Verified answer

Suppose that f(x) is irreducible.

Then, by the division algorithm, any p(x) in F[x] can be written as

p(x) = q(x) f(x) + r(x) for some q(x), r(x) in F[x], where deg(r(x)) < deg(f(x)) = n.

Hence, p(x) + <f(x)> = r(x) + <f(x)>.

Now, suppose that r(x) + <f(x)> = s(x) + <f(x)>, where r(x), s(x) have degree < n.

Then, r(x) - s(x) is a polynomial multiple of f(x).

Comparing degrees, we conclude that r(x) - s(x) = 0 ==> r(x) = s(x).

Hence, any element of F[x] / <f(x)> can be uniquely written as r(x) + <f(x)>, where

deg(r(x)) < n.

How many such r(x) are there?

Since each coefficient has q possibilities (since |F| = q), and r(x) is a F-linear combination of

1, x, ..., x^(n-1) [for n possible terms], there are q^n possibilities for r(x).

Hence, |F[x]/<f(x)>| = q^n.

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If f is not irreducible, then factor f as the product of irreducibles and the Chinese Remainder Theorem in conjunction with the result established above.

I hope this helps!