Showing existence of irreducible polynomial of degree 3 in $\mathbb{F}_p$

Question

I'am trying to show that for every p$ \in \mathbb{N}$ where p is prime, there is an irreducible polynomial of degree 3 in $\mathbb{F}_p$. I've found too general answers for that question, but I want to show it in the most simple way.

I know to do so for polynomial of degree 2: the function $x\mapsto x^2$ is not surjective, thus there is $a\in \mathbb{F}_p$ with $\forall b \in \mathbb{F}_p $ $b^2 - a \neq 0$ , which means $x^2-a$ has no roots.

But I can't do the reduction to my problem.

Thanks.

Answer 1

For something in the same spirit as the example for $2$: $x^3-x=(x+1)x(x-1)$ has three zeros (two for $p=2$), so it can't be surjective. Thus some $x^3-x-a$ has no zeros, and a cubic with no linear factors is irreducible.

This approach won't generalize any further; we need to account for more than just linear factors to verify that a polynomial of degree $\ge 4$ is irreducible.

Oh, and your example for quadratic polynomials isn't quite universal; both $x^2=x\cdot x$ and $x^2+1=(x+1)(x+1)$ factor for $p=2$. We need a special case $(x^2+x+1)$ for that.

Answer 2

Let $a_1$ and $a_2$ be distinct elements in $\mathbb{F}_p$. Next let $c \in \mathbb{F}_p$ be such that $a^3_1-a^3_2 = c(a_1 - a_2)$, there exists such a $c$ as $a_1 \not = a_2$. Then the mapping $f: x \mapsto x^3- cx$; $x \in \mathbb{F}_p$ is not surjective mapping from $\mathbb{F}_p$ onto $\mathbb{F}_p$, as there exist two distinct $a_1,a_2 \in \mathbb{F}_p$ satisfying $f(a_1)=f(a_2)$. So for this particular $c$, there exists an $A \in \mathbb{F}_p$ such that there is no $x \in \mathbb{F}_p$ that satisfies $x^3-cx = A$. Thus the polynomial $x^3-cx-A$ is irreducible in $\mathbb{F}_p$.

Answer 3

Count.

Find the number of monic (irreducible) polynomials of degree $1$.

Then count the number of monic polynomials of degree $2$ and subtract the number of reducible ones (they factor) to find the number of irreducible quadratics.

Now do the same for degree $3$. The reducible ones factor into three linear factors or one linear factor and an irreducible quadratic.

This method can be generalized.

Answer 4

If a degree $3$ polynomial is reducible over a field, then it has a root. So you (just) need a degree $3$ polynomial without a root.

There are $\frac{p^3-p}3$ monic irreducible polynomials of degree $3$, according to this argument.