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Let $\mathbb{F}_q$ be a finite field of order $q=p^n$ for some prime $p$ and $n>1$. Suppose both $f(x)=x^2-ax+b$ and $g(x)=x^2-a'x+b'$ are both irreducible.

If, assuming that either $a=a'=0$ or $a,a'\neq 0$, can we conclude $\exists\alpha \in$ Aut$(\mathbb{F}_q)$ such that $\alpha(a)=a'$ and $\alpha(b)=b'$ so that $g(x)=x^2-\alpha(a)x+\alpha(b)$?

Updated Questions

  1. If $g(x)=x^2+a'x+b'$ is irreducible, does there exist $f(x)=x^2+ax+b$ irreducible and a non trivial $\alpha$ such that $\alpha(a)=a'$ and $\alpha(b)=b'$?
  2. If $g(x)=x^2+a^kx+b^k$ is irreducible, then must $k=p^i$?
Mark Greer
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2 Answers2

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No.

There are $q^2$ degree $2$ monic polynomials, and $q(q+1)/2$ products of monic linear factors, which leaves $q(q-1)/2$ monic irreducible polynomials of degree $2$.

Meanwhile, the group of automorphisms of $\Bbb F_q$ has cardinal $n$, so there is no way that the nearly $p^{2n}/2$ different irreducible polynomials can be in a single orbit of size $n$.

Even if you restrict to $a=a'=0$ this leaves you with $(q-1)/2$ polynomials (for odd $q$, there is none if $p=2$ anyway), which is not much closer to $n$.

mercio
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  • Using the notation in http://math.stackexchange.com/questions/172980/galois-field-gf4 , the six irreducible monic polynomials (degree 2) are: $x^2+Bx+B,x^2+Dx+D,x^2+Bx+1,x^2+Dx+1,x^2+B,x^2+D$. Since $B^2=D$ and $D^2=B$, we have the desired result. What am I missing? – Mark Greer May 20 '16 at 15:37
  • if $f(x) = x^2+Bx+B$ and $g(x)=x^2+Bx+1$ you cannot conclude that there is an $\alpha$ such that $g = \alpha(f)$. So the answer to your question is no. – mercio May 21 '16 at 10:14
  • Of course! I've updated my question to clarify what I'm looking for. Thanks for your insight! – Mark Greer May 21 '16 at 13:07
  • @MarkGreer Check that list of irreducible quadratics in $GF(4)[x]$. The polynomials $x^2+B$ and $x^2+D$ have (double) zeros in $GF(4)$ and cannot be irreducible. Six is the correct number though. You are missing $x^2+x+B$ and $x^2+x+D$. – Jyrki Lahtonen May 23 '16 at 07:15
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Answering the updated parts.

  1. Yes. For each non-trivial $\alpha$ the polynomial $f(x)=x^2+\alpha^{-1}(a')x+\alpha^{-1}(b')$ is irreducible whenever $g(x)=x^2+a'x+b'$ is. Applying the same automorphism to all the coefficients of an irreducible polynomial is an automorphism of the polynomial ring, and hence maps any irreducible polynomial to another irreducible polynomial.
  2. No. If you really meant this to be about $g(x)$ alone, then it is obviously "No", because by using $k$ such that $\gcd(k,p^n-1)=1$ we can use surjectivity of the mapping $x\mapsto x^k$ to get any polynomial we want written in that form. If you really meant that both $f(x)=x^2+ax+b$ and $g(x)=x^2+a^kx+b^k$ should be irreducible then the answer is still "No". A simple example is when $p>2$, $a=0$, and $b$ is a non-square. Then the polynomial $f(x)=x^2-b$ is irreducible. As is $g(x)=x^2-b^k$ for any odd exponent $k$.
Jyrki Lahtonen
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  • I'm afraid I'm still not sure whether you have asked the question you really wanted to ask. – Jyrki Lahtonen May 23 '16 at 07:22
  • Thanks! $1.$ seemed obvious and you're of course right about both $g(x)$ and $f(x)$ being irreducible. If we assume $g(x)=x^2+a^{k^l}x+b^{k^l}$ is irreducible ($i.e.$ all powers of $k$), I'm still guessing we can't conclude that $k=p^i$. Is this correct? – Mark Greer May 23 '16 at 14:35