Mixing
Prove $mathbb{Z}[sqrt{5}]$ and $mathbb{Z}[sqrt{6}]$ are Euclidean Domains
-2
$begingroup$
I would like to know how to prove $mathbb{Z}[sqrt{5}]$ and $mathbb{Z}[sqrt{6}]$. I don't know where to start, the method used for $mathbb{Z}[sqrt{2}]$ and $mathbb{Z}[sqrt{3}]$ doesn't work anymore. I think I'm lacking the intuitive idea.
abstract-algebra euclidean-domain
asked Jan 1 at 18:37
SevenSeven
989
$endgroup$
|
show 4 more comments
-2
$begingroup$
I would like to know how to prove $mathbb{Z}[sqrt{5}]$ and $mathbb{Z}[sqrt{6}]$. I don't know where to start, the method used for $mathbb{Z}[sqrt{2}]$ and $mathbb{Z}[sqrt{3}]$ doesn't work anymore. I think I'm lacking the intuitive idea.
abstract-algebra euclidean-domain
asked Jan 1 at 18:37
SevenSeven
989
$endgroup$
3
$begingroup$
$Bbb Q[sqrt k]$ is a field....
$endgroup$
– Lord Shark the Unknown
Jan 1 at 18:44
$begingroup$
Then it is always an ED... I'm most confused now. How can I then prove $mathbb{Z}[sqrt{6}]$ is an ED ?
$endgroup$
– Seven
Jan 1 at 18:49
$begingroup$
Is it possible that the book/article meant that if $mathbb{Q}[sqrt{k}]$ is norm-Euclidean?
$endgroup$
– freakish
Jan 1 at 18:55
1
$begingroup$
@Seven Guess a possible Euclidean norm, and then prove it really is one.
$endgroup$
– Lord Shark the Unknown
Jan 1 at 19:04
1
$begingroup$
@Deitrich Yes, $,R = Bbb Z[sqrt 5]$ is not a UFD since it is not even integrally closed; said simpler a UFD satisfies the Rational Root Tests (same proof as in $Bbb Z)$ but that fails in $R$ since the monic $,x^2-x-1,$ has a proper fractional root $,varphi = (1+sqrt{5})/2.,$ But it we enlarge to its integral closure $,Bbb Z[varphi]$ then that too is norm Euclidean. Seven: which ring did you intend?
$endgroup$
– Bill Dubuque
Jan 1 at 21:21
|
show 4 more comments
-2
-2
-2
$begingroup$
I would like to know how to prove $mathbb{Z}[sqrt{5}]$ and $mathbb{Z}[sqrt{6}]$. I don't know where to start, the method used for $mathbb{Z}[sqrt{2}]$ and $mathbb{Z}[sqrt{3}]$ doesn't work anymore. I think I'm lacking the intuitive idea.
abstract-algebra euclidean-domain
asked Jan 1 at 18:37
SevenSeven
989
$endgroup$
I would like to know how to prove $mathbb{Z}[sqrt{5}]$ and $mathbb{Z}[sqrt{6}]$. I don't know where to start, the method used for $mathbb{Z}[sqrt{2}]$ and $mathbb{Z}[sqrt{3}]$ doesn't work anymore. I think I'm lacking the intuitive idea.
abstract-algebra euclidean-domain
abstract-algebra euclidean-domain
asked Jan 1 at 18:37
SevenSeven
989
asked Jan 1 at 18:37
SevenSeven
989
edited Jan 1 at 20:29
Seven
asked Jan 1 at 18:37
SevenSeven
989
asked Jan 1 at 18:37
SevenSeven
989
asked Jan 1 at 18:37
SevenSeven
989
989
3
$begingroup$
$Bbb Q[sqrt k]$ is a field....
$endgroup$
– Lord Shark the Unknown
Jan 1 at 18:44
$begingroup$
Then it is always an ED... I'm most confused now. How can I then prove $mathbb{Z}[sqrt{6}]$ is an ED ?
$endgroup$
– Seven
Jan 1 at 18:49
$begingroup$
Is it possible that the book/article meant that if $mathbb{Q}[sqrt{k}]$ is norm-Euclidean?
$endgroup$
– freakish
Jan 1 at 18:55
1
$begingroup$
@Seven Guess a possible Euclidean norm, and then prove it really is one.
$endgroup$
– Lord Shark the Unknown
Jan 1 at 19:04
1
$begingroup$
@Deitrich Yes, $,R = Bbb Z[sqrt 5]$ is not a UFD since it is not even integrally closed; said simpler a UFD satisfies the Rational Root Tests (same proof as in $Bbb Z)$ but that fails in $R$ since the monic $,x^2-x-1,$ has a proper fractional root $,varphi = (1+sqrt{5})/2.,$ But it we enlarge to its integral closure $,Bbb Z[varphi]$ then that too is norm Euclidean. Seven: which ring did you intend?
$endgroup$
– Bill Dubuque
Jan 1 at 21:21
|
show 4 more comments
3
$begingroup$
$Bbb Q[sqrt k]$ is a field....
$endgroup$
– Lord Shark the Unknown
Jan 1 at 18:44
$begingroup$
Then it is always an ED... I'm most confused now. How can I then prove $mathbb{Z}[sqrt{6}]$ is an ED ?
$endgroup$
– Seven
Jan 1 at 18:49
$begingroup$
Is it possible that the book/article meant that if $mathbb{Q}[sqrt{k}]$ is norm-Euclidean?
$endgroup$
– freakish
Jan 1 at 18:55
1
$begingroup$
@Seven Guess a possible Euclidean norm, and then prove it really is one.
$endgroup$
– Lord Shark the Unknown
Jan 1 at 19:04
1
$begingroup$
@Deitrich Yes, $,R = Bbb Z[sqrt 5]$ is not a UFD since it is not even integrally closed; said simpler a UFD satisfies the Rational Root Tests (same proof as in $Bbb Z)$ but that fails in $R$ since the monic $,x^2-x-1,$ has a proper fractional root $,varphi = (1+sqrt{5})/2.,$ But it we enlarge to its integral closure $,Bbb Z[varphi]$ then that too is norm Euclidean. Seven: which ring did you intend?
$endgroup$
– Bill Dubuque
Jan 1 at 21:21
3
3
$begingroup$
$Bbb Q[sqrt k]$ is a field....
$endgroup$
– Lord Shark the Unknown
Jan 1 at 18:44
$begingroup$
$Bbb Q[sqrt k]$ is a field....
$endgroup$
– Lord Shark the Unknown
Jan 1 at 18:44
$begingroup$
Then it is always an ED... I'm most confused now. How can I then prove $mathbb{Z}[sqrt{6}]$ is an ED ?
$endgroup$
– Seven
Jan 1 at 18:49
$begingroup$
Then it is always an ED... I'm most confused now. How can I then prove $mathbb{Z}[sqrt{6}]$ is an ED ?
$endgroup$
– Seven
Jan 1 at 18:49
$begingroup$
Is it possible that the book/article meant that if $mathbb{Q}[sqrt{k}]$ is norm-Euclidean?
$endgroup$
– freakish
Jan 1 at 18:55
$begingroup$
Is it possible that the book/article meant that if $mathbb{Q}[sqrt{k}]$ is norm-Euclidean?
$endgroup$
– freakish
Jan 1 at 18:55
1
1
$begingroup$
@Seven Guess a possible Euclidean norm, and then prove it really is one.
$endgroup$
– Lord Shark the Unknown
Jan 1 at 19:04
$begingroup$
@Seven Guess a possible Euclidean norm, and then prove it really is one.
$endgroup$
– Lord Shark the Unknown
Jan 1 at 19:04
1
1
$begingroup$
@Deitrich Yes, $,R = Bbb Z[sqrt 5]$ is not a UFD since it is not even integrally closed; said simpler a UFD satisfies the Rational Root Tests (same proof as in $Bbb Z)$ but that fails in $R$ since the monic $,x^2-x-1,$ has a proper fractional root $,varphi = (1+sqrt{5})/2.,$ But it we enlarge to its integral closure $,Bbb Z[varphi]$ then that too is norm Euclidean. Seven: which ring did you intend?
$endgroup$
– Bill Dubuque
Jan 1 at 21:21
$begingroup$
@Deitrich Yes, $,R = Bbb Z[sqrt 5]$ is not a UFD since it is not even integrally closed; said simpler a UFD satisfies the Rational Root Tests (same proof as in $Bbb Z)$ but that fails in $R$ since the monic $,x^2-x-1,$ has a proper fractional root $,varphi = (1+sqrt{5})/2.,$ But it we enlarge to its integral closure $,Bbb Z[varphi]$ then that too is norm Euclidean. Seven: which ring did you intend?
$endgroup$
– Bill Dubuque
Jan 1 at 21:21
|
show 4 more comments
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3
$begingroup$
$Bbb Q[sqrt k]$ is a field....
$endgroup$
– Lord Shark the Unknown
Jan 1 at 18:44
$begingroup$
Then it is always an ED... I'm most confused now. How can I then prove $mathbb{Z}[sqrt{6}]$ is an ED ?
$endgroup$
– Seven
Jan 1 at 18:49
$begingroup$
Is it possible that the book/article meant that if $mathbb{Q}[sqrt{k}]$ is norm-Euclidean?
$endgroup$
– freakish
Jan 1 at 18:55
1
$begingroup$
@Seven Guess a possible Euclidean norm, and then prove it really is one.
$endgroup$
– Lord Shark the Unknown
Jan 1 at 19:04
1
$begingroup$
@Deitrich Yes, $,R = Bbb Z[sqrt 5]$ is not a UFD since it is not even integrally closed; said simpler a UFD satisfies the Rational Root Tests (same proof as in $Bbb Z)$ but that fails in $R$ since the monic $,x^2-x-1,$ has a proper fractional root $,varphi = (1+sqrt{5})/2.,$ But it we enlarge to its integral closure $,Bbb Z[varphi]$ then that too is norm Euclidean. Seven: which ring did you intend?
$endgroup$
– Bill Dubuque
Jan 1 at 21:21