Prove that there is a quadratic irreducible in $\mathbb Z_p[x]$ for $p$ prime.

Question

I am kind of stuck at this problem. I know how to find irreducible polynomials in fields like $\mathbb Z_1,\mathbb Z_2$ or where the specific field is given.

I can't figure out a way to prove this. I am thinking to use theorems like: If an irreducible polynomial $p(x)$ belongs to a field $F$, then $F[x]/\langle p(x)\rangle$ is a field.

Or should I use arguments like if $\mathbb Z_p$ is $\mathbb Z_7$, then polynomials like $x^2 + 3$ are irreducible. So, there exists a quadratic irreducible in $\mathbb Z_p[x]$?

Or is there any other method that I should follow? Any suggestions will be helpful.

Answer 1

There are $p^2$ monic quadratic polynomials in $\Bbb Z/p \Bbb Z[x]$ but only $p+ \binom p2$ distinct products of monic linear polynomials, so since $p+ \binom p2 \lt p^2$, there is some monic quadratic polynomial that is not a product of linear factors. That quadratic polynomial must be irreducible.

To see that the inequality holds, $p+ \binom p2 = \binom p1 + \binom p2 =\binom{p+1}{2}$ and a little arithmetic shows that $p^2-\binom{p+1}{2}=\binom p2 \gt 0$ for all primes $p$.

Answer 2

Just a comment on the very nice answer of @Robert Shore:

There are exactly $\frac{p(p-1)}{2}$ irreducible polynomials of degree $2$ in $\mathbb{Z}/p[x]$. They are of the form (assume $p$ odd) $$(x-a)^2 - b$$ where $a\in \mathbb{Z}/p$ and $b$ is a quadratic non-residue $\mod p$. There are $p$ choices for $a$ and $\frac{p-1}{2}$ choices for $b$.

Answer 3

A quadratic polynomial is irreducible if and only if it has no roots.

This easily implies that $x^2+x+1$ is irreducible if $p=2$.

If $p\neq 2$, then for any nonzero $a\in \mathbf Z_p$, you have $a\neq -a$ and $a^2=(-a)^2$. Thus, the function $x\mapsto x^2$ is non-injective, so (because $\mathbf Z_p$ is finite), it is non-surjective. It follows that for some $a$, the polynomial $x^2-a$ has no root.

Note that this proof works for any finite field of characteristic $p>2$.