MHB 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 $$\phi$$ is injective, so by the First Isomorphism Theorem, $$R$$ is isomorphic to $$\text{Im } \phi$$. ... ... "
*** NOTE *** The function $$\phi$$ 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 $$R$$ is isomorphic to $$\text{Im } \phi$$.

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 $$\phi$$ is an injection by Lemma 6.2.5 ... and further ... obviously the map of $$R$$ to $$\text{Im } \phi$$ is onto, that is a surjection ... so $$R$$ is isomorphic to $$\text{Im } \phi$$ ... 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 $$R$$ is isomorphic to $$\text{Im } \phi$$.

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 $$\phi$$ is an injection by Lemma 6.2.5 ... and further ... obviously the map of $$R$$ to $$\text{Im } \phi$$ is onto, that is a surjection ... so $$R$$ is isomorphic to $$\text{Im } \phi$$ ... 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