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Let $\mathbb{F}_p$ be a finite field of order $p$ and $G$ be the general affine group of degree one over this finite field. Further let $V$ denote the quadratic polynomials over $\mathbb{F}_p$. I want to understand how the action of $G$ on $V$ decompose into irreducibles and construct a decomposition. What would be good references to look at?

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I assume you mean to decompose $V$ into orbits since $V$ is only a $G$-set. I'm also going to guess that the action of $\alpha\in G=\mathbb{F}_p$ on $V$ is $\alpha.f(x)=f(x+\alpha)$. In this case, every orbit is of size $p$ unless $p=2$ in which case there is one orbit of size 2 and two orbits of size 1.

To see this, note that by the generalized class equation every orbit has size either 1 or $p$. Now, there are $\displaystyle (p-1){p+1\choose 2}$ reducible quadratic polynomials of the form $f(x)=k(x-a)(x-b)$ with $0<k<p$ and $0\leq a\leq b<p$. A polynomial $g(x)=\ell(x-c)(x-d)$ is in the orbit of $f(x)$ (as above) if and only if $k=\ell$ and $b-a=d-c$. When $p>2$ this yields $(p-1)(p+1)/2$ orbits of size $p$. When $p=2$, we get one orbit of size 2 while $x^2+x$ determines an orbit of size 1.

Now, the remaining $\displaystyle \frac{(p-1)^2p}{2}$ polynomials in $V$ are irreducible of the form $h(x)=k(x^2+ax+b)$ (note that there are a total of $(p-1)p^2$ quadratic polynomials over $\mathbb{F}_p$). Then, $$\alpha.h(x)=k(x^2+(2\alpha+a)x+(a\alpha+\alpha^2+b))$$ for any $\alpha\in G$. In particular, $h$ is fixed by the generator $1\in G$ if and only if $2+a\equiv a$ (mod $p$) and $a+b+1\equiv b$ (mod $p$). This only happens for the polynomial $x^2+x+1\in\mathbb{F}_2$. Hence, for $p>3$, we get another $(p-1)^2/2$ orbits of size $p$.

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  • $\begingroup$ Note that if $f(x) \in \mathbb{F}_{p}[x]$ is a non-zero polynomial satisfying $f(x)=f(x+\alpha)$ for some $\alpha \in \mathbb{F}_{p}^{\times}$, then necessarily $\deg f \ge p$: This is because if $r$ is a root of $f$ (over the algebraic closure) then its translates $r+i,i\in \mathbb{F}_{p}$ must also be roots. $\endgroup$ Jul 13, 2015 at 21:41
  • $\begingroup$ That is a good point. $\endgroup$
    – David Hill
    Jul 13, 2015 at 21:45
  • $\begingroup$ I am a bit confused: Why is $V$ not a representation of $G$ over $\mathbb{F}_q$? $\endgroup$
    – Seppo
    Jul 14, 2015 at 15:35
  • $\begingroup$ @Thomas $V$ is not closed under addition. It is just a set. $\endgroup$
    – David Hill
    Jul 14, 2015 at 15:37

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