In every finite field there is 2 degree irreducible polynomial

Question

Given $F$ finite field, how to prove that there is $f \in F[x]$ such that $\deg(f) =2$ and $f$ is irreducible, meaning there is no two polynomial $g_1,g_2 \in F[x]$ such that $\deg(g_1) = 1 , \deg(g_2)=1$ and $f = g_1 g_2$ ?!

I thought that $x^2+1,x^2+2,\cdots , x^2+n-1$ given that the number of elements in $F$ is $n$, then at least one them is quadratic irreducible (but this way of thinking is more number theory than algebra)

So i also thought that if i could prove that there is more number of 2 degree polynomials than polynomial made of the product of 2 polynomial of degree 1, but could not proceed either.

Could any one provide an algebraic proof.

Thanks

Answer 1

If $F$ is not of characteristic $2$, then half-ish of the elements of $F$ are quadratic residues, so there are plenty of quadratic non-residues. (Yeah, I know, number theory, but one could hide the language.) If $a$ is a QNR, then $x^2-a$ is irreducible.

If $(x-b)(x-c) = x^2-a$ then $c=-b$ and $-a = bc = -b^2$, so $a$ is a QR, contradiction.

If $F$ is of characteristic $2$, then see Irreducible polynomial of degree $2$ over a finite field of characteristic $2$

Answer 2

Seems to me your "I also thought" works.

Say $F$ has $n$ elements. Then there are exactly $n^2$ monic quadratic polynomials $x^2+bx+c$. But the number of polynomials of the form $(x-a)(x-b)$ is less than $n^2$, because $(x-a)(x-b)=(x-b)(x-a)$.

Answer 3

There are lot of ways (more or less direct to answer to your question) but let me say that your second idea works well so I would take it as a demonstration. To finish your second proof just count explicitly the number of 2 degree polynomials.

If you want a more algebraic reason , just remember that for every finite field you have an extension of field (say $F \subset F'$) such that $[F':F]=2$ and every such extension comes from a root of some irreducible 2 degree polynomial with coefficients in $F$