$R$ is defined as \begin{bmatrix}a&b\\b&a\end{bmatrix} such that $a,b \in \mathbb{Z}$ where R is a subring of all 2-by-2 matrices with integer inputs.
There exists a homomorphism $θ :R→Z$ such that $θ(R) = ra + sb$.
You are asked to prove that must $r = 1$ and $s = ± 1$.
I know the 2 points that must be satisfied for a homomorphism to happen between two rings and don't think these help to proving the question. I think you need to start with using the idea that $\ker(θ)$ is an ideal of $R$ and the $im(θ)$ is a subring of $\mathbb{Z}$. Am I right or is this wrong?
Thank you.
Assume $\theta:R\to\Bbb Z$ is a homomorphism of unital rings, mapping $\theta\pmatrix{a&b\\b&a}=ra+sb$.
Then specifically it maps the identity element ($I=\pmatrix{1&0\\0&1}$, i.e. when $a=1,\ b=0$) to $1$, which means $r=1$.
On the other hand, observe that for $B:=\pmatrix{0&1\\1&0}$ we have $B^2=I$, so we must also have $\theta(B)^2=\theta(I)=1$, where $\theta(B)=s$.