# Irreducible polynomials and fields

Theorem

Let $F$ be a field. Let $f(x)$ be a polynomial in $F[x]$ and let $I = (f(x))$ be the principal ideal generated by $f(x)$. Then

$I$ is a maximal ideal of $F[x]$ if and only if $f(x)$ is irreducible in $F[x]$ if and only if $F[x]/I$ is a field.

## Creating fields using irreducible polynomials

• Creating a field $E$

Let $p(x)$ be an irreducible polynomial of degree $n$ over a field $F$. Let $I = (p(x))$, the principal ideal of $F[x]$ generated by $p(x)$. Since $p(x)$ is irreducible, the above theorem tells us that $I$ is a maximal ideal of $F[x]$. Let $E = F[x]/I$. Then $E$ is a field.

• The elements of $E$

A coset in $E$ is of the form $f(x)+I$ for some $f(x) \in F[x]$.

Write $f(x) = q(x)p(x) + r(x)$ with deg$\,r(x) <$ deg$\,p(x) = n$ so that $$r(x) = a_0 + a_1x + \cdots + a_{n-1}x^{n-1}$$ with $a_0, a_1, \ldots , a_{n-1} \in F$. So $f(x)+I = r(x)+I = a_0 +a_1(x+I) + \cdots + a_{n-1}(x+I)^{n-1}.$

Now let $\theta = x+I \in E$. Thus $f(x)+I = a_0 + a_1\theta + \cdots + a_{n-1}\theta^{n-1}.$ Hence

$E = \{a_0 + a_1\theta + \cdots + a_{n-1}\theta^{n-1} : a_i \in F\}.$

Back to Chapter 2: Polynomials