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.
Best Answer
For any commutative ring $R,$ there is a unique homomorphism $f : \Bbb Z\to R,$ as specifying that $1\mapsto 1$ tells you where each element of $\Bbb Z$ must map to. Concretely, for any positive $n\in\Bbb Z,$ you have $n = 1 + \dots + 1$ ($n$ times,) so that $$f(n) = f(1 + \dots + 1) = f(1) + \dots + f(1) = n\cdot f(1).$$ You also know that $f(0) = 0,$ and if $n\in\Bbb Z$ is negative, then $f(n) = f(-(-n)) = -f(-n).$ Thus, there is exactly one morphism $f : \Bbb Z\to R,$ 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 $\Bbb Z\subseteq\Bbb Z[x]$ is sent. What about $x$? Well, it turns out we can send $x$ to any element of $R$ that we want.
Suppose that $r\in R.$ 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*} f\left(\sum_{i = 0}^n a_i x^i\right) &= \sum_{i = 0}^nf\left( a_i x^i\right)\\ &= \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 $r\in R$ 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 $g : \Bbb Z\to R$ is the unique ring homomorphism from paragraph one. We still have $f(1) = g(1) = 1$ and for any $n,m\in\Bbb Z\subseteq\Bbb Z[x],$ 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 $m\leq n,$ 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*} f\left(\sum_{i = 0}^n a_i x^i + \sum_{i = 0}^n b_i x^i\right) &= f\left(\sum_{i = 0}^n (a_i + b_i) x^i\right)\\ &= \sum_{i = 0}^n g(a_i + b_i)r^i\qquad\textrm{(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\\ &= f\left(\sum_{i = 0}^n a_i x^i\right) + f\left(\sum_{i = 0}^n b_i x^i\right). \end{align*} You can check similarly that $$f\left(\left(\sum_{i = 0}^n a_i x^i\right)\cdot\left(\sum_{i = 0}^n b_i x^i\right)\right) = f\left(\sum_{i = 0}^n a_i x^i\right)\cdot f\left(\sum_{i = 0}^n b_i x^i\right),$$ 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 $r,s\in R$ gives a ring homomorphism with $x\mapsto r$ and $y\mapsto s.$
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.