How to show that the kernel of this ring homomorphism $phi: mathbbZ[i]tomathbbZ/5mathbbZ$ is the principal ideal $(2+i)$?

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In order to prove that $Z[i]/(2+i)cong mathbbZ/5mathbbZ$, I defined the homomorphism $phi(a+bi)=(a-2b)mod 5$. It is easy to show that $2+iinker(phi)$, but I'm unsure how to prove the reverse inclusion.

I have the following so far: Suppose $a+bimapsto 0$. Then $a-2bequiv 0mod 5$, so $5$ divides $a-2b$.

I'm wondering if I can use the fact that $2+i$ is irreducible in $mathbbZ[i]$.

abstract-algebra ring-theory gaussian-integers

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asked Jul 25 at 22:45

FakeAnalyst56

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In order to prove that $Z[i]/(2+i)cong mathbbZ/5mathbbZ$, I defined the homomorphism $phi(a+bi)=(a-2b)mod 5$. It is easy to show that $2+iinker(phi)$, but I'm unsure how to prove the reverse inclusion.

I have the following so far: Suppose $a+bimapsto 0$. Then $a-2bequiv 0mod 5$, so $5$ divides $a-2b$.

I'm wondering if I can use the fact that $2+i$ is irreducible in $mathbbZ[i]$.

abstract-algebra ring-theory gaussian-integers

share|cite|improve this question

asked Jul 25 at 22:45

FakeAnalyst56

175

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

In order to prove that $Z[i]/(2+i)cong mathbbZ/5mathbbZ$, I defined the homomorphism $phi(a+bi)=(a-2b)mod 5$. It is easy to show that $2+iinker(phi)$, but I'm unsure how to prove the reverse inclusion.

I have the following so far: Suppose $a+bimapsto 0$. Then $a-2bequiv 0mod 5$, so $5$ divides $a-2b$.

I'm wondering if I can use the fact that $2+i$ is irreducible in $mathbbZ[i]$.

abstract-algebra ring-theory gaussian-integers

share|cite|improve this question

asked Jul 25 at 22:45

FakeAnalyst56

175

In order to prove that $Z[i]/(2+i)cong mathbbZ/5mathbbZ$, I defined the homomorphism $phi(a+bi)=(a-2b)mod 5$. It is easy to show that $2+iinker(phi)$, but I'm unsure how to prove the reverse inclusion.

I have the following so far: Suppose $a+bimapsto 0$. Then $a-2bequiv 0mod 5$, so $5$ divides $a-2b$.

I'm wondering if I can use the fact that $2+i$ is irreducible in $mathbbZ[i]$.

abstract-algebra ring-theory gaussian-integers

share|cite|improve this question

asked Jul 25 at 22:45

FakeAnalyst56

175

share|cite|improve this question

share|cite|improve this question

asked Jul 25 at 22:45

FakeAnalyst56

175

asked Jul 25 at 22:45

FakeAnalyst56

175

asked Jul 25 at 22:45

FakeAnalyst56

175

175

add a comment | 

add a comment | 

2 Answers
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up vote
1
down vote



accepted

Hint: Going from where you left off, suppose that $a-2b = 5n$; then $a+bi = 2b+5n+bi$. Now, why is this a multiple of $2+i$?

share|cite|improve this answer

answered Jul 25 at 22:58

Daniel Schepler

6,6681513

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I got it now, thank you very much: I found that $(2+i)[b+n(2-i)]=2b+5n+bi$.
  – FakeAnalyst56
  Jul 25 at 23:04
  

  
  
  
  

add a comment | 

up vote
0
down vote

Now show that the quotient has five elements (for example, by considering $mathbb Z[i]$ as a free abelian group of rank two and the ideal $(2+i)$ as the subgroup generated by $(2,1)$ and $(-1,2)$, and doing row and column reduction i.e. Smith normal form). Any surjective or injective function between sets of the same cardinality is bijective.

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answered Jul 25 at 22:47

Pedro Tamaroff♦

93.8k10143290

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I'm somewhat confused, would you mind clarifying: is this answer intended to help me prove the isomorphism or is it related to showing that $ker(phi)subset (2+i)$? (My confusion is mostly with the latter) Thanks for your answer regardless.
  – FakeAnalyst56
  Jul 25 at 22:53
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @FakeAnalyst56 You know that $mathbb Z[i]/kerphi$ has five elements: the induced map is onto $mathbb Z/5$ so this is an isomorphism. The inclusion you gave provides you with a map from $mathbb Z[i]/kerphi to mathbb Z[i]/(2+i)$ which is onto. Check that this last quotient also has five elements to conlude this is an isomorphism, so $(2+i)=kerphi$.
  – Pedro Tamaroff♦
  Jul 25 at 22:58
  
  

  
  
  
  

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

Hint: Going from where you left off, suppose that $a-2b = 5n$; then $a+bi = 2b+5n+bi$. Now, why is this a multiple of $2+i$?

share|cite|improve this answer

answered Jul 25 at 22:58

Daniel Schepler

6,6681513

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I got it now, thank you very much: I found that $(2+i)[b+n(2-i)]=2b+5n+bi$.
  – FakeAnalyst56
  Jul 25 at 23:04
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

Hint: Going from where you left off, suppose that $a-2b = 5n$; then $a+bi = 2b+5n+bi$. Now, why is this a multiple of $2+i$?

share|cite|improve this answer

answered Jul 25 at 22:58

Daniel Schepler

6,6681513

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I got it now, thank you very much: I found that $(2+i)[b+n(2-i)]=2b+5n+bi$.
  – FakeAnalyst56
  Jul 25 at 23:04
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

Hint: Going from where you left off, suppose that $a-2b = 5n$; then $a+bi = 2b+5n+bi$. Now, why is this a multiple of $2+i$?

share|cite|improve this answer

answered Jul 25 at 22:58

Daniel Schepler

6,6681513

Hint: Going from where you left off, suppose that $a-2b = 5n$; then $a+bi = 2b+5n+bi$. Now, why is this a multiple of $2+i$?

share|cite|improve this answer

answered Jul 25 at 22:58

Daniel Schepler

6,6681513

share|cite|improve this answer

share|cite|improve this answer

answered Jul 25 at 22:58

Daniel Schepler

6,6681513

answered Jul 25 at 22:58

Daniel Schepler

6,6681513

answered Jul 25 at 22:58

Daniel Schepler

6,6681513

6,6681513

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I got it now, thank you very much: I found that $(2+i)[b+n(2-i)]=2b+5n+bi$.
  – FakeAnalyst56
  Jul 25 at 23:04
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I got it now, thank you very much: I found that $(2+i)[b+n(2-i)]=2b+5n+bi$.
  – FakeAnalyst56
  Jul 25 at 23:04
  

  
  
  
  

I got it now, thank you very much: I found that $(2+i)[b+n(2-i)]=2b+5n+bi$.
– FakeAnalyst56
Jul 25 at 23:04

I got it now, thank you very much: I found that $(2+i)[b+n(2-i)]=2b+5n+bi$.
– FakeAnalyst56
Jul 25 at 23:04

add a comment | 

up vote
0
down vote

Now show that the quotient has five elements (for example, by considering $mathbb Z[i]$ as a free abelian group of rank two and the ideal $(2+i)$ as the subgroup generated by $(2,1)$ and $(-1,2)$, and doing row and column reduction i.e. Smith normal form). Any surjective or injective function between sets of the same cardinality is bijective.

share|cite|improve this answer

answered Jul 25 at 22:47

Pedro Tamaroff♦

93.8k10143290

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I'm somewhat confused, would you mind clarifying: is this answer intended to help me prove the isomorphism or is it related to showing that $ker(phi)subset (2+i)$? (My confusion is mostly with the latter) Thanks for your answer regardless.
  – FakeAnalyst56
  Jul 25 at 22:53
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @FakeAnalyst56 You know that $mathbb Z[i]/kerphi$ has five elements: the induced map is onto $mathbb Z/5$ so this is an isomorphism. The inclusion you gave provides you with a map from $mathbb Z[i]/kerphi to mathbb Z[i]/(2+i)$ which is onto. Check that this last quotient also has five elements to conlude this is an isomorphism, so $(2+i)=kerphi$.
  – Pedro Tamaroff♦
  Jul 25 at 22:58
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

Now show that the quotient has five elements (for example, by considering $mathbb Z[i]$ as a free abelian group of rank two and the ideal $(2+i)$ as the subgroup generated by $(2,1)$ and $(-1,2)$, and doing row and column reduction i.e. Smith normal form). Any surjective or injective function between sets of the same cardinality is bijective.

share|cite|improve this answer

answered Jul 25 at 22:47

Pedro Tamaroff♦

93.8k10143290

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I'm somewhat confused, would you mind clarifying: is this answer intended to help me prove the isomorphism or is it related to showing that $ker(phi)subset (2+i)$? (My confusion is mostly with the latter) Thanks for your answer regardless.
  – FakeAnalyst56
  Jul 25 at 22:53
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @FakeAnalyst56 You know that $mathbb Z[i]/kerphi$ has five elements: the induced map is onto $mathbb Z/5$ so this is an isomorphism. The inclusion you gave provides you with a map from $mathbb Z[i]/kerphi to mathbb Z[i]/(2+i)$ which is onto. Check that this last quotient also has five elements to conlude this is an isomorphism, so $(2+i)=kerphi$.
  – Pedro Tamaroff♦
  Jul 25 at 22:58
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

up vote
0
down vote

Now show that the quotient has five elements (for example, by considering $mathbb Z[i]$ as a free abelian group of rank two and the ideal $(2+i)$ as the subgroup generated by $(2,1)$ and $(-1,2)$, and doing row and column reduction i.e. Smith normal form). Any surjective or injective function between sets of the same cardinality is bijective.

share|cite|improve this answer

answered Jul 25 at 22:47

Pedro Tamaroff♦

93.8k10143290

Now show that the quotient has five elements (for example, by considering $mathbb Z[i]$ as a free abelian group of rank two and the ideal $(2+i)$ as the subgroup generated by $(2,1)$ and $(-1,2)$, and doing row and column reduction i.e. Smith normal form). Any surjective or injective function between sets of the same cardinality is bijective.

share|cite|improve this answer

answered Jul 25 at 22:47

Pedro Tamaroff♦

93.8k10143290

share|cite|improve this answer

share|cite|improve this answer

answered Jul 25 at 22:47

Pedro Tamaroff♦

93.8k10143290

answered Jul 25 at 22:47

Pedro Tamaroff♦

93.8k10143290

answered Jul 25 at 22:47

Pedro Tamaroff♦

93.8k10143290

93.8k10143290

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I'm somewhat confused, would you mind clarifying: is this answer intended to help me prove the isomorphism or is it related to showing that $ker(phi)subset (2+i)$? (My confusion is mostly with the latter) Thanks for your answer regardless.
  – FakeAnalyst56
  Jul 25 at 22:53
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @FakeAnalyst56 You know that $mathbb Z[i]/kerphi$ has five elements: the induced map is onto $mathbb Z/5$ so this is an isomorphism. The inclusion you gave provides you with a map from $mathbb Z[i]/kerphi to mathbb Z[i]/(2+i)$ which is onto. Check that this last quotient also has five elements to conlude this is an isomorphism, so $(2+i)=kerphi$.
  – Pedro Tamaroff♦
  Jul 25 at 22:58
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I'm somewhat confused, would you mind clarifying: is this answer intended to help me prove the isomorphism or is it related to showing that $ker(phi)subset (2+i)$? (My confusion is mostly with the latter) Thanks for your answer regardless.
  – FakeAnalyst56
  Jul 25 at 22:53
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @FakeAnalyst56 You know that $mathbb Z[i]/kerphi$ has five elements: the induced map is onto $mathbb Z/5$ so this is an isomorphism. The inclusion you gave provides you with a map from $mathbb Z[i]/kerphi to mathbb Z[i]/(2+i)$ which is onto. Check that this last quotient also has five elements to conlude this is an isomorphism, so $(2+i)=kerphi$.
  – Pedro Tamaroff♦
  Jul 25 at 22:58
  
  

  
  
  
  

I'm somewhat confused, would you mind clarifying: is this answer intended to help me prove the isomorphism or is it related to showing that $ker(phi)subset (2+i)$? (My confusion is mostly with the latter) Thanks for your answer regardless.
– FakeAnalyst56
Jul 25 at 22:53

I'm somewhat confused, would you mind clarifying: is this answer intended to help me prove the isomorphism or is it related to showing that $ker(phi)subset (2+i)$? (My confusion is mostly with the latter) Thanks for your answer regardless.
– FakeAnalyst56
Jul 25 at 22:53

@FakeAnalyst56 You know that $mathbb Z[i]/kerphi$ has five elements: the induced map is onto $mathbb Z/5$ so this is an isomorphism. The inclusion you gave provides you with a map from $mathbb Z[i]/kerphi to mathbb Z[i]/(2+i)$ which is onto. Check that this last quotient also has five elements to conlude this is an isomorphism, so $(2+i)=kerphi$.
– Pedro Tamaroff♦
Jul 25 at 22:58

@FakeAnalyst56 You know that $mathbb Z[i]/kerphi$ has five elements: the induced map is onto $mathbb Z/5$ so this is an isomorphism. The inclusion you gave provides you with a map from $mathbb Z[i]/kerphi to mathbb Z[i]/(2+i)$ which is onto. Check that this last quotient also has five elements to conlude this is an isomorphism, so $(2+i)=kerphi$.
– Pedro Tamaroff♦
Jul 25 at 22:58

add a comment | 

 

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