Let $G=Sl\_2(\mathbb{F}\_p)$ and $M= \mathbb{F}\_p[x\_1,x\_2]$, where $p$ is a prime. 


$M$ is a $G$-module with $(A\cdot x\_1, A\cdot x\_2)=(x\_1,x\_2)\cdot A, (\forall) A \in Sl\_2(\mathbb{F}\_p)$.


I have to show that 


$$M^G = \mathbb{F}\_p[x\_1\cdot v, x\_1^{p\cdot(p-1)}+ v^{p-1}],$$


where 


$$v=x\_2\cdot(x\_2+x\_1)\cdot(x\_2+2\cdot x\_1)\cdots(x\_2+(p-1)\cdot x\_1 )$$ and 


$$M^G=\lbrace m\in M \mid g\cdot m= m, \forall g\in G\rbrace.$$


I posted [on M.O.](https://mathoverflow.net/questions/279197/how-to-show-mg-mathbbf-2x-1-cdot-v-x-12-v) and M.S.E. the particular case when $p=2$ but I didn't recive any idea.



How to prove that $M^G=\mathbb{F}_p[x_1\cdot v, x_1^{p\cdot(p-1)}+ v^{p-1}]$?
Let $R$ be the ring that you think is equal to $M^G$. You need to check that $R$ is fixed by $G$ and that $M$ is a free module of rank equal to $|G|$ over $R$. Then Galois theory tells you that the field of fractions $QR$ is $(QM)^G$. Moreover, as $R$ is a polynomial ring it is a unique factorisation domain and so is integrally closed in $QR$, and from that you can deduce that $R=M^G$. If you search for "Dickson invariants" you will find more details (usually written for $GL\_n(\mathbb{F}\_p)$, but the special linear cases is similar).


