First isomorphism theorem for rings

[rukawakaede]

Consider: $$\varphi:R\rightarrow S$$ is a homomorphism.

Also,$$\hat{\varphi}:\frac{R}{ker\varphi}\rightarrow \varphi(R)$$.

How can I show $$\hat{\varphi}$$ is bijective?

Most textbooks say it is obvious. I see surjectivity obvious but not injectivity.

Could anyone provide a proof for injectivity?

 

[lavinia]

if two elements of the ring differ by an element of the kernel then they are equal in the quotient.

 

[rukawakaede]

if two elements of the ring differ by an element of the kernel then they are equal in the quotient.

Thank you. I think I know the injectivity already. Could you please check if my argument is right?

Assume $$\varphi(r)=\varphi(r')$$. I= kernel.
1. if $$r=r'$$ then then this is obvious as they are in the same coset.
2. if $$r\neq r'$$ and if $$r-r'=a\in I$$ then $$r+I=r'+I$$ and so $$\hat\varphi(r+I)=\hat\varphi(r'+I)$$ satisfying assumption above.
3. if $$r\neq r'$$ and if $$r-r'=a\not\in I$$ then $$r+I\neq r'+I$$ and so $$\hat\varphi(r+I)\neq\hat\varphi(r'+I)$$ and so $$\varphi(r)\neq\varphi(r')$$ contradicting assumption. (?)

So for all $$\varphi(r)=\varphi(r')$$, we must have this condition $$r+I=r'+I$$ holds. hence it is injective.

 

[lavinia]

Thank you. I think I know the injectivity already. Could you please check if my argument is right?

Assume $$\varphi(r)=\varphi(r')$$. I= kernel.
1. if $$r=r'$$ then then this is obvious as they are in the same coset.
2. if $$r\neq r'$$ and if $$r-r'=a\in I$$ then $$r+I=r'+I$$ and so $$\hat\varphi(r+I)=\hat\varphi(r'+I)$$ satisfying assumption above.
3. if $$r\neq r'$$ and if $$r-r'=a\not\in I$$ then $$r+I\neq r'+I$$ and so $$\hat\varphi(r+I)\neq\hat\varphi(r'+I)$$ and so $$\varphi(r)\neq\varphi(r')$$ contradicting assumption.

So for all $$\varphi(r)=\varphi(r')$$, we must have this condition $$r+I=r'+I$$ holds. hence it is injective.

right but you could say this more simply as

if $$\varphi(r - r')= 0$$ then r - r' is in I and so is zero in the quotient.

 

[Fredrik]

I would do it essentially the same way as Lavinia. Define $I=\ker\varphi$.

$$\begin{align*}<br /> & \hat\varphi(r+I)=\hat\varphi(r'+I)\ \Rightarrow\ \varphi(r)=\varphi(r')\ \Rightarrow\ \varphi(r-r')=0\\<br /> & \Rightarrow\ r-r'\in\ker\varphi=I\ \Rightarrow\ r+I=r'+I<br /> \end{align*}$$

 

[rukawakaede]

right but you could say this more simply as

if $$\varphi(r - r')= 0$$ then r - r' is in I and so is zero in the quotient.

Thanks, lavinia!

Another question: I found it strange for 3. in my argument above. In particular:
$$r+I\neq r'+I$$ and so $$\hat\varphi(r+I)\neq\hat\varphi(r'+I)$$. This is not generally true, isn't it?

Do you know a way to avoid this? If I want to show that the statement: if $$r-r'\not\in I$$ then $$\varphi(r)\neq\varphi(r')$$?

This question might be silly, since we can argue the opposite (i.e. if $$r-r'\not\in I$$ then $$\varphi(r)=\varphi(r')$$) and obtain a contradiction from your/Fredrick argument above. But could we prove that directly?

 

[rukawakaede]

I would do it essentially the same way as Lavinia. Define $I=\ker\varphi$.

$$\begin{align*}<br /> & \hat\varphi(r+I)=\hat\varphi(r'+I)\ \Rightarrow\ \varphi(r)=\varphi(r')\ \Rightarrow\ \varphi(r-r')=0\\<br /> & \Rightarrow\ r-r'\in\ker\varphi=I\ \Rightarrow\ r+I=r'+I<br /> \end{align*}$$

Thanks Fredrik!

Here is the same question as in my previous post:

Another question: I found it strange for 3. in my argument above. In particular:
$$r+I\neq r'+I$$ and so $$\hat\varphi(r+I)\neq\hat\varphi(r'+I)$$. This is not generally true, isn't it?

Do you know a way to avoid this? If I want to show that the statement: if $$r-r'\not\in I$$ then $$\varphi(r)\neq\varphi(r')$$?

This question might be silly, since we can argue the opposite (i.e. if $$r-r'\not\in I$$ then $$\varphi(r)=\varphi(r')$$) and obtain a contradiction from your/lavinia argument above. But could we prove that directly?

 

[Fredrik]

Another question: I found it strange for 3. in my argument above. In particular:
$$r+I\neq r'+I$$ and so $$\hat\varphi(r+I)\neq\hat\varphi(r'+I)$$. This is not generally true, isn't it?

It is. If I understand you correctly, you're asking if the implication $r+I\neq r'+I\Rightarrow \hat\varphi(r+I)\neq\hat\varphi(r'+I)$ is true (or rather, how to see that it's not). This implication is equivalent to $\hat\varphi(r+I)=\hat\varphi(r'+I)\Rightarrow r+I=r'+I$, and this is exactly what I proved in my previous post.

Do you know a way to avoid this? If I want to show that the statement: if $$r-r'\not\in I$$ then $$\varphi(r)\neq\varphi(r')$$?

This is equivalent to showing that $\varphi(r)=\varphi(r')\Rightarrow r-r'\in I$. This is also a part of my proof in my previous post.

 

[rukawakaede]

It is. If I understand you correctly, you're asking if the implication $r+I\neq r'+I\Rightarrow \hat\varphi(r+I)\neq\hat\varphi(r'+I)$ is true (or rather, how to see that it's not). This implication is equivalent to $\hat\varphi(r+I)=\hat\varphi(r'+I)\Rightarrow r+I=r'+I$, and this is exactly what I proved in my previous post.


This is equivalent to showing that $\varphi(r)=\varphi(r')\Rightarrow r-r'\in I$. This is also a part of my proof in my previous post.

Thank you. I was asking something really silly