Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have two ring homomorphism $f,g\colon \mathbb{Q}\to X$. I know that $f=g$ on the integers, how can I show that $f$ and $g$ agree on the rationals?

My attempt:

let $x,y \in \mathbb{Z}$, $y\neq 0$. $$\begin{align*} f(x/y)&=f(x)f(1/y) &&\text{because }f\text{ is a ring homomorphism}\\ &=g(x)f(1/y) &&\text{because }f\text{ and }g\text{ agree on the integers} \end{align*}$$

I do not know how to rewrite $f(1/y)$ so that I can replace it with $g(1/y)$. Any hints will be much appreciated. Thanks.

share|cite|improve this question
up vote 20 down vote accepted

This can be done by using a "zig-zag": $$\begin{align*} f\left(\frac{x}{y}\right) &= f\left(\frac{1}{y}\cdot x\right)\\ &= f\left(\frac{1}{y}\right)\cdot f(x)\\ &= f\left(\frac{1}{y}\right)\cdot g(x)\\ &= f\left(\frac{1}{y}\right)\cdot g\left(xy\cdot \frac{1}{y}\right)\\ &= f\left(\frac{1}{y}\right)\cdot g(xy)\cdot g\left(\frac{1}{y}\right)\\ &= f\left(\frac{1}{y}\right)\cdot f(xy)\cdot g\left(\frac{1}{y}\right)\\ &= f\left(\frac{1}{y}\cdot xy\right)\cdot g\left(\frac{1}{y}\right)\\ &= f(x)\cdot g\left(\frac{1}{y}\right)\\ &= g(x)\cdot g\left(\frac{1}{y}\right)\\ &= g\left(x \cdot \frac{1}{y}\right)\\ &= g\left(\frac{x}{y}\right). \end{align*}$$

This proves, by the by, that the embedding $\mathbb{Z}\hookrightarrow\mathbb{Q}$ is a (non-surjective) epimorphism in the category of rings. The argument does not require $X$ to have a unit, or for homomorphisms to map the unity of $\mathbb{Q}$ to the unity of $X$ even when $X$ does have a unity.

Added. In fact, the argument does not even need $f$ and $g$ to be ring homomorphisms, only to be (multiplicative) semigroup homomorphisms. So $(\mathbb{Z},\cdot)\hookrightarrow (\mathbb{Q},\cdot)$ is an epimorphism in the category of semigroups.

The zig-zag is actually part of the characterization of when two semigroup homomorphisms that agree on a subsemigroup agree on an element:

Isbell's Zigzag Theorem for Semigroups

Let $S$ be a semigroup, $D$ a subsemigroup of $S$. Every pair of semigroup homomorphism with domain $S$ and common codomain that agree on $D$ agree on $s$ if and only if $s\in D$, or there is a sequence of factorizations of $s$ of the form $$s=a_1d_1=a_1e_1b_1 = a_2d_2b_1 = a_2e_2b_2 = \cdots = a_{n-1}d_{n-1}b_{n-1}=a_nb_{n-1},$$ where $d_i,e_j\in D$, $a_k,b_{\ell}\in S$, and $d_1=e_1b_1$, $a_{n-1}d_{n-1}=a_n$, and $$a_ie_i = a_{i+1}d_{i+1},\quad d_{i+1}b_i = e_{i+1}b_{i+1},\qquad\text{for }i=2,3,\ldots,n-2.$$

One direction (from the "zigzag" equations to the fact that $f$ and $g$ agree) is easy. There is a nice proof of the other direction in A short proof of Isbell's Zigzag Theorem by P. M. Higgins, Pacific Journal of Mathematics 144 no. 1 (1990), pages 47-50.

The collection $$\bigl\{ s\in S\mid \forall T\;\forall f,g\colon S\to T\ ( f|_D=g|_D\rightarrow f(s)=g(s)\;)\bigr\}$$ is called "the dominion of $D$ in $S$". It can be defined for any category of algebras, though in many standard categories the dominion of a subalgebra is always equal to the subalgebra itself.

The basic reference is the sequence of papers by John Isbell:

  • J.R. Isbell, Epimorphisms and dominions. 1966 Proc. Conf. Categorical Algebra (La Jolla, Calif., 1965) pp 232-246; Springer-Verlag, MR0209202 (35 #105a) (Note: the statement of the zigzag lemma for rings in this paper is incorrect; the correction appears in a later paper).

  • J.R. Isbell and John M. Howie, Epimorphisms and dominions, II. J. Algebra 6 (1967), pp. 7-21, MR0209203 (35 #105b)

  • J.R. Isbell, Epimorphisms and dominions, III. Amer. J. Math. 90 (1968), pp. 1025-1030, MR0237596 (38 #5877)

  • J.R. Isbell, Epimorphisms and dominions, IV. J. London Math. Soc. Ser. 2 1 (1969), pp. 265-273, MR0257120 (41 #1774)

  • J.R. Isbell, Epimorphisms and dominions V. Algebra Universalis 3 (1973), pp. 318-320, MR0349536 (50 #2029)

There is also a nice survey by Peter M. Higgins, Epimorphisms and Amalgams, Colloq. Math. 56 (1988) no. 1, pp. 1-17, MR0980507 (89m:20083).

share|cite|improve this answer
+1 I like the simplicity of this argument :) – Zev Chonoles Sep 9 '11 at 4:54
I assumed X had a 1, but in fact it may not have such an element. Thanks for pointing that out. – Edison Sep 9 '11 at 4:59

Hint: Because $f$ and $g$ are homomorphisms, $$f(y)f(\tfrac{1}{y})=f(y\cdot\tfrac{1}{y})=f(1)=1\qquad \text{ and }\qquad g(y)g(\tfrac{1}{y})=g(y\cdot\tfrac{1}{y})=g(1)=1.$$ Use the fact that multiplicative inverses are unique, when they exist.

share|cite|improve this answer
I see it, thanks! – Edison Sep 9 '11 at 4:52
Since there is no requirement that $X$ be a ring with $1$ or that homomorphisms take $1$ to $1$, how would you proceed without those assumptions? – Arturo Magidin Sep 9 '11 at 4:55
I hadn't seen the "zig-zag" trick before, so I would not have proceeded very far without having to sit and think a bit. @ElG, I recommend you accept Arturo's answer. – Zev Chonoles Sep 9 '11 at 5:01

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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