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let $F$ be a finite field of order $q$ and let $f(x)$ be a polynomial in $F[x]$ of degree $n \geq 1$. Prove that $F[x]/(f(x))$ has $q^n$ elements. I know that if a field is finite then the order is prime so that means q is prime, but what does that says about $F[x]/(f(x))$. Can anyone try to help me understand this problem because i just recently learn about field extension. thanks

user146269
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    Presumably $f(x)$ should be an irreducible polynomial. Also the order of a finite field does not have to be prime, but rather a power of a prime. So $q = p^r$ for some positive integer $r$. – bzc Apr 09 '15 at 17:58

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An element of your factor ring can be uniquely described as $a_0 + a_1 \overline{X} + ... + a_{n-1}\overline{X}^{n-1}$ where $a_j \in F$. Therefore there are $q\times q \times ... \times q = q^n$ elements.

Where $\overline{X} = X + (f(X)) \in F[X]/(f(X)) = k$. Take any polynomial $g(X)$ and divide it by $f(X)$ to get $g(X) = q(X)f(X) + r(X)$ where $\deg r(X) < \deg f(X)$. When you view this equation as in the field $k$, then $q(X)f(X)$ becomes zero. Therefore, every element in that field is tried by a polynomial of degree $n-1$.

Nicolas Bourbaki
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