorphic to $\mathbb{Q}$, it is a nontrivial field extension of $\mathbb{Q}$. In particular, $k$ can be considered as a $\mathbb{Q}$-vector space of dimension greater than $1$. Picking a basis, there is then a $\mathbb{Q}$-linear map $\phi:k\to k$ which is the identity on all but one basis vector but maps one basis vector to $0$. This $\phi$ cannot be $k$-linear, since any $k$-linear map $k\to k$ is either $0$ or injective.

(If you consider fields of arbitrary characteristic, similar arguments show that every abelian group homomorphism of $k$-vector spaces is $k$-linear iff $k$ is a prime field.)

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  • $\begingroup$ thank you very much for that solid answer! $\endgroup$ Aug 24, 2020 at 0:30
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Consider the case of conjugation $f=(-)^*: \mathbb{C} \to \mathbb{C}$, where $\mathbb{C}$ is a 1-dimensional vector space over itself (and the scaling is ordinary complex multiplication, of course). This is an additive homomorphism since $(z+w)^* =z^* + w^*$. However, $f(cz) = (cz)^* = c^*z^*$ is not necessarily equal to $cf(z)=cz^*$, so it is not a $\mathbb{C}-$vector space map. (Of course they will be the same for all $z$ only when $c$ is real.)

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    $\begingroup$ thank you, an important, simple counterexample! $\endgroup$ Aug 24, 2020 at 0:29

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