Rings of Fractions .... Lovett, Section 6.2 ....

[Math Amateur]

I am reading Stephen Lovett's book, "Abstract Algebra: Structures and Applications" and am currently focused on Section 6.2: Rings of Fractions ...

I need some help with some remarks following Definition 6.2.4 ... ... ...

The remarks following Definition 6.2.4 reads as follows:https://www.physicsforums.com/attachments/6461In the above text from Lovett we read the following:

" ... ... it is not hard to show that if we had taken \(\displaystyle D = { \mathbb{Z} }^{ \gt 0 }\) we would get a ring of fractions that is that is isomorphic to \(\displaystyle \mathbb{Q}\). ... ... "Can someone please help me to understand this statement ... how is such an isomorphism possible ... in particular, how does one achieve a one-to-one and onto homomorphism from the positive integers to the negative elements of \(\displaystyle \mathbb{Q}\) as well as the positive elements ...

Hope someone can help ... ...

Peter=================================================

To enable readers to understand Lovett's approach to the rings of fraction construction, I am providing Lovett Section 6.2 up to an including the remarks following Definition 6.2.4 ... as follows:https://www.physicsforums.com/attachments/6462
https://www.physicsforums.com/attachments/6463
https://www.physicsforums.com/attachments/6464

 

[johng1]

The simple reason is that any element $q\in Q$ can be realized as $q={a\over b}$ with $b>0$. If a denominator is negative, just multiply numerator and denominator by -1.

Formally, let $Q_1$ be the ring constructed with denominators positive. It's then easy to verify that the "identity" map from $Q_1$ to $Q$ is an isomorphism.

 

[Math Amateur]

The simple reason is that any element $q\in Q$ can be realized as $q={a\over b}$ with $b>0$. If a denominator is negative, just multiply numerator and denominator by -1.

Formally, let $Q_1$ be the ring constructed with denominators positive. It's then easy to verify that the "identity" map from $Q_1$ to $Q$ is an isomorphism.

Thanks johng ... appreciate the help ...

Peter