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Fundamental Theorem on Homomorphisms - Application

1

$begingroup$

I want to show

1.) For $n,min mathbb Z$ with $m|n$ there exists a ring homomorphism between $mathbbZ/nmathbbZ rightarrow mathbbZ/mmathbbZ$.

I know that there is the canonical homomorphism $f: mathbbZ rightarrow mathbbZ/mmathbbZ$. The Fundamental theorem on homomorphisms for rings now allows me, because $nmathbbZ$ is an ideal in $mathbb Z$, to conclude the existence of a ring homomorphism $mathbbZ/nmathbbZ rightarrow mathbbZ/mmathbbZ$. But here is the problem: $mathbbZ/nmathbbZ$ has to be a subset of $ker(f)$ for that conclusion. Is that the case and if so: why?

2.) A restriction of the above Homomorphism on the unity groups $(mathbb Z/nmathbb Z)^* rightarrow (mathbb Z/mmathbb Z)^*$ is a group homomorphism.

I believe that one can solve this easily if 1.) is solved.

abstract-algebra ring-theory modules ideals

share|cite|improve this question

edited Jan 3 at 15:39

Antonios-Alexandros Robotis

10.6k41741

asked Jan 3 at 15:25

KingDingelingKingDingeling

1857

$endgroup$

add a comment | 

1

$begingroup$

I want to show

1.) For $n,min mathbb Z$ with $m|n$ there exists a ring homomorphism between $mathbbZ/nmathbbZ rightarrow mathbbZ/mmathbbZ$.

I know that there is the canonical homomorphism $f: mathbbZ rightarrow mathbbZ/mmathbbZ$. The Fundamental theorem on homomorphisms for rings now allows me, because $nmathbbZ$ is an ideal in $mathbb Z$, to conclude the existence of a ring homomorphism $mathbbZ/nmathbbZ rightarrow mathbbZ/mmathbbZ$. But here is the problem: $mathbbZ/nmathbbZ$ has to be a subset of $ker(f)$ for that conclusion. Is that the case and if so: why?

2.) A restriction of the above Homomorphism on the unity groups $(mathbb Z/nmathbb Z)^* rightarrow (mathbb Z/mmathbb Z)^*$ is a group homomorphism.

I believe that one can solve this easily if 1.) is solved.

abstract-algebra ring-theory modules ideals

share|cite|improve this question

edited Jan 3 at 15:39

Antonios-Alexandros Robotis

10.6k41741

asked Jan 3 at 15:25

KingDingelingKingDingeling

1857

$endgroup$

add a comment | 

$begingroup$

I want to show

1.) For $n,min mathbb Z$ with $m|n$ there exists a ring homomorphism between $mathbbZ/nmathbbZ rightarrow mathbbZ/mmathbbZ$.

I know that there is the canonical homomorphism $f: mathbbZ rightarrow mathbbZ/mmathbbZ$. The Fundamental theorem on homomorphisms for rings now allows me, because $nmathbbZ$ is an ideal in $mathbb Z$, to conclude the existence of a ring homomorphism $mathbbZ/nmathbbZ rightarrow mathbbZ/mmathbbZ$. But here is the problem: $mathbbZ/nmathbbZ$ has to be a subset of $ker(f)$ for that conclusion. Is that the case and if so: why?

2.) A restriction of the above Homomorphism on the unity groups $(mathbb Z/nmathbb Z)^* rightarrow (mathbb Z/mmathbb Z)^*$ is a group homomorphism.

I believe that one can solve this easily if 1.) is solved.

abstract-algebra ring-theory modules ideals

share|cite|improve this question

edited Jan 3 at 15:39

Antonios-Alexandros Robotis

10.6k41741

asked Jan 3 at 15:25

KingDingelingKingDingeling

1857

$endgroup$

share|cite|improve this question

edited Jan 3 at 15:39

Antonios-Alexandros Robotis

10.6k41741

asked Jan 3 at 15:25

KingDingelingKingDingeling

1857

share|cite|improve this question

edited Jan 3 at 15:39

Antonios-Alexandros Robotis

10.6k41741

asked Jan 3 at 15:25

KingDingelingKingDingeling

1857

edited Jan 3 at 15:39

Antonios-Alexandros Robotis

10.6k41741

edited Jan 3 at 15:39

Antonios-Alexandros Robotis

10.6k41741

asked Jan 3 at 15:25

KingDingelingKingDingeling

1857

asked Jan 3 at 15:25

KingDingelingKingDingeling

1857

2 Answers
2

active

oldest

votes

2

$begingroup$

Why not study the canonical map $mathbbZxrightarrowvarphi mathbbZ/mmathbbZ$ given by the projection $nmapsto [n]_m$? The first isomorphism theorem tells us that $mathbbZ/kervarphicong mathbbZ/mmathbbZ$. Because $m|n$, we get that for $kin nmathbbZ$, i.e. $k=ell n=ell rm$ for some $kin mathbbZ$, we have
$$varphi(k) =[k]_m=[ell r m]_m=0 pmodm.$$
So, $nmathbbZsubseteqkervarphi$ and the factorization theorem tells us that we get an induced map $widetildevarphi:nmathbbZto mmathbbZ$. This is defined by $widetildevarphi(a+nmathbbZ)=varphi(a)=[a]_m.$

share|cite|improve this answer

edited Jan 3 at 15:40

answered Jan 3 at 15:34

Antonios-Alexandros RobotisAntonios-Alexandros Robotis

10.6k41741

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for your answer and for taking the time :)
  $endgroup$
  – KingDingeling
  Jan 3 at 16:02
  

  
  
  
  

add a comment | 

0

$begingroup$

If $f : Gto G'$ be a homomorphic and onto then $h : dfracGK to G'$ is an isomorphism and onto where $K = ker f$.

share|cite|improve this answer

edited Mar 31 at 12:22

Max

1,0041319

answered Mar 31 at 12:10

Yamini Singh Yamini Singh

1

$endgroup$

add a comment | 

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2

$begingroup$

Why not study the canonical map $mathbbZxrightarrowvarphi mathbbZ/mmathbbZ$ given by the projection $nmapsto [n]_m$? The first isomorphism theorem tells us that $mathbbZ/kervarphicong mathbbZ/mmathbbZ$. Because $m|n$, we get that for $kin nmathbbZ$, i.e. $k=ell n=ell rm$ for some $kin mathbbZ$, we have
$$varphi(k) =[k]_m=[ell r m]_m=0 pmodm.$$
So, $nmathbbZsubseteqkervarphi$ and the factorization theorem tells us that we get an induced map $widetildevarphi:nmathbbZto mmathbbZ$. This is defined by $widetildevarphi(a+nmathbbZ)=varphi(a)=[a]_m.$

share|cite|improve this answer

edited Jan 3 at 15:40

answered Jan 3 at 15:34

Antonios-Alexandros RobotisAntonios-Alexandros Robotis

10.6k41741

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for your answer and for taking the time :)
  $endgroup$
  – KingDingeling
  Jan 3 at 16:02
  

  
  
  
  

add a comment | 

2

$begingroup$

Why not study the canonical map $mathbbZxrightarrowvarphi mathbbZ/mmathbbZ$ given by the projection $nmapsto [n]_m$? The first isomorphism theorem tells us that $mathbbZ/kervarphicong mathbbZ/mmathbbZ$. Because $m|n$, we get that for $kin nmathbbZ$, i.e. $k=ell n=ell rm$ for some $kin mathbbZ$, we have
$$varphi(k) =[k]_m=[ell r m]_m=0 pmodm.$$
So, $nmathbbZsubseteqkervarphi$ and the factorization theorem tells us that we get an induced map $widetildevarphi:nmathbbZto mmathbbZ$. This is defined by $widetildevarphi(a+nmathbbZ)=varphi(a)=[a]_m.$

share|cite|improve this answer

edited Jan 3 at 15:40

answered Jan 3 at 15:34

Antonios-Alexandros RobotisAntonios-Alexandros Robotis

10.6k41741

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for your answer and for taking the time :)
  $endgroup$
  – KingDingeling
  Jan 3 at 16:02
  

  
  
  
  

add a comment | 

$begingroup$

Why not study the canonical map $mathbbZxrightarrowvarphi mathbbZ/mmathbbZ$ given by the projection $nmapsto [n]_m$? The first isomorphism theorem tells us that $mathbbZ/kervarphicong mathbbZ/mmathbbZ$. Because $m|n$, we get that for $kin nmathbbZ$, i.e. $k=ell n=ell rm$ for some $kin mathbbZ$, we have
$$varphi(k) =[k]_m=[ell r m]_m=0 pmodm.$$
So, $nmathbbZsubseteqkervarphi$ and the factorization theorem tells us that we get an induced map $widetildevarphi:nmathbbZto mmathbbZ$. This is defined by $widetildevarphi(a+nmathbbZ)=varphi(a)=[a]_m.$

share|cite|improve this answer

edited Jan 3 at 15:40

answered Jan 3 at 15:34

Antonios-Alexandros RobotisAntonios-Alexandros Robotis

10.6k41741

$endgroup$

share|cite|improve this answer

edited Jan 3 at 15:40

answered Jan 3 at 15:34

Antonios-Alexandros RobotisAntonios-Alexandros Robotis

10.6k41741

answered Jan 3 at 15:34

Antonios-Alexandros RobotisAntonios-Alexandros Robotis

10.6k41741

answered Jan 3 at 15:34

Antonios-Alexandros RobotisAntonios-Alexandros Robotis

10.6k41741

0

$begingroup$

If $f : Gto G'$ be a homomorphic and onto then $h : dfracGK to G'$ is an isomorphism and onto where $K = ker f$.

share|cite|improve this answer

edited Mar 31 at 12:22

Max

1,0041319

answered Mar 31 at 12:10

Yamini Singh Yamini Singh

1

$endgroup$

add a comment | 

0

$begingroup$

If $f : Gto G'$ be a homomorphic and onto then $h : dfracGK to G'$ is an isomorphism and onto where $K = ker f$.

share|cite|improve this answer

edited Mar 31 at 12:22

Max

1,0041319

answered Mar 31 at 12:10

Yamini Singh Yamini Singh

1

$endgroup$

add a comment | 

$begingroup$

If $f : Gto G'$ be a homomorphic and onto then $h : dfracGK to G'$ is an isomorphism and onto where $K = ker f$.

share|cite|improve this answer

edited Mar 31 at 12:22

Max

1,0041319

answered Mar 31 at 12:10

Yamini Singh Yamini Singh

1

$endgroup$

share|cite|improve this answer

edited Mar 31 at 12:22

Max

1,0041319

answered Mar 31 at 12:10

Yamini Singh Yamini Singh

1

edited Mar 31 at 12:22

Max

1,0041319

edited Mar 31 at 12:22

Max

1,0041319

answered Mar 31 at 12:10

Yamini Singh Yamini Singh

1

answered Mar 31 at 12:10

Yamini Singh Yamini Singh

1