# Explicit construction of a finite field with $p^n$ elements

I'm trying to understand the following explicit construction of a finite field with $p^n$ elements where $p$ is prime, $n \geq 2$:

Take any irreducible polynomial $f(X) \in \mathbb{F}_p$ of degree $n$. Then $\mathbb{F}_p[X] / (f(X))$ is a field with $p^n$ elements.

Now, I understand that if $f$ is irreducible, the ideal generated by $f$ is maximal, so $\mathbb{F}_p[X] / (f(X))$ is indeed a field.

I don't understand the following points:

• Why does for every $n \geq 2$ an irreducible $f(X) \in \mathbb{F}_p[X]$ of degree $n$ exist?
• Why does the field $\mathbb{F}_p[X] / (f(X))$ have $p^n$ elements?

Thanks in advance for any help!

For the second question, consider the elements of $\mathbb{F}_{q}[x]/(f(X))$; they have as representatives the polynomials of degree at most $n-1$, since a polynomial of degree $n$ or larger can have multiples of $f(x)$ subtracted from it to reduce the degree. Counting these polynomials gives the order of the new field.
• Thank you for your answer! I still don't understand though why any polynomial of degree bigger or equal to $n$ is divisible by $f(X)$. Could you elaborate on that? – Tom Bombadil Sep 13 '15 at 16:39
• The nice thing is, it doesn't have to be divisible by $f(x)$ for it to be reduced. For example, in $\mathbb{F}_{3}[x]/(x^2+1)$; consider the polynomial $x^3$. We have that $x^3 = x(x^2+1) -x$, so we can reduce $x^3$ to $-x = 2x$ (a polynomial of degree less than 2). – Morgan Rodgers Sep 13 '15 at 17:21
• (as a side note, if the polynomial is divisible by $f(X)$ it will reduce to $0$ in the new field; otherwise, you replace it with the remainder when you divide by $f(X)$, which always has degree less than $f(X)$.) – Morgan Rodgers Sep 13 '15 at 17:30