I'm reading about a function field analogue of the following conjecture
Let $f \in \mathbb{Z}[x]$ be a square-free polynomial of degree $k$. Then there are infinitely many square-free values taken by $f(n)$. In fact, the set $\{n\in \mathbb{N}:f(n) \text{ square-free}\}$ has density $$c_f:=\prod_{p\in \mathcal{P}} \left(1-\frac{\rho(p^2)}{p^2}\right).$$
The function field analogue is the following.
Assume $f\in \mathbb{F}_q[t]$ is separable. Then the cardinality of the set $\mathcal{N}_f(n)$ of monic polynomials $a(t)$ of degree n such that $f(a(t)) \in \mathbb{F}_q[t]$ is square-free, is: $$\#\mathcal{N}_f(n) = c_fq^n + O_{f,q}\left(\frac{q^n}{n}\right),\quad \text{as }n\to \infty,$$ with $$c_f = \prod_{P\in \mathcal{P}}\left(1-\frac{\rho_f(P^2)}{|P|^2}\right),$$ the product over prime polynomial $P$. The density $c_f$ is positive if and only if there is some $a\in \mathbb{F}_q[t]$ such that f(a) is square-free.
My question Why do we need $f$ to be separable? I suppose it is the analogue of $f$ being a square-free polynomial in the first example? Can anyone clarify with an example?