My question is related to this question. For convenience I'm giving the question :
How many ring homomorphisms there is between $\mathbb{Z}[x,y]/(x^3+y^2-1)$ and $\mathbb{Z_7}$? Here $\mathbb{Z_7}$ denote ring of integers mod 7. I don't know how to approach this problem,so far I've only worked with one variable ring so I can't tell the properties of $\mathbb{Z}[x,y]/(x^3+y^2-1)$. Thanks in advance.
In the answer of this question, this answer is pretty straight forward and I was able to understand it. The answer given by Arthur is as follows :
Note that any homomorphism $\mathbb{Z}[x,y]/(x^3+y^2-1) \to \Bbb Z_7$ induces by composition a unique canonical homomorphism $\Bbb Z[x, y]\to \Bbb Z_7$, so we can start by looking at those, because that's a lot easier.
Let's say we have a homomorphism $f$. The ring $\mathbb{Z}[x,y]$ has three generators, $1, x$ and $y$. The homomorphism has to send $1$ to $1$, which leaves in total $49$ possibilities for $f(x)$ and $f(y)$.
That would be the final answer if we were interested in maps from $\Bbb Z[x, y]$ to $\Bbb Z_7$. However, we are interested in maps from $\mathbb{Z}[x,y]/(x^3+y^2-1)$ to $\Bbb Z_7$. By the universal property of quotient rings, this is equivalent to counting the homomorphisms $\Bbb Z[x, y]\to \Bbb Z_7$ whose kernel contains the ideal $(x^3 + y^2 -1)$.
That specifically means that we want $f(x)$ and $f(y)$ to satisfy the relation $f(x)^3 + f(y)^2 - 1 = 0$, which limits the possibilities greatly. For instance, if $f(x) = 0$, then we must have $f(y)^2 = 1$, which has two solutions: $1$ and $6$. Thus there are two possible homomorphisms with $f(x) = 0$. Do this for the $6$ remaining possible $f(x)$, and you should have your answer.
But I have a question regarding the answer.
- How can I be sure that those 49 mappings are in fact ring homomorphisms by only saying that it takes the multiplicative identity to the multiplicative identity ?
I was thinking that all ring homomorphisms must be group homomorphisms, but this didn't help me a lot.
Any insight. Thank you.