A cubic polynomial over a field is irreducible if it has no roots

Question

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?

Answer 1

Yes, a polynomial of degree $3$ will be irreducible if and only if it has no root.

The only way it could factor in a non-trivial is as a product of a degree $2$ and a degree $1$, or as three degree $1$ polynomials.

Either way there is a degree $1$ factor and this means there'd be a root.

This also works for degree two polynomials yet not for other degrees.

Answer 2

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 $k$. $\Box$

Answer 3

Yes, a polynomial (over a field) with degree less than 4 is irreducible if and only if it has no roots. This is because having roots is equivalent to having linear factors.

Answer 4

Yes, but you need to justify why this works. If $\pi$ were reducible, you could write $\pi(\alpha) = \pi_1(\alpha) \cdot \pi_2(\alpha)$ in a non-trivial way. This would imply that the possible degree for $\pi_1$ is either one or two, in which case $\pi_2$ would be of degree one. In any case, since one of the factors must be of degree one, this means that if $\pi$ is reducible, it must have a root and so your method works.

In general, if $\pi$ wouldn't be of degree three, your method could fail.