Second Isomorphism Theorem for Rings .... Bland Theorem 3.3.1

[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:

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

 

[andrewkirk]

I can see that ##f## is a ring homomorphism ... but how do we prove that the kernel is ##I_1/I_2## ... ... ?

First prove ##I_1/I_2## is in the kernel. Take an arbitrary ##x+I_2\in I_1/I_2##. That means that ##x\in I_1##. Then

##f(x+I_2)\triangleq x+I_1=I_1## (since ##x\in I_1)\ =0_{R/I_1}##.

So ##x+I_2## is in the kernel.

Now prove that the kernel is in ##I_1/I_2##. Take an arbitrary element ##x+I_2## of the kernel. Then we have ##f(x+I_2)\triangleq x+I_1=0_{R/I_1}=I_1##. Hence ##x\in I_1##. Hence ##x+I_1\in I_1/I_2##.

 

[Math Amateur]

First prove ##I_1/I_2## is in the kernel. Take an arbitrary ##x+I_2\in I_1/I_2##. That means that ##x\in I_1##. Then

##f(x+I_2)\triangleq x+I_1=I_1## (since ##x\in I_1)\ =0_{R/I_1}##.

So ##x+I_2## is in the kernel.

Now prove that the kernel is in ##I_1/I_2##. Take an arbitrary element ##x+I_2## of the kernel. Then we have ##f(x+I_2)\triangleq x+I_1=0_{R/I_1}=I_1##. Hence ##x\in I_1##. Hence ##x+I_1\in I_1/I_2##.

Thanks for the help, Andrew ...

But ... just a minor point of clarification ...

You write ... " ... ... Take an arbitrary ##x+I_2\in I_1/I_2##. That means that ##x\in I_1## ... ... "

Can you explain why ##x+I_2\in I_1/I_2## implies that ##x\in I_1## ... ... ?

Peter

 

[andrewkirk]

You write ... " ... ... Take an arbitrary ##x+I_2\in I_1/I_2##. That means that ##x\in I_1## ... ... "

Can you explain why ##x+I_2\in I_1/I_2## implies that ##x\in I_1## ... ... ?

Peter

It is true by definition.

##I_1/I_2## is defined to be the following collection of cosets of ##I_2##
$$I_1/I_2 \triangleq \{x+I_2\ :\ x\in I_1\}$$

 

[Math Amateur]

Hi Andrew ...

Hmmm ... yes ... of course ... you're right ...

Thanks again for your help ...

Peter