Extend the absolute value on an integral domain

Question

Let $A$ be an integral domain, denote by $FF(A)$, the field of $A$: the set of all fractions of the form $\frac{a} {b}$ for $a, b \in A, b\neq 0$, where $\frac{a} {b} =\frac{c} {d}$ iff $ad-bc=0$. Assume that | | is an absolute value function on $A$. Show that we can extend it uniquely to an absolute value function on $FF(A)$ by $|\frac{a}{b}| =\frac{|a|}{|b|}$. i.e

a.prove that the function we defined is well defined, defines an absolute value function on $FF(A)$ and extends the origional function on $A$.

b.prove that every absolute value function on $FF(A)$ that extends the origional function on $A$ equals the function we defined.

In a, when saying "to show that the function we defined is well defined", does this mean if $a, b \in A, b\neq 0$ then $\frac{a} {b}$ is actually a real number (because an absolute value function is defined from the field to real numbers) ?

But isn't this obvious, since $|a|, |b|$ are real number as | | is absolute value on $A$ and thus $|a|/|b|$ is also a real number, right?

Now, the defined function is an absolute value on $FF(A)$:

1. If $a, b \in A, b\neq 0$ then $|a/b|=|a|/|b| \geq 0$ since $|a|\geq 0$ as | | is an absolute value on $A$ and so $|b|>0$ (as $b\neq 0$).

2. Let $a, b, c, d\in A, b, d\neq 0$ then $|\frac{a} {b} \frac{c} {d} |$=(by def)=$\frac{|ac|} {|bd|} $=(| | is absolute value) =$\frac{|a||c|}{|b||d|}$=(by def) =$|\frac{a} {b} ||\frac{c} {d} |$.

3. Let $a, b, c, d \in A, b, d \neq 0$ we need to show that $|\frac{a}{b} +\frac{c}{d}|\leq |\frac{a}{b}|+|\frac{c}{d}|$: Well,

$|\frac{a}{b}+\frac{c}{d}|=|\frac{ad+bc}{bd}|$=(by def)= $\frac{|ad+bc|}{|bd|}\leq$ (| | is absolute value) $\frac{(|ad|+|bc|)}{|bd|}$ (| | is absolute value) $\frac{(|a||d|+|b||c|) }{|b||d|}= \frac{|a|}{|b|}+\frac{|c|}{|d|}$= (by def) =$|\frac{a}{b}|+|\frac{c}{d}|$.

Is that okay?

What does it mean "extends the origional function on $A$"?

Does it mean that for $a\in A, b=1\in A$ ($A$ is an integral domain so $1\in A$) we have that the absolute value on $FF(A)$ satisfies $|\frac{a}{1}|= $(by def)$\frac{|a|}{|1|}=|a|$ (since $|1|=1$) and it is indeed the | | on $A$?

Can you please provide any ideas for part b.

Many thanks for helping me to understand this.

Answer 1

In a, when saying "to show that the defined function is well defined", does this mean, if $a, b \in A, b\neq 0$ then $\frac{a} {b}$ is actually a real number (because an absolute function is defined from the field to real numbers) ?

I assume you mean $\frac{|a|}{|b|}$ is a real number, and indeed this is obvious (given the fact that $b\not=0_A$, hence $|b|\not=0$). To show the function is well-defined, you need to show that if $\frac{a}{b}=\frac{c}{d}$, then $\frac{|a|}{|b|}=\frac{|c|}{|d|}$, that is the definition doesn't depend on the specific choices of the denominator and numerator, but only the fraction.

What does it mean "extends the origional function on $A$"?

Does it mean that for $a\in A, b=1\in A$ ($A$ is an integral domain so $1\in A$) we have that the absolute value on $FF(A)$ satisfies $|a/1|= $(by def)$|a|/|1|=|a|$ (since $|1|=1$) and it is indeed the | | on $A$?

Yes. While technically $A$ is not part of $FF(A)$, we can show that $a\mapsto \frac{a}{1}$ is an injective ring homomorphism from $A$ to $FF(A)$. Hence we may regard $A$ as a subring of $FF(A)$.

Your proof is basically right... but please improve the formatting.

As for Part (b), if $|\cdot|$ on $FF(A)$ extends the original absolute value $|\cdot|'$ on $A$, then we have $$|a|'=|\frac{a}{1}|=|\frac{a}{b} \cdot \frac{b}{1}|= |\frac{a}{b}| |\frac{b}{1}|=|\frac{a}{b}||b|'$$ Hence $$|\frac{a}{b}|=\frac{|a|'}{|b|'}$$