TITLE: How many homomorphisms from $\frac{\mathbb{Z}[x,y]}{(x^3+y^2-1)}$ to $\frac{\mathbb{Z}}{(7)}$?
QUESTION [5 upvotes]: How many homomorphisms from $\displaystyle\frac{\mathbb{Z}[x,y]}{(x^3+y^2-1)}$ to $\displaystyle\frac{\mathbb{Z}}{(7)}$?
I already checked that $x^3+y^2-1$ has these five integer roots: $(-2,3),(-2,-3),(0,1),(0,-1)$ and $(1,0)$, so it is not an irreducible polynomial in $\mathbb{Z}$.
The ring $\displaystyle \frac{\mathbb{Z}}{(7)}$ is the same as $\displaystyle\frac{\mathbb{Z}}{7\mathbb{Z}}$, right? Since $7$ will generate $7\mathbb{Z}$, so this ring $\displaystyle \frac{\mathbb{Z}}{(7)}$ is a finite ring of $7$ elements. But then how can I find the number of homomorphisms? I am also failing to analyze the order of elements that will generate the domain ring.
Thanks.

REPLY [6 votes]: The homomorphisms from $\Bbb Z[x,y]/(f(x,y))$ to $\Bbb Z/p\Bbb Z$
are the homomorphisms from $\Bbb Z[x,y]$ to $\Bbb Z/p\Bbb Z$
which send to $f(x,y)$ to zero. 
The homomorphisms from $\Bbb Z[x,y]$ to $\Bbb Z/p\Bbb Z$ are the maps
$$\phi_{a,b}:h(x,y)\mapsto h(a,b)$$
where $a$, $b\in\Bbb Z/p\Bbb Z$. So $\phi_{a,b}$ kills $f(x,y)$
iff $f(a,b)=0$. So the number of  homomorphisms from $\Bbb Z[x,y]/(f(x,y))$
to $\Bbb Z/p\Bbb Z$ is the number of solutions of $f(a,b)=0$ over
$\Bbb Z/p\Bbb Z$.
In your example, you are counting the number of solutions of $x^3+y^2-1=0$
modulo $7$, so counting points on an elliptic curve.