Let $h$ be a monic polynomial in $\mathbb{F}[x]$ of degree $m\geq 1$, where $\mathbb{F}$ is a field. $\mathbb{F}[x]/(h)$ is a field if and only if $(h)$ is irreducible.
Forward direction:
Suppose $\mathbb{F}[x]/(h)$ is a field then $(h)$ is a maximal ideal. Suppose $h=fg$, for polynomials $f$ and $g$ in $\mathbb{F}[x]$. Then, $(h)\subseteq (f)\subseteq \mathbb{F}[x]$. Since $A$ is maximal, $(h)=(f)$ or $(f)=\mathbb{F}[x]$. If $(h)=(f)$ then by uniquness $h=f$ and so the degree of $g$ must be $0$. If $(f)=\mathbb{F}[x]$ then $f$ is a unit and so must have degree $0$. Hence $h$ is irreducible.
How do I show the backwards direction? Can someone do it without using gcd?
Best Answer
The backwards direction goes like this:
Let $h$ be irreducible and assume we have an ideal $\mathfrak{a} = (a)$ in $\mathbb{F}[x]$ (this works because $\mathbb{F}[x]$ is a PID), such that $(h) \subset \mathfrak{a} \subset \mathbb{F}[x]$. Then there exists some $c \in \mathbb{F}[x]$ such that $h=ac$. Since $h$ is irreducible, we have $a \in (\mathbb{F}[x])^\times$ or $c \in (\mathbb{F}[x])^\times$. In the first case, we get $\mathfrak{a}=\mathbb{F}[x]$ and in the second case we get $\mathfrak{a}=(h)$. So $(h)$ is maximal, hence $\mathbb{F}[x]/(h)$ is a field. I leave it up to you to fill in any details.
I hope this was helpful!