Is there a bijection between the set of prime ideals of norm $~ q~$, and the set of ring maps to $~F_q~$? OK, for each given ring morphism,
$\mathfrak{o}\rightarrow~F_{q}~$,
there exists another distinct arrow after composing with the Frobenius map (or a power of it).
So, if $q = p^{n}$, then there are n distinct endomorphisms $F_{q}\rightarrow F_{q}$., and hence at least $n$ distinct ring homomorphisms from $\mathfrak{o}$ to $F_{q}$.
But, from a highly technical text on etale cohomology (so I'd be inclined to believe it), the book simply says, the primes (or, prime ideals) $\mathfrak{p}$, such that,
$\mathfrak{o}/\mathfrak{p}\approx F_{q}$,  correspond bijectively to the ring maps,
$\mathfrak{o}\rightarrow F_{q}$, which given the first (?) isomorphism theorem made sense intuitively.
However, to get a grasp on this, I tried an example,
$\mathbb{Z}_{3} [x]$ as my ring $\mathfrak{o}$, the three prime ideals,
$(x^2+1), (x^2+x+2),$ and $(x^2+2x+2) =\mathfrak{p}$, so then I obtained the isomorphisms, $\mathfrak{o}/\mathfrak{p} \approx F_{9}$, but I instead find SIX ring maps $\mathfrak{o} \rightarrow F_9$!
The three given from the three prime ideals, and their composition with the non trivia Frobenius map.
So finally, my question:
Are the sets bijective "up to endomorphism", where the maps are paired in an equivalence relation, or am I missing something?
I have four books on abstract algebra, and naturally each isomorphism yields an ideal as the kernel, like I said I'm getting 6 not just 3 maps, and can't find any reference on such a bijective correspondence.
Thanks in advance.

Prime ideals of norm $q$ are in bijection with surjections $R \to \mathbb{F}_q$, up to postcomposition with Frobenius, for exactly the reason you state. I don't know whether your text has a typo or whether you have misread it, but it is not true that prime ideals of norm $q$ are in bijection with surjections $R \to \mathbb{F}_q$.
