First isomorphism theorem for rings

In summary, when considering a homomorphism \varphi:R\rightarrow S and its induced homomorphism \hat\varphi:\frac{R}{\ker\varphi}\rightarrow \varphi(R), it is important to show that \hat\varphi is bijective. While surjectivity may be obvious, injectivity can be proven by showing that if \varphi(r) = \varphi(r'), then r-r' is in the kernel I and thus equal in the quotient. This can be done by considering the two cases where r-r' is in I or not in I, and using the fact that \hat\varphi is a homomorphism.
  • #1
rukawakaede
59
0
Consider: [tex]\varphi:R\rightarrow S[/tex] is a homomorphism.

Also,[tex]\hat{\varphi}:\frac{R}{ker\varphi}\rightarrow \varphi(R)[/tex].

How can I show [tex]\hat{\varphi}[/tex] is bijective?

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

Could anyone provide a proof for injectivity?
 
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  • #2
if two elements of the ring differ by an element of the kernel then they are equal in the quotient.
 
  • #3
lavinia said:
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 [tex]\varphi(r)=\varphi(r')[/tex]. I= kernel.
1. if [tex]r=r'[/tex] then then this is obvious as they are in the same coset.
2. if [tex]r\neq r'[/tex] and if [tex]r-r'=a\in I[/tex] then [tex]r+I=r'+I[/tex] and so [tex]\hat\varphi(r+I)=\hat\varphi(r'+I)[/tex] satisfying assumption above.
3. if [tex]r\neq r'[/tex] and if [tex]r-r'=a\not\in I[/tex] then [tex]r+I\neq r'+I[/tex] and so [tex]\hat\varphi(r+I)\neq\hat\varphi(r'+I)[/tex] and so [tex]\varphi(r)\neq\varphi(r')[/tex] contradicting assumption. (?)

So for all [tex]\varphi(r)=\varphi(r')[/tex], we must have this condition [tex]r+I=r'+I[/tex] holds. hence it is injective.
 
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  • #4
rukawakaede said:
Thank you. I think I know the injectivity already. Could you please check if my argument is right?

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

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

right but you could say this more simply as

if [tex]\varphi(r - r')= 0[/tex] then r - r' is in I and so is zero in the quotient.
 
  • #5
I would do it essentially the same way as Lavinia. Define [itex]I=\ker\varphi[/itex].

[tex]
\begin{align*}
& \hat\varphi(r+I)=\hat\varphi(r'+I)\ \Rightarrow\ \varphi(r)=\varphi(r')\ \Rightarrow\ \varphi(r-r')=0\\
& \Rightarrow\ r-r'\in\ker\varphi=I\ \Rightarrow\ r+I=r'+I
\end{align*}
[/tex]
 
  • #6
lavinia said:
right but you could say this more simply as

if [tex]\varphi(r - r')= 0[/tex] 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:
[tex]r+I\neq r'+I[/tex] and so [tex]\hat\varphi(r+I)\neq\hat\varphi(r'+I)[/tex]. 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 [tex]r-r'\not\in I[/tex] then [tex]\varphi(r)\neq\varphi(r')[/tex]?

This question might be silly, since we can argue the opposite (i.e. if [tex]r-r'\not\in I[/tex] then [tex]\varphi(r)=\varphi(r')[/tex]) and obtain a contradiction from your/Fredrick argument above. But could we prove that directly?
 
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  • #7
Fredrik said:
I would do it essentially the same way as Lavinia. Define [itex]I=\ker\varphi[/itex].

[tex]
\begin{align*}
& \hat\varphi(r+I)=\hat\varphi(r'+I)\ \Rightarrow\ \varphi(r)=\varphi(r')\ \Rightarrow\ \varphi(r-r')=0\\
& \Rightarrow\ r-r'\in\ker\varphi=I\ \Rightarrow\ r+I=r'+I
\end{align*}
[/tex]
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:
[tex]r+I\neq r'+I[/tex] and so [tex]\hat\varphi(r+I)\neq\hat\varphi(r'+I)[/tex]. 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 [tex]r-r'\not\in I[/tex] then [tex]\varphi(r)\neq\varphi(r')[/tex]?

This question might be silly, since we can argue the opposite (i.e. if [tex]r-r'\not\in I[/tex] then [tex]\varphi(r)=\varphi(r')[/tex]) and obtain a contradiction from your/lavinia argument above. But could we prove that directly?
 
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  • #8
rukawakaede said:
Another question: I found it strange for 3. in my argument above. In particular:
[tex]r+I\neq r'+I[/tex] and so [tex]\hat\varphi(r+I)\neq\hat\varphi(r'+I)[/tex]. This is not generally true, isn't it?
It is. If I understand you correctly, you're asking if the implication [itex]r+I\neq r'+I\Rightarrow \hat\varphi(r+I)\neq\hat\varphi(r'+I)[/itex] is true (or rather, how to see that it's not). This implication is equivalent to [itex]\hat\varphi(r+I)=\hat\varphi(r'+I)\Rightarrow r+I=r'+I[/itex], and this is exactly what I proved in my previous post.

rukawakaede said:
Do you know a way to avoid this? If I want to show that the statement: if [tex]r-r'\not\in I[/tex] then [tex]\varphi(r)\neq\varphi(r')[/tex]?
This is equivalent to showing that [itex]\varphi(r)=\varphi(r')\Rightarrow r-r'\in I[/itex]. This is also a part of my proof in my previous post.
 
  • #9
Fredrik said:
It is. If I understand you correctly, you're asking if the implication [itex]r+I\neq r'+I\Rightarrow \hat\varphi(r+I)\neq\hat\varphi(r'+I)[/itex] is true (or rather, how to see that it's not). This implication is equivalent to [itex]\hat\varphi(r+I)=\hat\varphi(r'+I)\Rightarrow r+I=r'+I[/itex], and this is exactly what I proved in my previous post.


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

Thank you. I was asking something really silly :-p
 

Related to First isomorphism theorem for rings

1. What is the First Isomorphism Theorem for Rings?

The First Isomorphism Theorem for Rings is a fundamental result in abstract algebra that states that if f is a ring homomorphism from one ring R to another ring S, then there exists a unique isomorphism between R/ker(f) and im(f), where ker(f) is the kernel of f and im(f) is the image of f.

2. What is the significance of the First Isomorphism Theorem for Rings?

The First Isomorphism Theorem for Rings allows us to understand the structure and relationships between different rings by studying their homomorphisms. It also helps us prove properties and results in ring theory and other areas of mathematics.

3. How is the First Isomorphism Theorem for Rings related to other isomorphism theorems?

The First Isomorphism Theorem for Rings is a generalization of the First Isomorphism Theorem for Groups and the First Isomorphism Theorem for Vector Spaces. It is also related to the Fundamental Theorem of Homomorphisms, which states that every ring homomorphism can be factored uniquely through the quotient of the kernel and the image.

4. Can the First Isomorphism Theorem for Rings be extended to other algebraic structures?

Yes, the First Isomorphism Theorem can be extended to other algebraic structures such as modules, algebras, and fields. However, the specific form and conditions of the theorem may vary depending on the structure being studied.

5. How is the First Isomorphism Theorem for Rings applied in real-world situations?

The First Isomorphism Theorem for Rings has applications in many areas of mathematics, including abstract algebra, number theory, and algebraic geometry. It also has practical applications in computer science, coding theory, and cryptography.

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