Rings of Fractions .... Lovett, Section 6.2, Proposition 6.2.6 .... ....

[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 the proof of Proposition 6.2.6 ... ... ...

Proposition 6.2.6 and its proof read as follows:
View attachment 6465
View attachment 6466In the above proof by Lovett we read the following:

" ... ... By Lemma 6.2.5, the function \(\displaystyle \phi\) is injective, so by the First Isomorphism Theorem, \(\displaystyle R\) is isomorphic to \(\displaystyle \text{Im } \phi\). ... ... "
*** NOTE *** The function \(\displaystyle \phi\) is defined in Lemma 6.2.5 which I have provided below ... ..
My questions are as follows:Question 1

I am unsure of exactly how the First Isomorphism Theorem establishes that \(\displaystyle R\) is isomorphic to \(\displaystyle \text{Im } \phi\).

Can someone please show me, rigorously and formally, how the First Isomorphism Theorem applies in this case ... Question 2

I am puzzled as to why the First Isomorphism Theorem is needed in the first place as \(\displaystyle \phi\) is an injection by Lemma 6.2.5 ... and further ... obviously the map of \(\displaystyle R\) to \(\displaystyle \text{Im } \phi\) is onto, that is a surjection ... so \(\displaystyle R\) is isomorphic to \(\displaystyle \text{Im } \phi\) ... BUT ... why is Lovett referring to the First Isomorphism Theorem ... I must be missing something ... hope someone can clarify this issue ...Peter===================================================

In the above, Lovett refers to Lemma 6.2.5 and the First Isomorphism Theorem ... so I am providing copies of both ...Lemma 6.2.5 reads as follows:
https://www.physicsforums.com/attachments/6467
The First Isomorphism Theorem reads as follows:
https://www.physicsforums.com/attachments/6468

 

[Euge]

Question 1

I am unsure of exactly how the First Isomorphism Theorem establishes that \(\displaystyle R\) is isomorphic to \(\displaystyle \text{Im } \phi\).

Can someone please show me, rigorously and formally, how the First Isomorphism Theorem applies in this case ...

Since $\phi$ is injective, its kernel is trivial, and hence $R \approx \text{Im } \phi$ by the first isomorphism theorem. Perhaps you're forgetting that $R/{0} \approx R$.

Question 2

I am puzzled as to why the First Isomorphism Theorem is needed in the first place as \(\displaystyle \phi\) is an injection by Lemma 6.2.5 ... and further ... obviously the map of \(\displaystyle R\) to \(\displaystyle \text{Im } \phi\) is onto, that is a surjection ... so \(\displaystyle R\) is isomorphic to \(\displaystyle \text{Im } \phi\) ... BUT ... why is Lovett referring to the First Isomorphism Theorem ... I must be missing something ... hope someone can clarify this issue ...

Having a surjection from one ring onto another does not imply that the rings are isomorphic. The image of $\phi$ is a subring of $D^{-1}R$, and is also isomorphic to $R$ by the first isomorphism theorem. So the first conclusion of Proposition 6.2.6 holds.

 

[Math Amateur]

Since $\phi$ is injective, its kernel is trivial, and hence $R \approx \text{Im } \phi$ by the first isomorphism theorem. Perhaps you're forgetting that $R/{0} \approx R$.

Having a surjection from one ring onto another does not imply that the rings are isomorphic. The image of $\phi$ is a subring of $D^{-1}R$, and is also isomorphic to $R$ by the first isomorphism theorem. So the first conclusion of Proposition 6.2.6 holds.

Thanks Euge ...

Appreciate your helpPeter