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.

Solution 1:

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).

Solution 2:

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.

Solution 3:

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.