group action of $\mathbb{F}_q$ on a $\mathbb{F}_p$-vector space. Question: Let $p, q$ be distinct prime numbers. Let $V$ be a $\mathbb{F}_p$-vector space.  Does there exists a non-trivial $\mathbb{F}_p$-linear action of $\mathbb{F}_q$ on $V$?
My guess in that there doesn't exists a non-trivial action and here is a proof which I have thought of (but at one point I am not so sure):
Suppose the action is non-trivial then for $1 \in \mathbb{F}_q$ there exists a $v \in V$ such that $1\cdot v = w$, where $w \neq v$. Now consider the action $1\cdot (pv) = p(1\cdot v)= p\cdot v$ (since the action is $\mathbb{F}_p$-linear). This implies $p\cdot v = 0$, which is a contradiction because action of $p \in \mathbb{F}_q$ is a vector space isomorphism and by assumption $v \neq 0$.
Is this reasoning correct ?
Thanks in advance !!
 A: A crucial mistake in your argument is the claim that
$p\cdot v= 0$, where $p$ in the left would be an element of the field $\Bbb{F}_q$. The step
$$
1\cdot (pv)=p(1\cdot v)
$$
does follow from the linearity (though $pv=0$ as well as $p(1\cdot v)=0$, so this is not very interesting). But $p(1\cdot v)$ is unrelated to $p\cdot v$.
To see the error consider the case $p=2$. We have for all $v\in V$
$$
2\cdot v=(1+1)\cdot v=1\cdot(1\cdot v),
$$
and there is no way to combine the $1$s on the right hand side.
May be the mistake is that you falsely expected the group operation of $\Bbb{F}_q$, the addition, to magically correspond to the addition in $V$?
In fact, in addition to the rule $0\cdot x=x$ (the neutral element acts trivially) all that the action prescribes is
$$
(a+b)\cdot v= a\cdot (b\cdot v)
$$
for all $a,b\in\Bbb{F}_q$ and all $v\in V$. The requirement that the action should be $\Bbb{F}_p$-linear means that
$$
a\cdot(\alpha v+\beta w)=\alpha(a\cdot v)+\beta (a\cdot w)
$$
for all $a\in\Bbb{F}_q$, $\alpha,\beta\in\Bbb{F}_p$ and $v,w\in V$.

There exist non-trivial linear actions of the additive group $\Bbb{F}_q$ on the vector space $V=\Bbb{F}_p^n$ if the dimension $n$ is large enough. We really only want a cyclic group of order $q$ to act on $V$ by $\Bbb{F}_p$-linear transformations.
Let $d$ be the smallest positive integer such that $q\mid p^d-1$. By Little Fermat
$p^{q-1}\equiv1\pmod q$, so $d$ will be a factor of $q-1$. I claim that $n\ge d$ is all we need.
Consider the extension field $K=\Bbb{F}_{p^d}$. Its multiplicative group is cyclic of order $p^d-1$ so there exists an element $\mu\in K^*$ such that $\mu\neq1$ and $\mu^q=1$.
We can define an $\Bbb{F}_p$-linear action of $\Bbb{F}_q$ on $K$ by declaring that for
all $a\in\{0,1,2,\ldots,q-1\}$ we have
$$\overline{a}\cdot x=\mu^a x.$$
Because $\mu^q=1$, this is well-defined. Because $\mu^{a+b}=\mu^a\mu^b$ it defines an action of $(\Bbb{F}_q,+)$. Because multiplication by any element of $K$ obviously yields an $\Bbb{F}_p$-linear transformation $K\to K$, the action is $\Bbb{F}_p$-linear.
If $n\ge d$, we can choose $V=K\oplus \Bbb{F}_p^{n-d}$, and let $\Bbb{F}_q$ act on $K$ as above, and trivially on $\Bbb{F}_p^{n-d}$.
It is not difficult to show that the condition $n\ge d$ is also necessary for the space $\Bbb{F}_p^n$ to have a linear transformation of order $q$.
