TITLE: How to check that a cubic polynomial is irreducible?
QUESTION [7 upvotes]: I want to argue that argue that $\pi(\alpha)=\alpha^3+3\alpha+3$ is an irreducible polynomial over the finite field with 5 elements $\mathbb{F}_5$. My approach was just to check that $\pi$ has no roots in $\mathbb{F}_5$. Is this right?

REPLY [2 votes]: One has the:

Proposition. Let $k$ be a field and $p\in k[X]$ be a polynomial of degree $2$ or $3$. $p$ is irreducible over $k$ if and only if $p$ has no roots in $k$.

Proof. We distinguish the two following cases:

If $\deg(p)=2$. Assume that $p$ has no roots in $k$ and assume their exists $(f,g)\in k[X]^2$ such that: $$p=fg.$$
Taking the degree in this equality leads to: $$(\deg(f),\deg(g))\in\{(0,2),(1,1),(2,0)\}.$$
Hence, since $p$ has no roots in $k$, $(\deg(f),\deg(g))\in\{(0,2),(2,0)\}$. In other words, $p\in k[X]^\times$ or $q\in k[X]^{\times}$ and $p$ is irreducible over $k$.
If $\deg(p)=3$. Assume that $p$ has no roots in $k$ and assume their exists $(f,g)\in k[X]^2$ such that: $$p=fg.$$
Taking the degree in this equality leads to: $$(\deg(f),\deg(g))\in\{(0,3),(1,2),(2,1),(3,0)\}.$$
Hence, since $p$ has no roots in $k$, $(\deg(f),\deg(g))\in\{(0,3),(3,0)\}$. In other words, $p\in k[X]^\times$ or $q\in k[X]^{\times}$ and $p$ is irreducible over $k$. $\Box$