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.

  • 1
    $\begingroup$ It isn't that those $49$ maps are homomorphisms because the identity gets mapped to the identity, but that homomorphisms will be determined by how they act on the generators of the ring. The remark that $1$ goes to $1$ is just showing that you have no choice when it comes to that generator. You are correct that you need to care about when such choices of how to map generators results in a homomorphism, but that is a finite problem in your case, and you should be able to handle it. $\endgroup$ Dec 21, 2018 at 3:03

1 Answer 1


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.

  • $\begingroup$ Your answer is quite understandable. But will you help me in getting that "Conversely part". I am stuck at showing from converse that $f$ will be a ring homomorphism. @Stahl $\endgroup$ Dec 21, 2018 at 3:27
  • $\begingroup$ @hiren_garai There are three things that must be true of $f$ to be a ring homomorphism. 1. $f(1) = 1,$ 2. $f(a + b) = f(a) + f(b),$ 3. $f(ab) = f(a)f(b)$ for any $a,b$ in the domain. My argument shows that if you specify that $f(1) = 1,$ $f(x) = r,$ then applying the rules $f(a + b) = f(a) + f(b)$ and $f(ab) = f(a)f(b)$ gives a well defined ring homomorphism. First, examine what $f(p(x))$ is for any polynomial $p(x)$ (you have $f(p(x)) = p(r)$). Next, show that $f(p(x) + q(x)) = f(p(x)) + f(q(x))$ for any polynomials $p$ and $q$ - I did this for you in my computation. $\endgroup$
    – Stahl
    Dec 21, 2018 at 3:32
  • $\begingroup$ Finally, show that $f(p(x)q(x)) = f(p(x))f(q(x)).$ I didn't do this in the answer, but the computation follows the same strategy as showing $f(p(x) + q(x)) = f(p(x)) + f(q(x)).$ First, simplify $p(x)q(x)$ into an explicit polynomial $P(x)$, and then replace $x$ by $r.$ This tells you $f(p(x)q(x)) = f(P(x)) = P(r).$ Then, unwind the definition of the product of two polynomials to show that $f(p(x))f(q(x)) = p(r)q(r)$ is equal to $P(r).$ $\endgroup$
    – Stahl
    Dec 21, 2018 at 3:34
  • $\begingroup$ I get it, but how $f$ moved from outside to the inside of the summation in computing $f( p(x) + q(x)) = f(p(x)) + f(q(x))$ in the second line? $\endgroup$ Dec 21, 2018 at 3:38
  • 1
    $\begingroup$ Because I know that if I want $f$ to be a homomorphism, I need to define it on individual monomials $ax^n$ as $f(ax^n) = f(a) f(x^n) = f(a)f(x)^n.$ This tells you how you need to define it on individual polynomials, because you must have $f(ax^n + bx^m) = f(ax^n) + f(bx^m),$ and so this means that for a single polynomial you must have $f(p(x)) = p(r).$ So you may define $f(p(x)) = p(r),$ and from there it is a direct computation (as above) to show that this is a homomorphism. $\endgroup$
    – Stahl
    Dec 21, 2018 at 3:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.