Does there exists an additive group homomorphism between two $K$-vector space that is not $K$-linear

Question

My question is: Give me a field $K$. Can we always find two $K$-vector space $V_{1}$, $V_{2}$ and a map $f:V_{1}\rightarrow V_{2}$ such that:

(1) If we view $V_{1}$, $V_{2}$ as additive group, then $f$ is a group homomorphism.

(2) If we view $V_{1}$, $V_{2}$ as $K$-vector space, then $f$ is not a vector space homomorphism.

A similar but not same question was seemingly discussed here:

Group homomorphisms and vector space homomorphisms

My current thoughts:

(1) First, release the condition that $\mathbb{K}$ is a field, instead if we consider $\mathbb{K}$ to be a ring, $V_{1},~V_{2}$ to be $\mathbb{K}$-modules then there certainly exists situation that we cannot do so, especially when $\mathbb{K}$ is not commutative.

(2) For $K=\mathbb{F}_{p}\cong \mathbb{Z}/p\mathbb{Z}$. We can't. Given two $\mathbb{F}_{p}$-vector space $V_{1},~V_{2}$, we can show every additive group homomorphism $f:V_{1}\rightarrow V_{2}$ is $\mathbb{F}_{p}$-linear:

Let $a\in\mathbb{F}_{p},~v\in V_{1}$

$f(a v)=f(v+v+\cdots+v)$, where addition is carried out $n$ times, $n\equiv a$ mod $p$.

$f(v+v+\cdots+v)=f(v)+f(v)+\cdots+f(v)=nf(v)$. We have $pf(v)=f(v)+\cdots+f(v)$ $p$ times $=f(v+\cdots+v)=f(p v)=f(0)=0$.

Hence $f(av)=af(v)$ is $\mathbb{F}_{p}$ linear.

(3) (Inspired by Mariano Suárez-Alvarez♦) For $\mathbb{K}=\mathbb{Q}$. We can't either:

$a_{2}f(\frac{a_{1}}{a_{2}}v)=f(a_{1}v) = a_1 f(v) \Rightarrow f(\frac{a_{1}}{a_{2}}v)=\frac{a_{1}}{a_{2}}f(v)$

For other kind of $\mathbb{K}$, see answers below.

Answer 1

If the field is a prime field (one of the $F_p$ with $p$ prime, or $\mathbb Q$) then you can't, and this for exactly the reason you mention (which works also for the rationals). If not, you can.

Indeed, in that case the field $K$ contains a proper subfield $L$, and you can find an $L$-linear map which is not $K$-linear.

Answer 2

Inspired by Mariano Suárez-Alvarez♦'s:

Just like Mariano Suárez-Alvarez♦ comments: Given me a non-prime field $K$. $K$ has some proper subfield $L$ and $K$ can be viewed as an $L$-vector space.

Every vector space has a basis. Say $\{a_{i}:i\in I\}$ is a basis of $K$ over $L$. Consider a map $f:\{a_{1},a_{2}\}\rightarrow \mathbb{K},~a_{1}\mapsto 0, a_{2}\mapsto 1$. Such $f$ will be extended to a $L$ linear map $\bar{f}$. However $\bar{f}$ won't be $\mathbb{K}$-linear since $\bar{f}(\frac{a_{2}}{a_{1}}a_{1})=\bar{f}(a_{2})=1\neq\frac{a_{2}}{a_{1}}f(a_{1})=0$.