Ring Homomorphisms from the Ring of Rational Numbers are Determined by the Values at Integers

Problems and solutions of ring theory in abstract algebra

Problem 318

Let $R$ be a ring with unity.
Suppose that $f$ and $g$ are ring homomorphisms from $\Q$ to $R$ such that $f(n)=g(n)$ for any integer $n$.

Then prove that $f=g$.

LoadingAdd to solve later


Let $a/b \in \Q$ be an arbitrary rational number with integers $a, b$.
Then we have
&=f(a)f\left(\frac{1}{b}\right) && \text{ (since $f$ is a ring homomorphism)}\\
&=g(a)f\left(\frac{1}{b}\right) && \text{ (since $a$ is an integer)}\\
&=g\left(\frac{a}{b}\cdot b\right) f\left(\frac{1}{b}\right)\\
&=g\left(\frac{a}{b}\right) g(b) f\left(\frac{1}{b}\right) && \text{ (since $g$ is a ring homomorphism)}\\
&=g\left(\frac{a}{b}\right) f(b) f\left(\frac{1}{b}\right) && \text{ (since $b$ is an integer)}\\
&=g\left(\frac{a}{b}\right) f\left(b\cdot \frac{1}{b}\right) && \text{ (since $f$ is a ring homomorphism)}\\
&=g\left(\frac{a}{b}\right) f(1) \\
&=g\left(\frac{a}{b}\right) \cdot 1\\

Therefore, we proved
\[f\left(\frac{a}{b}\right)=g\left(\frac{a}{b}\right),\] for any rational number $a/b\in \Q$.
Hence we have $f=g$.


In the language of category theory, this shows that the inclusion $\Z\to \Q$ is epi in the category of rings.
Also, note that this inclusion is not epi in the category of abelian groups ($\Z$-mod).

LoadingAdd to solve later

More from my site

You may also like...

Leave a Reply

Your email address will not be published. Required fields are marked *

More in Ring theory
Problems and solutions of ring theory in abstract algebra
Generators of the Augmentation Ideal in a Group Ring

Let $R$ be a commutative ring with $1$ and let $G$ be a finite group with identity element $e$. Let...