Division of polynomials
I put what is unclear to me in between three asterisks bounding the unclear lines...
$\equiv_{P}$ is defined as ($\forall Q, S \in F[t]$) $Q\equiv_{P} S \iff P|Q-S$
("|"-divides...)
On to the proof:
"As proven earlier $F[t]|_{\equiv_P}$ is a commutative, associative ring with unity. Let us prove that this is a finite integral domain (thereby proving it is indeed a field). $$F[t]|_{\equiv_P}=\{\ [Q]\ \ |\ \ Q\in F[t]\ \}$$ In the class of equivalence $[Q]$ is the actual remainder $R$ (when dividing $Q$ by $P$) $R$ is of the shape: $$R=b_0+...+b_kt^{k}, k \leq n-1$$
The number of all remainders is $***|F|^n***$ (finite field, so there are a finite amount.) Let us prove that the non-existence of divisors of zero (integral domain defining characteristic.) Let: $$0=[Q][S]=[Q \ast S]$$ (proven before)**$$\implies P| Q\ast S$$ **Now because P is irreducible $***\implies P|Q \lor P| S \implies [Q]=0 \lor [S]=0***$. Therefore $F[t]|_{\equiv_P}$ is a finite integral domain, and therefore a field.
If $P$ had a factor in common with $Q$ and another factor in common with $S$ then the two factors would provide a factorisation of $P$ - but $P$ is irreducible.
For the number of remainders, there are $f$ possibilities for each of the $n$ coefficients, where $f=|F|$ is the number of elements in the field - giving $f^n$ ways of choosing them.
And, if you need it, the degree of the difference between any pair of polynomials obtained in this way is too small for the difference to be divisible by $P$ so the remainders are distinct.