# For vector spaces $V,W$ over $\mathbf{k}$, is every additive $\phi: V \to W$ also $\mathbf{k}$-linear?

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!

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.)
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.)