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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.

For any commutative ring $R,$ there is a unique homomorphism $f : Bbb Zto R,$ as specifying that $1mapsto 1$ tells you where each element of $Bbb Z$ must map to. Concretely, for any positive $ninBbb Z,$ you have $n = 1 + dots + 1$ ($n$ times,) so that $$f(n) = f(1 + dots + 1) = f(1) + dots + f(1) = ncdot f(1).$$
You also know that $f(0) = 0,$ and if $ninBbb Z$ is negative, then $f(n) = f(-(-n)) = -f(-n).$ Thus, there is exactly one morphism $f : Bbb Zto R,$ because the image of any element is determined by where $1$ is sent and the ring homomorphism rules.

Now, let's examine what it takes to define a morphism $Bbb Z[x]to R.$ We already know that we don't have a choice for where $Bbb ZsubseteqBbb Z[x]$ is sent. What about $x$? Well, it turns out we can send $x$ to any element of $R$ that we want.

Suppose that $rin R.$ If $f : Bbb Z[x]to R$ is a ring homomorphism sending $x$ to $r,$ then the ring homomorphism properties imply that we must have
begin{align*}
fleft(sum_{i = 0}^n a_i x^iright) &= sum_{i = 0}^nfleft( a_i x^iright)\
&= sum_{i = 0}^n f(a_i) f(x^i)\
&= sum_{i = 0}^n f(a_i) f(x)^i\
&= sum_{i = 0}^n f(a_i) r^i.
end{align*}
Every element of $Bbb Z[x]$ is of the form $sum_{i = 0}^n a_i x^i$ for some $n$ and some collection of integers $a_i,$ so we see that specifying where $x$ is sent determines the entire homomorphism. In particular, it is given by
$$
p(x) = sum_{i = 0}^n a_i x^i mapsto sum_{i = 0}^n f(a_i) r^i = p(r).
$$

Conversely, setting $f(x) = r$ and extending in the above way is always a ring homomorphism. That is, let $rin R$ and define
begin{align*}
f : Bbb Z[x]&to R\
sum_{i = 0}^n a_i x^i &mapsto sum_{i = 0}^n g(a_i) r^i,
end{align*}
where $g : Bbb Zto R$ is the unique ring homomorphism from paragraph one. We still have $f(1) = g(1) = 1$ and for any $n,minBbb ZsubseteqBbb Z[x],$ we have $$f(n + m) = g(n + m) = g(n) + g(m) = f(n) + f(m).$$ Suppose we have two arbitrary polynomials $sum_{i = 0}^n a_i x^i,$ $sum_{i = 0}^m b_i x^i.$ Then without loss of generality $mleq n,$ and we can say that $sum_{i = 0}^m b_i x^i = sum_{i = 0}^n b_i x^i,$ where we define $b_j = 0$ for $j > m.$ Then we have
begin{align*}
fleft(sum_{i = 0}^n a_i x^i + sum_{i = 0}^n b_i x^iright) &= fleft(sum_{i = 0}^n (a_i + b_i) x^iright)\
&= sum_{i = 0}^n g(a_i + b_i)r^iqquadtextrm{(by definition)}\
&= sum_{i = 0}^n left(g(a_i) + g(b_i)right)r^i\
&= sum_{i = 0}^n (g(a_i)r^i + g(b_i)r^i)\
&= sum_{i = 0}^n g(a_i)r^i + sum_{i = 0}^n g(b_i)r^i\
&= fleft(sum_{i = 0}^n a_i x^iright) + fleft(sum_{i = 0}^n b_i x^iright).
end{align*}
You can check similarly that $$fleft(left(sum_{i = 0}^n a_i x^iright)cdotleft(sum_{i = 0}^n b_i x^iright)right) = fleft(sum_{i = 0}^n a_i x^iright)cdot fleft(sum_{i = 0}^n b_i x^iright),$$
so that the map defined is indeed a ring homomorphism.

The case of two variables is similar - a ring homomorphism is completely determined by where each variable is sent, and any choice of $r,sin R$ gives a ring homomorphism with $xmapsto r$ and $ymapsto s.$

What's happening is that $Bbb Z[x,y]$ is the free commutative ring on two generators $x$ and $y,$ which means essentially what I stated above - a ring homomorphism from $Bbb Z[x,y]$ is given by a choice of where $x$ and $y$ will be sent, and any choices will work. To make sure that this defines a map on the quotient you want, you need to further specify that the images of $x$ and $y$ satisfy the given relation - i.e., because $x^3 + y^2 - 1 = 0$ in $Bbb Z[x,y]/(x^3 + y^2 - 1),$ we must have $f(x)^3 + f(y)^2 - 1 = 0$ as well.