How to show evaluation map $\mathbb{Z}[x,y] \rightarrow R$ given by $\text{ev}_{(r,s)}(xy)=rs$ is a ring homomorphism? $R$ is commutative Show evaluation map $\mathbb{Z}[x,y] \rightarrow R$, $R$ is commutative, given by $\text{ev}_{(r,s)}(xy)=rs$ is a ring homomorphism? Where $r,s \in R$
I am confused on how to show this I would try to do this, but it doesn't seem right to me. I have seen an answer that is too complicated for the tools I have at hand.
Attempt: $$\text{ev}_{(r,s)}((f+g)(x,y))=\text{ev}_{(r,s)}(f(x,y)+g(x,y))=f(r,s)+g(r,s)=\text{ev}_{(r,s)}(f(x,y))+\text{ev}_{(r,s)}(g(x,y))$$
and $$\text{ev}_{(r,s)}((fg)(x,y))=\text{ev}_{(r,s)}(f(xy)g(xy))=f(rs)g(rs)=\text{ev}_{(r,s)}f(x,y)\text{ev}_{(r,s)}g(x,y)$$
 A: Let $f \in \mathbb Z[x, y]$, then you can write $f = \sum_{i = 0}^n \sum_{j = 0}^m a_{ij}x^iy^j$. We can view $f$ as a polynomial in $R[x, y]$ by sending $f$ to the polynomial $\tilde{f} = \sum_{i = 0}^n \sum_{j = 0}^m (a_{ij} \cdot 1)x^iy^j$. Note this is what you suggested doing in one of your earlier comments.
Then you can write $\mathrm{ev}_{(r, s)} : \mathbb Z[x, y] \to R$ as the composition $\mathrm{ev}_{(r, s)} = \widetilde{\mathrm{ev}}_{(r, s)} \circ \varphi$, where $\varphi : \mathbb Z[x, y] \to R[x, y]$ is the map such that $\varphi(f) = \tilde{f}$ and $\widetilde{\mathrm{ev}}_{(r, s)} : R[x, y] \to R$ is the map sending a polynomial $g \in R[x, y]$ to $\widetilde{\mathrm{ev}}_{(r, s)}(g) = g(r, s)$.
Since the map $\mathbb Z \to R$ sending $a \to a \cdot 1$ is a ring homomorphism, you can verify that $\varphi$ is a ring homomorphism (since everything happens coefficientwise).
To verify that $\widetilde{\mathrm{ev}}_{(r, s)}$ is a ring homomorphism, note that for the constant polynomial $1 \in R[x, y]$ you have $$\widetilde{\mathrm{ev}}_{(r, s)}(1) = 1,$$
and for $g, g' \in R[x, y]$ you have $$\widetilde{\mathrm{ev}}_{(r, s)}(g + g') = (g + g')(r, s) = g(r, s) + g'(r, s) = \widetilde{\mathrm{ev}}_{(r, s)}(g) + \widetilde{\mathrm{ev}}_{(r, s)}(g').$$
I'll leave verifying that $\varphi$ is a ring homomorphism and that $\widetilde{\mathrm{ev}}_{(r, s)}(g g') = \widetilde{\mathrm{ev}}_{(r, s)}(g)\widetilde{\mathrm{ev}}_{(r, s)}(g')$ to you.
Once you have shown this, you can conclude that since $\mathrm{ev}_{(r, s)}$ is the composition of the ring homomorphisms $\widetilde{\mathrm{ev}}_{(r, s)}$ and $\varphi$, it is a ring homomorphism.
