On analogy between $\Bbb Z$ and $\Bbb F_q[x]$

Question

There are objects and operations analogous between $\Bbb Z$ and $\Bbb F_q[x]$. For example primes in $\Bbb Z$ and irreducibles in $\Bbb F_q[x]$ are analogous and so is multiplication operation.

Consider objects $\Bbb Z/p$ for $p$ a prime and $\Bbb F_q[x]/f(x)$ where $f(x)$ is irreducible.

For $a\in\Bbb Z/p,b\in\Bbb Z/(p-1)$ we can define $a^b\in\Bbb Z/p$.

Similarly for $a(x)\in\Bbb F_q[x]/f(x),b(x)\in\Bbb F_q[x]/(f(x)-1)$ can we define $a(x)^{b(x)}\in\Bbb F_q[x]/f(x)$?

Answer 1

There is a Fermat's little theorem in this setting, and it says the following: if $f$ is irreducible over $\mathbb{F}_p[x]$ of degree $d$, then

$$g(x)^{p^d} \equiv g(x) \bmod f(x).$$

So exponents are well-defined $\bmod p^d - 1$ (the order of the cyclic group $\left( \mathbb{F}_p[x]/f(x) \right)^{\times}$). In the analogy between the integers and $\mathbb{F}_p[x]$, exponents stay integers: they play a totally different role than bases.

Answer 2

For $a,b\in\Bbb Z/p$, we cannot define $a^b\in\Bbb Z/p$: we can only do so if we take $b\in\Bbb Z/(p-1)$.

For example, $2^7 \equiv 2 \bmod 7$ but $2^{0} \equiv 1 \bmod 7$ even though $ 7 \equiv 0 \bmod 7$.