Homomorphic images of $mathbbZ$

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Suppose there is a homomorphism $f:mathbbZto R$ where $R$ is a non-zero ring.
By the definition of homomorphism $f(n)=f(1+1+1+1.....+1)=f(1)+f(1)+f(1)+....+f(1)$
Now suppose $f(1)=x$ then $f(n)=nx$ by definition. By varying x we can say that homomorphic images of $mathbbZ$ is $0,mathbbZ,mathbb2Z,mathbb3Z$...etc.

Am I right? Please provide some deep intuitions if any.

abstract-algebra proof-verification ring-theory

share|cite|improve this question

edited Jul 19 at 5:22

Vim

7,77931144

asked Jul 19 at 4:25

MathCosmo

298214

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Image isn't really related to $mathbbZ$ since ring $R$ might not even contain any integers.
  – coffeemath
  Jul 19 at 4:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Well in that case if R is a ring with unity then Z is embedded inside it. Question is what will be the homomorphic image is R has no unity?
  – MathCosmo
  Jul 19 at 4:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Then is your question about what can be said about $f(mathbbZ)$ ? It is an additive subgroup of the additive part of $R$ generated by $f(1)$ and this doesn't require $R$ to have a unity. It may also be a subring but I didn't check that.
  – coffeemath
  Jul 19 at 4:37
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes sir I want to know the image of Z. Give me an example where the image is other than 0,Z,2Z,3Z etc.
  – MathCosmo
  Jul 19 at 4:39
  

  
  
  
  

add a comment | 

up vote
1
down vote

favorite

1

Suppose there is a homomorphism $f:mathbbZto R$ where $R$ is a non-zero ring.
By the definition of homomorphism $f(n)=f(1+1+1+1.....+1)=f(1)+f(1)+f(1)+....+f(1)$
Now suppose $f(1)=x$ then $f(n)=nx$ by definition. By varying x we can say that homomorphic images of $mathbbZ$ is $0,mathbbZ,mathbb2Z,mathbb3Z$...etc.

Am I right? Please provide some deep intuitions if any.

abstract-algebra proof-verification ring-theory

share|cite|improve this question

edited Jul 19 at 5:22

Vim

7,77931144

asked Jul 19 at 4:25

MathCosmo

298214

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Image isn't really related to $mathbbZ$ since ring $R$ might not even contain any integers.
  – coffeemath
  Jul 19 at 4:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Well in that case if R is a ring with unity then Z is embedded inside it. Question is what will be the homomorphic image is R has no unity?
  – MathCosmo
  Jul 19 at 4:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Then is your question about what can be said about $f(mathbbZ)$ ? It is an additive subgroup of the additive part of $R$ generated by $f(1)$ and this doesn't require $R$ to have a unity. It may also be a subring but I didn't check that.
  – coffeemath
  Jul 19 at 4:37
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes sir I want to know the image of Z. Give me an example where the image is other than 0,Z,2Z,3Z etc.
  – MathCosmo
  Jul 19 at 4:39
  

  
  
  
  

add a comment | 

up vote
1
down vote

favorite

1

up vote
1
down vote

favorite

1

1

Suppose there is a homomorphism $f:mathbbZto R$ where $R$ is a non-zero ring.
By the definition of homomorphism $f(n)=f(1+1+1+1.....+1)=f(1)+f(1)+f(1)+....+f(1)$
Now suppose $f(1)=x$ then $f(n)=nx$ by definition. By varying x we can say that homomorphic images of $mathbbZ$ is $0,mathbbZ,mathbb2Z,mathbb3Z$...etc.

Am I right? Please provide some deep intuitions if any.

abstract-algebra proof-verification ring-theory

share|cite|improve this question

edited Jul 19 at 5:22

Vim

7,77931144

asked Jul 19 at 4:25

MathCosmo

298214

Suppose there is a homomorphism $f:mathbbZto R$ where $R$ is a non-zero ring.
By the definition of homomorphism $f(n)=f(1+1+1+1.....+1)=f(1)+f(1)+f(1)+....+f(1)$
Now suppose $f(1)=x$ then $f(n)=nx$ by definition. By varying x we can say that homomorphic images of $mathbbZ$ is $0,mathbbZ,mathbb2Z,mathbb3Z$...etc.

Am I right? Please provide some deep intuitions if any.

abstract-algebra proof-verification ring-theory

share|cite|improve this question

edited Jul 19 at 5:22

Vim

7,77931144

asked Jul 19 at 4:25

MathCosmo

298214

share|cite|improve this question

share|cite|improve this question

edited Jul 19 at 5:22

Vim

7,77931144

edited Jul 19 at 5:22

Vim

7,77931144

edited Jul 19 at 5:22

Vim

7,77931144

7,77931144

asked Jul 19 at 4:25

MathCosmo

298214

asked Jul 19 at 4:25

MathCosmo

298214

asked Jul 19 at 4:25

MathCosmo

298214

298214

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Image isn't really related to $mathbbZ$ since ring $R$ might not even contain any integers.
  – coffeemath
  Jul 19 at 4:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Well in that case if R is a ring with unity then Z is embedded inside it. Question is what will be the homomorphic image is R has no unity?
  – MathCosmo
  Jul 19 at 4:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Then is your question about what can be said about $f(mathbbZ)$ ? It is an additive subgroup of the additive part of $R$ generated by $f(1)$ and this doesn't require $R$ to have a unity. It may also be a subring but I didn't check that.
  – coffeemath
  Jul 19 at 4:37
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes sir I want to know the image of Z. Give me an example where the image is other than 0,Z,2Z,3Z etc.
  – MathCosmo
  Jul 19 at 4:39
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Image isn't really related to $mathbbZ$ since ring $R$ might not even contain any integers.
  – coffeemath
  Jul 19 at 4:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Well in that case if R is a ring with unity then Z is embedded inside it. Question is what will be the homomorphic image is R has no unity?
  – MathCosmo
  Jul 19 at 4:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Then is your question about what can be said about $f(mathbbZ)$ ? It is an additive subgroup of the additive part of $R$ generated by $f(1)$ and this doesn't require $R$ to have a unity. It may also be a subring but I didn't check that.
  – coffeemath
  Jul 19 at 4:37
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes sir I want to know the image of Z. Give me an example where the image is other than 0,Z,2Z,3Z etc.
  – MathCosmo
  Jul 19 at 4:39
  

  
  
  
  

1

1

Image isn't really related to $mathbbZ$ since ring $R$ might not even contain any integers.
– coffeemath
Jul 19 at 4:29

Image isn't really related to $mathbbZ$ since ring $R$ might not even contain any integers.
– coffeemath
Jul 19 at 4:29

Well in that case if R is a ring with unity then Z is embedded inside it. Question is what will be the homomorphic image is R has no unity?
– MathCosmo
Jul 19 at 4:32

Well in that case if R is a ring with unity then Z is embedded inside it. Question is what will be the homomorphic image is R has no unity?
– MathCosmo
Jul 19 at 4:32

Then is your question about what can be said about $f(mathbbZ)$ ? It is an additive subgroup of the additive part of $R$ generated by $f(1)$ and this doesn't require $R$ to have a unity. It may also be a subring but I didn't check that.
– coffeemath
Jul 19 at 4:37

Then is your question about what can be said about $f(mathbbZ)$ ? It is an additive subgroup of the additive part of $R$ generated by $f(1)$ and this doesn't require $R$ to have a unity. It may also be a subring but I didn't check that.
– coffeemath
Jul 19 at 4:37

Yes sir I want to know the image of Z. Give me an example where the image is other than 0,Z,2Z,3Z etc.
– MathCosmo
Jul 19 at 4:39

Yes sir I want to know the image of Z. Give me an example where the image is other than 0,Z,2Z,3Z etc.
– MathCosmo
Jul 19 at 4:39

add a comment | 

1 Answer
1

active

oldest

votes

up vote
1
down vote



accepted

Take $R=mathbfZ/nmathbfZ$ and define $fcolon mathbfZto mathbfZ/nmathbfZ$ to be the homomorphism $f(k) = kpmod n$. The image is a ring with finitely many elements and meets your requirement.

share|cite|improve this answer

answered Jul 19 at 4:51

P Vanchinathan

13.9k12035

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yeah I just realized that if R has unity and the map is one one then Z is embedded inside R and if the map is not one one then the images are Zn(upto isomorphism ), Now if R has no unity then the map is trivial, I have to show this now...any hints?
  – MathCosmo
  Jul 19 at 4:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  If you are disregarding unity, just treat Z as a group, focus on group homomorphisms defined there, and the images of such homomorphisms.
  – P Vanchinathan
  Jul 19 at 5:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Okay I got my answer.. Ideals of Z are Z,(0),nZ then applying first isomorphism theorem image will be isomorphic to Z,0,Zn(upto isomorphism)
  – MathCosmo
  Jul 19 at 5:15
  

  
  
  
  

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

Take $R=mathbfZ/nmathbfZ$ and define $fcolon mathbfZto mathbfZ/nmathbfZ$ to be the homomorphism $f(k) = kpmod n$. The image is a ring with finitely many elements and meets your requirement.

share|cite|improve this answer

answered Jul 19 at 4:51

P Vanchinathan

13.9k12035

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yeah I just realized that if R has unity and the map is one one then Z is embedded inside R and if the map is not one one then the images are Zn(upto isomorphism ), Now if R has no unity then the map is trivial, I have to show this now...any hints?
  – MathCosmo
  Jul 19 at 4:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  If you are disregarding unity, just treat Z as a group, focus on group homomorphisms defined there, and the images of such homomorphisms.
  – P Vanchinathan
  Jul 19 at 5:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Okay I got my answer.. Ideals of Z are Z,(0),nZ then applying first isomorphism theorem image will be isomorphic to Z,0,Zn(upto isomorphism)
  – MathCosmo
  Jul 19 at 5:15
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

Take $R=mathbfZ/nmathbfZ$ and define $fcolon mathbfZto mathbfZ/nmathbfZ$ to be the homomorphism $f(k) = kpmod n$. The image is a ring with finitely many elements and meets your requirement.

share|cite|improve this answer

answered Jul 19 at 4:51

P Vanchinathan

13.9k12035

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yeah I just realized that if R has unity and the map is one one then Z is embedded inside R and if the map is not one one then the images are Zn(upto isomorphism ), Now if R has no unity then the map is trivial, I have to show this now...any hints?
  – MathCosmo
  Jul 19 at 4:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  If you are disregarding unity, just treat Z as a group, focus on group homomorphisms defined there, and the images of such homomorphisms.
  – P Vanchinathan
  Jul 19 at 5:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Okay I got my answer.. Ideals of Z are Z,(0),nZ then applying first isomorphism theorem image will be isomorphic to Z,0,Zn(upto isomorphism)
  – MathCosmo
  Jul 19 at 5:15
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

Take $R=mathbfZ/nmathbfZ$ and define $fcolon mathbfZto mathbfZ/nmathbfZ$ to be the homomorphism $f(k) = kpmod n$. The image is a ring with finitely many elements and meets your requirement.

share|cite|improve this answer

answered Jul 19 at 4:51

P Vanchinathan

13.9k12035

Take $R=mathbfZ/nmathbfZ$ and define $fcolon mathbfZto mathbfZ/nmathbfZ$ to be the homomorphism $f(k) = kpmod n$. The image is a ring with finitely many elements and meets your requirement.

share|cite|improve this answer

answered Jul 19 at 4:51

P Vanchinathan

13.9k12035

share|cite|improve this answer

share|cite|improve this answer

answered Jul 19 at 4:51

P Vanchinathan

13.9k12035

answered Jul 19 at 4:51

P Vanchinathan

13.9k12035

answered Jul 19 at 4:51

P Vanchinathan

13.9k12035

13.9k12035

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yeah I just realized that if R has unity and the map is one one then Z is embedded inside R and if the map is not one one then the images are Zn(upto isomorphism ), Now if R has no unity then the map is trivial, I have to show this now...any hints?
  – MathCosmo
  Jul 19 at 4:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  If you are disregarding unity, just treat Z as a group, focus on group homomorphisms defined there, and the images of such homomorphisms.
  – P Vanchinathan
  Jul 19 at 5:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Okay I got my answer.. Ideals of Z are Z,(0),nZ then applying first isomorphism theorem image will be isomorphic to Z,0,Zn(upto isomorphism)
  – MathCosmo
  Jul 19 at 5:15
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yeah I just realized that if R has unity and the map is one one then Z is embedded inside R and if the map is not one one then the images are Zn(upto isomorphism ), Now if R has no unity then the map is trivial, I have to show this now...any hints?
  – MathCosmo
  Jul 19 at 4:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  If you are disregarding unity, just treat Z as a group, focus on group homomorphisms defined there, and the images of such homomorphisms.
  – P Vanchinathan
  Jul 19 at 5:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Okay I got my answer.. Ideals of Z are Z,(0),nZ then applying first isomorphism theorem image will be isomorphic to Z,0,Zn(upto isomorphism)
  – MathCosmo
  Jul 19 at 5:15
  

  
  
  
  

Yeah I just realized that if R has unity and the map is one one then Z is embedded inside R and if the map is not one one then the images are Zn(upto isomorphism ), Now if R has no unity then the map is trivial, I have to show this now...any hints?
– MathCosmo
Jul 19 at 4:58

Yeah I just realized that if R has unity and the map is one one then Z is embedded inside R and if the map is not one one then the images are Zn(upto isomorphism ), Now if R has no unity then the map is trivial, I have to show this now...any hints?
– MathCosmo
Jul 19 at 4:58

If you are disregarding unity, just treat Z as a group, focus on group homomorphisms defined there, and the images of such homomorphisms.
– P Vanchinathan
Jul 19 at 5:01

If you are disregarding unity, just treat Z as a group, focus on group homomorphisms defined there, and the images of such homomorphisms.
– P Vanchinathan
Jul 19 at 5:01

Okay I got my answer.. Ideals of Z are Z,(0),nZ then applying first isomorphism theorem image will be isomorphic to Z,0,Zn(upto isomorphism)
– MathCosmo
Jul 19 at 5:15

Okay I got my answer.. Ideals of Z are Z,(0),nZ then applying first isomorphism theorem image will be isomorphic to Z,0,Zn(upto isomorphism)
– MathCosmo
Jul 19 at 5:15

add a comment | 

 

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