Let $A=(a_{ij})\in\operatorname{SL}_2(\mathbb{F}_p)$. Consider the ring map $A:\mathbb{F}_p[x,y]\to\mathbb{F}_p[x,y]$ defined by

$$A(x)=a_{11}x+a_{21}y$$ $$A(y)=a_{12}x+a_{22}y$$

and extended multiplicatively. Are there non-constant polynomials $f(x_1,x_2)\in\mathbb{F}_p[x,y]$ such that $f(x,y)=A(f(x,y))$ for all $A\in\operatorname{SL}_2(\mathbb{F}_p)$?

I tried to solve this problem by writing $f(x,y)=\sum\lambda_{ij}x^iy^j$, and discovering different restrictions on the $\lambda_{ij}$ for particularly nice choices of $A$. This quickly got complicated and so I'm hoping this has been studied before and there is a more elegant solution.

Edit: After working through this for some small primes, I believe that the following two polynomials are fixed by all $A$:

$$f(x,y)=x^py-xy^p$$ $$g(x,y)=\sum_{i=0}^px^{(p-i)(p-1)}y^{i(p-1)}$$

However, I can't prove that $g(x,y)$ is fixed for all $p$, and it's not clear to me that these two polynomials generate the entire subalgebra of $\mathbb{F}_p(x,y)$ invariant under the action of $\operatorname{SL}_2(\mathbb{F}_p)$. Is there an invariant $h(x,y)$ not in the subalgebra generated by $f$ and $g$?

  • $\begingroup$ If instead of $\Bbb{F}_p$ we had an algebraically closed field of the same characteristic the representation theory of $SL_2$ would give a lot of answers. I need to think about this case, but the following observations can be made immediately. 1) Subspaces $V_n$ of homogeneous polynomials (of a fixed degree $n$) are stable under this action, so it suffices to find the homogeneous invariants $f(x,y)=\sum_{i=0}^n\lambda_ix^{n-i}y^i$. 2) A diagonal matrix with entries $(a,1/a)$ acts on $x^{n-i}y^i$ by a scalar $a^{n-2i}$. For this scalar to be $=1$ for all $a\in\Bbb{F}_p^*$ it is necessary $\endgroup$ – Jyrki Lahtonen Sep 26 '15 at 7:25
  • $\begingroup$ (cont'd) and sufficient that $n-2i$ is a multiple of $p-1$ whenever $\lambda_i\neq0$. This narrows down the search considerably. IIRC trivial representations (=invariants) appear as composition factors of $V_n$ for some $n$, but not necessarily as submodules, which is what you are looking for. I gotta rush now. Hopefully more later. $\endgroup$ – Jyrki Lahtonen Sep 26 '15 at 7:28
  • $\begingroup$ @JyrkiLahtonen: Thanks for your suggestions. If you're still interested, note the edit. $\endgroup$ – Jared Sep 27 '15 at 1:17

These are the Dickson's polynomials. If we let $$h(x,y)=x^{p^2}y-y^{p^2}x$$ then $h$ is also invariant under $SL(2,p)$ action. It then follows that $$g(x,y)=\frac{h(x,y)}{f(x,y)}$$ is invariant.

For a general $n$, let

$$L_{n,s}=\left|\begin{array}{cccc} x_1&x_2&\cdots&x_n\\ x_1^{p}&x_2^{p}&\cdots&x_n^p\\ \cdots&\cdots&\cdots&\cdots\\ \widehat{x_1^{p^s}}&\widehat{x_2^{p^s}}&\cdots&\widehat{x_n^{p^s}}\\ \cdots&\cdots&\cdots&\cdots\\ x_1^{p^n}&x_2^{p^n}&\cdots&x_n^{p^n} \end{array}\right|$$

where the $p^s$ row is omitted. Then the Dickson's invariant for $SL(n,p)$ are $\dfrac{L_{n,s}}{L_{n,n}}$.

And you are correct, these two generate the whole invariant algebra. I don't know the details but they are likely in Dickson's works.


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