Let $p$ be a prime number, $m, s \in \mathbb{N}$ and $q = p^s$.
Let $f(t) = \sum_{i=0}^m \lambda_i t^{p^{i}} \in \mathbb{F}_q[t]$.
I'm asked to prove that $f(a\alpha+b\beta) = af(\alpha)+bf(\beta)$, whenever $a, b \in \mathbb{F}_p, \alpha,\beta \in \mathbb{F}_q$ (straightforward) and consequently $\{ \rho \in \mathbb{F}_q : f(\rho) = 0 \}$ is a vector subspace of $\mathbb{F}_q$ (seen as a vector space over $\mathbb{F}_p$).
The reason I mention this is that it seems logical that it can be used in the following (third part of the) exercise.
Show that, if $\lambda_0 \neq 0$, then $f(t)$ doesn't have any multiple root over a splitting field.
I know this is equivalent to show that $gcd(f,f')=1$, but then I get stuck.
In my tentatives, I considered $h$ a common divisor of $f$ and $f'$. Since $f'(0) = \lambda_0 \neq 0$, it follows that $h(t) = \gamma_n t^n + \cdots + \gamma_1 t + \gamma_0$, with $\gamma_0 \neq 0$ and then I'd like to prove that $h(t) = \gamma_0$, that is, $\gamma_1 = \cdots = \gamma_n = 0$, or equivalently, $h'(t) = 0$; but here I cannot proceed...
Other approaches included factoring $f = h \cdot h_1$ and $f' = h \cdot h_2$ and then compare with $f' = h'\cdot h_1 + h \cdot h_1'$ and juggle around trying to conclude that $h'=0$, but I go around circles.
I also don't seem to find any good use for the previous results and it doesn't seem adequate to compute $gcd(f,f')$, so I must be missing some important detail.
Can anyone help me on this?
Write $f$ “explicitly”: $$ f(t)=\lambda_0t+\lambda_1t^p+\lambda_2t^{p^2}+\dotsb $$ so $$ f'(t)=\lambda_0+\lambda_1 pt^{p-1}+\lambda_2p^2t^{p^2-1}+\dotsb =\lambda_0 $$ If $\lambda_0\ne0$, $\lambda_0=f'(t)$ is a divisor of $1$, so $\gcd(f,f')=1$. If $\lambda_0=0$, $\gcd(f,f')=f$ has multiple roots as soon as $m>0$.