Homomorphism $\mathbb{Z}[x,y] \rightarrow \mathbb{Z}_{7}$.

I have question to this answer, in homomorphism $\mathbb{Z}[x,y] \rightarrow \mathbb{Z}_7$. I undestand that we have to send $1 \mapsto 1$ to have homomorphism. Author of answer said that if we send $1\mapsto 1$ we have $49$ options of $f(x)$ and $f(y)$. But I do not understand why we can send $x \mapsto 0$ or $y\mapsto 0$.

$0$ is not a generator of $\mathbb{Z}_7$. I suspect that If we are sending $1\mapsto 1$ we get whole $\mathbb{Z}_7$ in image of $f$ but I can not see why it is a homomorphism then. So why we can send $x \mapsto 0$ or $y\mapsto 0$?

Solution 1:

In general, there is a 1-1 correspondence, for rings $A$ and $B$:

\begin{align*} \{\text{ring homomorphisms }A[x]\to B\}&\leftrightarrow\{\text{ring homomorphisms }A\to B\}\times B\\ f&\mapsto(f|_A,f(x))\\ \big(\sum_{i=0}^ka_ix^i\mapsto\sum_{i=0}^k\varphi(a_i)b^i\big)&\leftarrow(\varphi,b). \end{align*} More generally, there is a 1-1 correspondence between ring homomorphisms $A[x_1,\dots,x_n]\to B$ and ring homomorphisms $A\to B$, together with an $n$-tuple of elements of $B$.

For any ring $A$ there is always a unique homomorphism $\mathbb Z\to A$ (i.e., $\mathbb Z$ is the initial object), so that ring homomorphisms $\mathbb Z[x_1,\dots,x_n]\to A$ correspond to $n$-tuples of elements of $A$.

This phenomenon can be taken further. For example, ring homomorphisms $\mathbb Z[x,y]/(x^2+y^2-1)\to A$ correspond to pairs $(a,b)\in A$ with $a^2+b^2=1$. Ring homomorphisms $\mathbb Z[\sqrt2]\cong\mathbb Z[x]/(x^2-2)\to A$ correspond to elements $a\in A$ with $a^2=2$.