Is the space $R^2$ for a ring $R=\mathbb{Z}/p\mathbb{Z}$ the sum of two invariant lines? I didn't do a very good job of summarising the problem in the title, sorry, but here is the full question:
Let us define, for every ring $R$, the set
$S(R)=$ {$\rho : \mathbb{Z}/2\mathbb{Z} \to GL(2,R) | \rho (0+2\mathbb{Z}) \ne \rho (1+2\mathbb{Z}) $} 
of representation of the group $\mathbb{Z}/2\mathbb{Z}$ on $R^2$, where $R^\times$ is the subset of elements of $R$ with a multiplicative inverse, and that $GL(2, R)$ is the group of size 2 square matrices with entries in $R$, which are invertible, and whose inverse has coefficients in $R$ as well.
Consider the following property:
$(P(\rho,R))$ The space $R^2$ is a sum of two invariant lines, that is:
$\exists  v, w \in R^2$ such that $\forall g \in \mathbb{Z}/2\mathbb{Z}, \rho(g)(R\cdot v)=R \cdot v $ and $\rho(g)(R\cdot w)=R \cdot w$ and that $R \cdot v + R \cdot w = R^2.$
For every $R$ of the form $\mathbb{Z}/p\mathbb{Z}$ with $p$ a prime, does property $(P(\rho,R))$ hold for all $\rho$, some $\rho$ or no $\rho$ in $S(R)$?
I am struggling to make progress with this but I have started by looking at the problem when $p=3$, and letting $\rho(0)=\begin{bmatrix}1 & 0\\1 & 1\end{bmatrix}$, $\rho(1)=\begin{bmatrix}1 & 1\\0 & 1\end{bmatrix}$, $v=(v_1,v_2)$ and $w=(w_1,w_2)$. I am not sure how a matrix multiplies a ring so I am not sure how to proceed with the following; 
$\begin{bmatrix}1 & 0\\1 & 1\end{bmatrix} R \begin{bmatrix}1 & 0\\1 & 1\end{bmatrix}(v_1,v_2)$
I would greatly appreciate any help.
 A: The answer is all if $p\neq 2$.
Consider the representation $\rho(0)=\pmatrix{1& 0\\0& 1}$ and $\rho(1)=\pmatrix{-1& -1\\ 0& 1}$
Then the lines $L_1=\langle\pmatrix{-1\\ 2}\rangle$ and $L_1=\langle\pmatrix{1\\ 0}\rangle$ (These are basically the eigenvectors of $\rho(1)$ with eigen values $1$ and $-1$ respectively) are both invariant under both $\rho(0)$ and $\rho(1)$. The only problem is when $R=\mathbb Z/2\mathbb Z$ because both the eigenvectors are the same in this ring.
To solve the problem completely, notice that $\rho$ is determined by which order two element of $GL_2(\mathbb Z/p \mathbb Z )$ that $1$ is mapped to. You need to find all order two elements of $GL_2(\mathbb Z/p \mathbb Z )$ and find their eigenvectors and show that there are two distinct linearly independant ones. Since any matrix must satisfy its own characteristic polynomial , it follows that $x^2-1$ divides the characteristic polynomial of $\rho(1)$, which must be monic and have degree two, so it must be $x^2-1$ itself. This polynomial has identical roots if and only if $1=-1$ which only happens if $R=Z/2\mathbb Z$. So if $p\neq 2$, the matrix has distinct eigenvalues and hence two linearly independant eigenvectors whose span will yield $L_1$ and $L_2$ as above.
If $R=Z/2\mathbb Z$, there are only three non-identity elements of order two. Can you find out which these are and check their eigenvectors to finish?
