# I Rings of Fractions ... Lovett, Section 6.2, Proposition 6.2

1. Mar 11, 2017

### 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:

In 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 ...

Hoe that someone can help ... ...

Peter

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In the above, Lovett refers to Lemma 6.2.5 and the First Isomorphism Theorem ... so I am providing copies of both ...

The First Isomorphism Theorem reads as follows:

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• ###### Lovett - Theorem 5.6.8 - First Isomorphism Theorem ... ... .png
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2. Mar 12, 2017

### willem2

question1.
since φ is injective, Ker φ only contains the identity of R, so R is isomorphic to φ(R) = Im φ according to the isomorphism theorem.
question2.
no Idea. You can easily see that φ i a bijection, so R must be isomorphic to Im φ.

3. Mar 12, 2017

### Math Amateur

Hi willem2,

Thanks for the help ...

... basically ...

$\phi injective$

$\Longrightarrow \text{Ker } \phi = {0}$ ... ... that is $\text{Ker } \phi$ contains only additive identity of R

$\Longrightarrow R/ \text{Ker } \phi = R/ {0} = R \cong \text{Im } \phi$

Is that correct?

... BUT ...

Can you explain in simple terms, why $R/ {0} = R$ ... ...???

Peter

4. Mar 12, 2017

### Staff: Mentor

$R/\{0\} = \{r + \{0\}\,\vert \,r \in R\} = \{\{r\}\,\vert \,r \in R\} \cong \{r\,\vert \,r \in R\} = R$.
Did you mean something like this? Remember that $R/I$ is the set of all cosets $r+I$.

5. Mar 12, 2017

### Math Amateur

Thanks fresh_42 ... yes, exactly what I meant ...

Would the answer be the same for $R/ \{a \}$ where $a$ was a given particular element of $R$? Indeed what would the various cosets be?

Peter

6. Mar 13, 2017

### Staff: Mentor

You could formally build cosets $r+\{a\}$ but unless you take the ideal generated by $a$, i.e. $Ra$, you won't get a structure of interest on $R/\{a\}$. E.g. if $[x]_a$ denotes the coset $x+\{a\}$, then $[0]_a \neq [a+a]_a$ because $a+a \notin \{a\}$ and on the other hand $[a]_a + [a]_a = [0]_a + [0]_a.$ Thus $[a+a]_a = [a]_a + [a]_a$ cannot be concluded, which means you cannot add anymore unless $a=0$.

So better forget about this idea of cosets of single elements other than zero. At least we want to have a subring; ideal would be better.

If you want to learn something about cosets, you could prove the following as an exercise:
Given a subgroup $U$ of a finite group $G$. Then $G/U$ is well defined and $U$ partitions $G$ into equally large subsets of $G$, the cosets $xU$.
But $G/U$ carries a group structure again, if and only if $U$ is a normal subgroup of $G$.

This is the reason why we consider normal subgroups instead of only subgroups. I haven't done the math, but I assume it's similar with ideals and subrings. So factoring along a single point set only makes sense, if this point is the neutral element of the underlying group. And rings only build a group with addition, because multiplication isn't invertible.

Last edited: Mar 13, 2017
7. Mar 29, 2017

### WWGD

I wonder if in a sense we can consider this as an allowable case of division by 0.

8. Mar 29, 2017

### Staff: Mentor

You look like a quotient
Walk like a quotient
Talk like a quotient
But I got wise
Oh, yes, you are

You fooled me with your slashes
You cheated and you schemed
Heaven knows how you lied to me
You're not the way you seemed
...

9. Apr 2, 2017

### WWGD

Still, you may mod out by a non-Abelian group, in which case it does not come down to addition.