# $F[x]/\langle f(x) \rangle$ has $q^n$ elements

today I have a problem which I see in book Abstract Algebra of David. This is problem:

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 $F[x]/\langle f(x) \rangle$ has $q^n$ elements.

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I am having second thoughts about whether this is a close enough duplicate of this one. All potential "close"-voters, please form your opinion without trusting mine. Sorry about being a tad trigger happy today. –  Jyrki Lahtonen Nov 20 '12 at 14:32
@Firmino: You should accept some answers to your questions! –  tomasz Nov 20 '12 at 15:37

Hint: It is a vector space of dimension $n$ over the field $F$.

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Hint: When you mod out a polynomial ring by a polynomial $f(x)$, it is as if you are saying that "$f(x)=0$, and using that in your computations.

Now suppose you solve that equation for the highest degree term in $f(x)$, which we will call $\alpha x^n$. You get something that looks like this: $\alpha x^n=f(x)-\alpha x^n$. The thing on the right hand side has degree strictly less than $n$. You can further solve it by dividing by $\alpha$, so: $x^n=(f(x)-\alpha x^n)/\alpha$.

This means that in all your computations, if you see $x^n$, or any higher power, you can replace it with a polynomial of degree less than $n$ (whatever the degree of $(f(x)-\alpha x^n)/\alpha$ happens to be.) After doing this repeatedly, you will have no powers of $x$ greater than the $n$th power. That means everything in the quotient can be represented as a polynomial of degree less than $n$, and conversely all polynomials like that will appear in the quotient.

So, how many such polynomials are there?

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