# Showing two ring homomorphisms that agree on the integers must agree on the rationals

I have two ring homomorphisms $$f,g\colon \mathbb{Q}\to X$$. 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.

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).

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

• 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

There is an exercise (Exercise III.1.18) in Hungerford's Algebra which is the same as this question.

Let $$\Bbb{Q}$$ be the field of rational numbers and $$R$$ any ring. If $$f, g:\Bbb{Q}\to R$$ are homomorphisms of rings such that $$f|\Bbb{Z}=g|\Bbb{Z}$$, then $$f=g$$. [Hint: show that for $$n\in \Bbb{Z}$$ ($$n\neq 0$$), $$f(1/n)g(n)=g(1)$$, whence $$f(1/n)=g(1/n)$$.]

My solution: Since $$\Bbb{Q}$$ is a field (the only ideals are $$0$$ and $$\Bbb{Q}$$) and $$\ker{f}, \ker{g}$$ are ideals of $$\Bbb{Q}$$, we have $$\ker{f}$$ and $$\ker{g}$$ are in $$\{0, \Bbb{Q}\}$$. Note that $$\ker{f}=\Bbb{Q}\Leftrightarrow \ker{g}=\Bbb{Q}$$. In this case, $$f(q)=g(q)=0$$ for all $$q\in \Bbb{Q}$$.

Now, we assume $$\ker{g}=0$$. Then by the First Isomorphism Theorem for Rings, $$\Bbb{Q}\cong \Bbb{Q}/0=\Bbb{Q}/\ker{g}\cong \text{Im }g=g(\Bbb{Q}).$$ Thus, $$g(n)$$ has an inverse in $$g(\Bbb{Q})$$.

On the other hand, by the hint, $$f(1/n)g(n)=f(1/n)f(n)=f(1/n\cdot n)=f(1)=g(1)=g(1/n\cdot n)=g(1/n)g(n).$$ It follows that $$f(1/n)g(n)-g(1/n)g(n)=0$$ and $$[f(1/n)-g(1/n)]g(n)=0$$. We get $$f(1/n)-g(1/n)=0$$ by multiplying $$g(n)^{-1}$$ on both sides.