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No gcd$left(6, 3+3sqrt-5right)$ in $mathbbZ[sqrt-5]$ [duplicate]

Showing that the elements $6$ and $2+2sqrt5$ in $mathbbZ[sqrt5]$ have no gcdProve that $6$ and $2(1+sqrt-5)$ do not have a gcd in $mathbbZ[sqrt-5]$Integral domain with two elements that do not have a gcdShow that a space $A$ is an integral domain.Would love some criticism on a couple divisibility proofsProve $gcdleft(fraca^m - 1a -1,a -1right) = gcd(m,a-1)$Find all ideals of $mathbbZ_n$if $pmid a$ and $pmid b$ then $pmid gcd(a,b)$Prove: if $d=textgcd(m,n)$ so $textgcdleft(fracmd,fracndright)=1$GCD of two polynomials without Euclidean algorithmGCD of $2$ and $4 + sqrt10$ in $mathbbZ[sqrt10]$.Prove $a mid m$ and $b mid m implies fraca cdot btextgcd(a, b) mid m$

0

$begingroup$

This question already has an answer here:

• 
  Showing that the elements $6$ and $2+2sqrt5$ in $mathbbZ[sqrt5]$ have no gcd
  
  2 answers
  
  


• 
  Prove that $6$ and $2(1+sqrt-5)$ do not have a gcd in $mathbbZ[sqrt-5]$ [closed]
  
  2 answers
  
  

Let $R$ be an integral domain, $a,b,d in R$. We call $d$ a greatest common divisor of $a$ and $b$ (notation: $d = textgcd(a,b)$) if we have

I) $(a) + (b) subseteq (d);$

II) $(forall xin R)((a)+(b) subseteq (x) Rightarrow (d) subseteq (x))$.

Now let $R = mathbbZleft[sqrt-5right]$, $a = 6, b = 3+3sqrt-5 in R$.

To demonstrate: $a$ and $b$ have no greatest common divisor.

Attempt: As $6 = left(1+sqrt-5right)left(1-sqrt-5right)$, both $x_1 = 3$ and $x_2 = 1+sqrt-5$ are common divisors of $a$ and $b$. Now suppose $d = d_1 + sqrt-5d_2$ satisfies $(d) subseteq (x_1)$ and $(d) subseteq (x_2)$. Then if we would assume that $d mid 6$, it should not be true that also $6 mid 3+3sqrt-5$.

After tinkering around with huge numbers of variables trying to find a contradiction here, I'm starting to wonder whether this is the way to go. It looked really promising in the beginning to find it this way, but after 2 hours of failure I'm starting to wonder... What I also haven't used yet (because I wouldn't know how) is that gcd$left(1+sqrt-5, 1-sqrt-5right) = 1.$ I've proved this before so it can be considered a given.

abstract-algebra ring-theory greatest-common-divisor

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asked Mar 12 at 19:08

Jos van NieuwmanJos van Nieuwman

1409

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marked as duplicate by Dietrich Burde abstract-algebra
Users with the  abstract-algebra badge can single-handedly close abstract-algebra questions as duplicates and reopen them as needed.

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Mar 12 at 19:11

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Actually, Bill's proof at this duplicate is more detailed.
  $endgroup$
  – Dietrich Burde
  Mar 12 at 19:13
  

  
  
  
  

add a comment | 

0

$begingroup$

This question already has an answer here:

• 
  Showing that the elements $6$ and $2+2sqrt5$ in $mathbbZ[sqrt5]$ have no gcd
  
  2 answers
  
  


• 
  Prove that $6$ and $2(1+sqrt-5)$ do not have a gcd in $mathbbZ[sqrt-5]$ [closed]
  
  2 answers
  
  

Let $R$ be an integral domain, $a,b,d in R$. We call $d$ a greatest common divisor of $a$ and $b$ (notation: $d = textgcd(a,b)$) if we have

I) $(a) + (b) subseteq (d);$

II) $(forall xin R)((a)+(b) subseteq (x) Rightarrow (d) subseteq (x))$.

Now let $R = mathbbZleft[sqrt-5right]$, $a = 6, b = 3+3sqrt-5 in R$.

To demonstrate: $a$ and $b$ have no greatest common divisor.

Attempt: As $6 = left(1+sqrt-5right)left(1-sqrt-5right)$, both $x_1 = 3$ and $x_2 = 1+sqrt-5$ are common divisors of $a$ and $b$. Now suppose $d = d_1 + sqrt-5d_2$ satisfies $(d) subseteq (x_1)$ and $(d) subseteq (x_2)$. Then if we would assume that $d mid 6$, it should not be true that also $6 mid 3+3sqrt-5$.

After tinkering around with huge numbers of variables trying to find a contradiction here, I'm starting to wonder whether this is the way to go. It looked really promising in the beginning to find it this way, but after 2 hours of failure I'm starting to wonder... What I also haven't used yet (because I wouldn't know how) is that gcd$left(1+sqrt-5, 1-sqrt-5right) = 1.$ I've proved this before so it can be considered a given.

abstract-algebra ring-theory greatest-common-divisor

share|cite|improve this question

asked Mar 12 at 19:08

Jos van NieuwmanJos van Nieuwman

1409

$endgroup$

marked as duplicate by Dietrich Burde abstract-algebra
Users with the  abstract-algebra badge can single-handedly close abstract-algebra questions as duplicates and reopen them as needed.

StackExchange.ready(function()
if (StackExchange.options.isMobile) return;

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Mar 12 at 19:11

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Actually, Bill's proof at this duplicate is more detailed.
  $endgroup$
  – Dietrich Burde
  Mar 12 at 19:13
  

  
  
  
  

add a comment | 

$begingroup$

This question already has an answer here:

• 
  Showing that the elements $6$ and $2+2sqrt5$ in $mathbbZ[sqrt5]$ have no gcd
  
  2 answers
  
  


• 
  Prove that $6$ and $2(1+sqrt-5)$ do not have a gcd in $mathbbZ[sqrt-5]$ [closed]
  
  2 answers
  
  

Let $R$ be an integral domain, $a,b,d in R$. We call $d$ a greatest common divisor of $a$ and $b$ (notation: $d = textgcd(a,b)$) if we have

I) $(a) + (b) subseteq (d);$

II) $(forall xin R)((a)+(b) subseteq (x) Rightarrow (d) subseteq (x))$.

Now let $R = mathbbZleft[sqrt-5right]$, $a = 6, b = 3+3sqrt-5 in R$.

To demonstrate: $a$ and $b$ have no greatest common divisor.

Attempt: As $6 = left(1+sqrt-5right)left(1-sqrt-5right)$, both $x_1 = 3$ and $x_2 = 1+sqrt-5$ are common divisors of $a$ and $b$. Now suppose $d = d_1 + sqrt-5d_2$ satisfies $(d) subseteq (x_1)$ and $(d) subseteq (x_2)$. Then if we would assume that $d mid 6$, it should not be true that also $6 mid 3+3sqrt-5$.

After tinkering around with huge numbers of variables trying to find a contradiction here, I'm starting to wonder whether this is the way to go. It looked really promising in the beginning to find it this way, but after 2 hours of failure I'm starting to wonder... What I also haven't used yet (because I wouldn't know how) is that gcd$left(1+sqrt-5, 1-sqrt-5right) = 1.$ I've proved this before so it can be considered a given.

abstract-algebra ring-theory greatest-common-divisor

share|cite|improve this question

asked Mar 12 at 19:08

Jos van NieuwmanJos van Nieuwman

1409

$endgroup$

share|cite|improve this question

asked Mar 12 at 19:08

Jos van NieuwmanJos van Nieuwman

1409

share|cite|improve this question

asked Mar 12 at 19:08

Jos van NieuwmanJos van Nieuwman

1409

asked Mar 12 at 19:08

Jos van NieuwmanJos van Nieuwman

1409

asked Mar 12 at 19:08

Jos van NieuwmanJos van Nieuwman

1409