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