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.



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 $2$

  • 1
    $\begingroup$ $F = \Bbb Z_2$ has QNR? $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Mar 24 '18 at 14:48
  • $\begingroup$ In fact surely any field of characteristic $2$ gives a counterexample: The map $x\mapsto x^2$ is injective, hence surjective. (???) $\endgroup$ – David C. Ullrich Mar 24 '18 at 14:52
  • $\begingroup$ @GNUSupporter I said, "ish". $\endgroup$ – B. Goddard Mar 24 '18 at 15:49
  • $\begingroup$ Learnt a new Engl-ish word. Btw, I didn't upvote nor downvote. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Mar 24 '18 at 15:55

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 $(x-a)(x-b)=(x-b)(x-a)$.


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 $F$

  • $\begingroup$ The first idea doesn't work in characteristic $2$, does it? It certainly doesn't work in $\Bbb Z_2$... $\endgroup$ – David C. Ullrich Mar 24 '18 at 14:53
  • $\begingroup$ Regarding the "more algebraic reason": I don't know any algebra. Is there a simple way to prove the existence of that extension, other than by showing there is an irreducible quadratic? $\endgroup$ – David C. Ullrich Mar 24 '18 at 14:56
  • $\begingroup$ You're completely right. The first method clearly does not work in arbitrary charateristic. Sorry for the wrong answer $\endgroup$ – Tommaso Scognamiglio Mar 24 '18 at 16:26
  • $\begingroup$ About the algebraic reason:what do you know about finite fields? Because apart from the finite fields $\mathbb{Z_p}$ for some p , the other ones are constructed as algebraic finite extension of finite field of that simple type $\endgroup$ – Tommaso Scognamiglio Mar 24 '18 at 16:28

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