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Ring of integers of $mathbb{Q}(i,sqrt{5})$

6

$begingroup$

I'm trying to find the ring of integers $A_L$ of $mathbb{Q}(i,sqrt{5})$. I know that the ring of integers of $mathbb{Q}(i)$ is $mathbb{Z}[i]$ and that the one of $mathbb{Q}(sqrt{5})$ is $mathbb{Z}left[frac{1+sqrt{5}}{2}right]$.
I would like to say that $A_L=mathbb{Z}left[frac{1+sqrt{5}}{2},iright]$.

From Integral basis of the ring of integers of an extension, given integral bases of the rings of integers of subfields I understood it is possible to say $A_L=mathbb{Z}left[frac{1+sqrt{5}}{2}right]mathbb{Z}left[iright]$ using the fact the discriminants of the two integer basis are coprime (Here we use a result that is possible to find in Marcus' book).

Is there a way to find $A_L$ in a more direct way without using that result?

number-theory algebraic-number-theory integer-rings

share|cite|improve this question

asked Mar 22 at 12:41

TomiriTomiri

313

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  For what it's worth, multiplying units of the intermediate rings could give you the defining polynomial, e.g., $$frac{-i}{2} - frac{sqrt 5}{2}$$ has polynomial $x^4 + 3x^2 + 1$, from which you can find its entry in the LMFDB: lmfdb.com/NumberField/4.0.400.1
  $endgroup$
  – The Short One
  Mar 23 at 21:41
  

  
  
  
  

add a comment | 

6

$begingroup$

I'm trying to find the ring of integers $A_L$ of $mathbb{Q}(i,sqrt{5})$. I know that the ring of integers of $mathbb{Q}(i)$ is $mathbb{Z}[i]$ and that the one of $mathbb{Q}(sqrt{5})$ is $mathbb{Z}left[frac{1+sqrt{5}}{2}right]$.
I would like to say that $A_L=mathbb{Z}left[frac{1+sqrt{5}}{2},iright]$.

From Integral basis of the ring of integers of an extension, given integral bases of the rings of integers of subfields I understood it is possible to say $A_L=mathbb{Z}left[frac{1+sqrt{5}}{2}right]mathbb{Z}left[iright]$ using the fact the discriminants of the two integer basis are coprime (Here we use a result that is possible to find in Marcus' book).

Is there a way to find $A_L$ in a more direct way without using that result?

number-theory algebraic-number-theory integer-rings

share|cite|improve this question

asked Mar 22 at 12:41

TomiriTomiri

313

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  For what it's worth, multiplying units of the intermediate rings could give you the defining polynomial, e.g., $$frac{-i}{2} - frac{sqrt 5}{2}$$ has polynomial $x^4 + 3x^2 + 1$, from which you can find its entry in the LMFDB: lmfdb.com/NumberField/4.0.400.1
  $endgroup$
  – The Short One
  Mar 23 at 21:41
  

  
  
  
  

add a comment | 

$begingroup$

I'm trying to find the ring of integers $A_L$ of $mathbb{Q}(i,sqrt{5})$. I know that the ring of integers of $mathbb{Q}(i)$ is $mathbb{Z}[i]$ and that the one of $mathbb{Q}(sqrt{5})$ is $mathbb{Z}left[frac{1+sqrt{5}}{2}right]$.
I would like to say that $A_L=mathbb{Z}left[frac{1+sqrt{5}}{2},iright]$.

From Integral basis of the ring of integers of an extension, given integral bases of the rings of integers of subfields I understood it is possible to say $A_L=mathbb{Z}left[frac{1+sqrt{5}}{2}right]mathbb{Z}left[iright]$ using the fact the discriminants of the two integer basis are coprime (Here we use a result that is possible to find in Marcus' book).

Is there a way to find $A_L$ in a more direct way without using that result?

number-theory algebraic-number-theory integer-rings

share|cite|improve this question

asked Mar 22 at 12:41

TomiriTomiri

313

$endgroup$

share|cite|improve this question

asked Mar 22 at 12:41

TomiriTomiri

313

share|cite|improve this question

asked Mar 22 at 12:41

TomiriTomiri

313

asked Mar 22 at 12:41

TomiriTomiri

313

asked Mar 22 at 12:41

TomiriTomiri

313

1 Answer
1

active

oldest

votes

2

$begingroup$

Write $omega = frac{1 + sqrt{5}}2$. Then all elements $alpha = a + bi + comega + diomega$ where $a, b, c, d$ are integers. If the ring of integers is larger, there must be algebraic integers of the form
$alpha/2$ or $alpha/5$ since $2$ and $5$ are the only prime divisors of the discriminant of the subring generated by $i$ and $omega$. Now all you have to do is show that if $alpha/2$ is an algebraic integer, then $a$, $b$, $c$, $d$ are divisible by $2$, and then do the same with respect to the prime $5$.

share|cite|improve this answer

answered Mar 24 at 9:25

franz lemmermeyerfranz lemmermeyer

7,53922048

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  This must be in Marcus. If $K$ is any number field generated by an algebraic integer $alpha$, and if $alpha$ has discriminant $D$, then any algebraic integer in $K$ is a fraction with denominator $D$, whose numerator is a ${mathbb Z}$-linear combination of powers of $alpha$.
  $endgroup$
  – franz lemmermeyer
  Mar 24 at 16:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm sorry I deleted the question just two minutes after your comment because I didn't see it and I managed to solve the problem manually. I show it in our case: Let $ y= a+bα+cbeta+dalphabeta in A_L$ with $ a,b,c,d in mathbb{Q} $. We know that $$ D(a,bα,cbeta,dalphabeta)= (abcd)^2 cdot D(1,α,beta,αbeta).$$ It follows that $ (abcd)^2 = D(a,bα,cbeta,dalphabeta) D(1,α,beta,αbeta)^{-1}$ and $ D(1,α,beta,αbeta)in mathbb{Z} $. So we have trivially the result. I hope this is right and useful!
  $endgroup$
  – Tomiri
  Mar 24 at 17:49
  
  
  

  
  
  
  

add a comment | 

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2

$begingroup$

Write $omega = frac{1 + sqrt{5}}2$. Then all elements $alpha = a + bi + comega + diomega$ where $a, b, c, d$ are integers. If the ring of integers is larger, there must be algebraic integers of the form
$alpha/2$ or $alpha/5$ since $2$ and $5$ are the only prime divisors of the discriminant of the subring generated by $i$ and $omega$. Now all you have to do is show that if $alpha/2$ is an algebraic integer, then $a$, $b$, $c$, $d$ are divisible by $2$, and then do the same with respect to the prime $5$.

share|cite|improve this answer

answered Mar 24 at 9:25

franz lemmermeyerfranz lemmermeyer

7,53922048

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  This must be in Marcus. If $K$ is any number field generated by an algebraic integer $alpha$, and if $alpha$ has discriminant $D$, then any algebraic integer in $K$ is a fraction with denominator $D$, whose numerator is a ${mathbb Z}$-linear combination of powers of $alpha$.
  $endgroup$
  – franz lemmermeyer
  Mar 24 at 16:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm sorry I deleted the question just two minutes after your comment because I didn't see it and I managed to solve the problem manually. I show it in our case: Let $ y= a+bα+cbeta+dalphabeta in A_L$ with $ a,b,c,d in mathbb{Q} $. We know that $$ D(a,bα,cbeta,dalphabeta)= (abcd)^2 cdot D(1,α,beta,αbeta).$$ It follows that $ (abcd)^2 = D(a,bα,cbeta,dalphabeta) D(1,α,beta,αbeta)^{-1}$ and $ D(1,α,beta,αbeta)in mathbb{Z} $. So we have trivially the result. I hope this is right and useful!
  $endgroup$
  – Tomiri
  Mar 24 at 17:49
  
  
  

  
  
  
  

add a comment | 

2

$begingroup$

Write $omega = frac{1 + sqrt{5}}2$. Then all elements $alpha = a + bi + comega + diomega$ where $a, b, c, d$ are integers. If the ring of integers is larger, there must be algebraic integers of the form
$alpha/2$ or $alpha/5$ since $2$ and $5$ are the only prime divisors of the discriminant of the subring generated by $i$ and $omega$. Now all you have to do is show that if $alpha/2$ is an algebraic integer, then $a$, $b$, $c$, $d$ are divisible by $2$, and then do the same with respect to the prime $5$.

share|cite|improve this answer

answered Mar 24 at 9:25

franz lemmermeyerfranz lemmermeyer

7,53922048

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  This must be in Marcus. If $K$ is any number field generated by an algebraic integer $alpha$, and if $alpha$ has discriminant $D$, then any algebraic integer in $K$ is a fraction with denominator $D$, whose numerator is a ${mathbb Z}$-linear combination of powers of $alpha$.
  $endgroup$
  – franz lemmermeyer
  Mar 24 at 16:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm sorry I deleted the question just two minutes after your comment because I didn't see it and I managed to solve the problem manually. I show it in our case: Let $ y= a+bα+cbeta+dalphabeta in A_L$ with $ a,b,c,d in mathbb{Q} $. We know that $$ D(a,bα,cbeta,dalphabeta)= (abcd)^2 cdot D(1,α,beta,αbeta).$$ It follows that $ (abcd)^2 = D(a,bα,cbeta,dalphabeta) D(1,α,beta,αbeta)^{-1}$ and $ D(1,α,beta,αbeta)in mathbb{Z} $. So we have trivially the result. I hope this is right and useful!
  $endgroup$
  – Tomiri
  Mar 24 at 17:49
  
  
  

  
  
  
  

add a comment | 

$begingroup$

Write $omega = frac{1 + sqrt{5}}2$. Then all elements $alpha = a + bi + comega + diomega$ where $a, b, c, d$ are integers. If the ring of integers is larger, there must be algebraic integers of the form
$alpha/2$ or $alpha/5$ since $2$ and $5$ are the only prime divisors of the discriminant of the subring generated by $i$ and $omega$. Now all you have to do is show that if $alpha/2$ is an algebraic integer, then $a$, $b$, $c$, $d$ are divisible by $2$, and then do the same with respect to the prime $5$.

share|cite|improve this answer

answered Mar 24 at 9:25

franz lemmermeyerfranz lemmermeyer

7,53922048

$endgroup$

share|cite|improve this answer

answered Mar 24 at 9:25

franz lemmermeyerfranz lemmermeyer

7,53922048

answered Mar 24 at 9:25

franz lemmermeyerfranz lemmermeyer

7,53922048

answered Mar 24 at 9:25

franz lemmermeyerfranz lemmermeyer

7,53922048