MHB Second Isomorphism Theorem for Rings .... Bland Theorem 3.3.15 .... ....

[Math Amateur]

I am reading "The Basics of Abstract Algebra" by Paul E. Bland ... ...

I am currently focused on Chapter 3: Sets with Two Binary Operations: Rings ... ...

I need help with Bland's proof of the Second Isomorphism Theorem for rings ...

Bland's Second Isomorphism Theorem for rings and its proof read as follows:
View attachment 7969
https://www.physicsforums.com/attachments/7970
In the above proof by Bland we read the following:

" ... ... This map is easily shown to be a well defined ring homomorphism with kernel $$I_1/I_2$$. ... ... "I can see that $$f$$ is a ring homomorphism ... but how do we prove that the kernel is $$I_1/I_2$$ ... ... ?Hope someone can help ...

Peter

 

[castor28]

Hi Peter,

The identity of $R/I_1$ is $I_1$. $\ker f$ contains the cosets $x+I_2$ such that $x+I_1 = I_1$, which means that $x\in I_1$.

This shows that $\ker f$ is the set of cosets $\{(x+I_2)\mid x\in I_1\}$, which is $I_1/I_2$.

 

[Math Amateur]

Hi Peter,

The identity of $R/I_1$ is $I_1$. $\ker f$ contains the cosets $x+I_2$ such that $x+I_1 = I_1$, which means that $x\in I_1$.

This shows that $\ker f$ is the set of cosets $\{(x+I_2)\mid x\in I_1\}$, which is $I_1/I_2$.

Thanks for the help, castor28 ...

But just a point of clarification ...

You write: " ... ... The identity of $R/I_1$ is $I_1$ ... ... "Surely the identity of $$R/I_1$$ is the coset $$1_R + I_1$$ because $$(a + I_1) ( 1_R + I_1 ) = a 1_R + I_1 = a + I_1$$ ... and similarly $$( 1_R + I_1 ) ( a + I_1) = a + I_1$$ ...

Can you please clarify ...

Peter

 

[castor28]

Thanks for the help, castor28 ...

But just a point of clarification ...

You write: " ... ... The identity of $R/I_1$ is $I_1$ ... ... "Surely the identity of $$R/I_1$$ is the coset $$1_R + I_1$$ because $$(a + I_1) ( 1_R + I_1 ) = a 1_R + I_1 = a + I_1$$ ... and similarly $$( 1_R + I_1 ) ( a + I_1) = a + I_1$$ ...

Can you please clarify ...

Peter

Hi Peter,

Quotient rings are defined in terms of the additive group; that is why cosets of an ideal $I$ are written as $x+I$. The identity of the additive group is $0$, and the identity of $R/I$ is the coset containing $0$, which is obviously $I$. The kernel is defined as the inverse image of the identity of the additive group.

The coset $1+I$ is the multiplicative identity of the quotient ring $R/I$.

 

[Math Amateur]

Hi Peter,

Quotient rings are defined in terms of the additive group; that is why cosets of an ideal $I$ are written as $x+I$. The identity of the additive group is $0$, and the identity of $R/I$ is the coset containing $0$, which is obviously $I$. The kernel is defined as the inverse image of the identity of the additive group.

The coset $1+I$ is the multiplicative identity of the quotient ring $R/I$.

Thanks castor28 ...

Hmm ... beginning to understand what you are saying ...

Still concerned and a bit confused ...

Surely $$R/I$$ is a ring under the addition and multiplication of cosets ...
and hence has a (multiplicative) identity as I described ..

But the definition of cosets of course involves only addition ...

Is that correct?

Peter

 

[castor28]

Thanks castor28 ...

Hmm ... beginning to understand what you are saying ...

Still concerned and a bit confused ...

Surely $$R/I$$ is a ring under the addition and multiplication of cosets ...
and hence has a (multiplicative) identity as I described ..

But the definition of cosets of course involves only addition ...

Is that correct?

Peter

Hi Peter,

Yes, that is correct.

If $I$ is a subgroup (necessarily normal) of the additive group of $R$, the quotient group $R/I$ is an additive group, and there is a canonical group homomorphism $f:R\to R/I$ that sends $x\in R$ to the coset $x+I$. The additive identity of $R/I$ is the coset $0+I=I$. $\ker f$ is the inverse image of that coset.

If we add the stronger condition that $I$ is an ideal of $R$, then it can be shown that multiplication of cosets is well-defined, and $f$ is also a ring homomorphism.

Note, however, that $R$ is not a group under multiplication (unless it is trivial); therefore, the (group-related) concept of kernel cannot be based on the multiplicative structure.

 

[steenis]

The well-definedness of the ring-homomorphism is explained to you in this thread:
https://mathhelpboards.com/linear-abstract-algebra-14/quotient-rings-remarks-adkins-weintraub-23836.html?highlight=adkins