Mrthdrb

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20

4

$begingroup$

Consider a field $K$, let's say $K subseteq mathbb R$. We can consider the 'plane' $K times K$. I am wondering in which cases the distance function $d: K times K to mathbb R$, defined as is normal by $d(x, y) = sqrtx^2 + y^2$, takes values in $K$.

Certainly this is not true for $mathbb Q$: we have $d(1, 1) = sqrt2 notin mathbb Q$. If we take any $K$ which is closed under taking square roots of non-negative numbers, then certainly $d$ will take values in $K$.

However, a priori it might still be true that $a in K$ positive has no square root, yet this does not provide an obstruction because there is no way to write $a = x^2 + y^2$. Thus I am wondering:

Are there fields $K subseteq mathbb R$ which do not have all square roots of positive numbers, yet are closed under $d$?

abstract-algebra field-theory

share|cite|improve this question

edited yesterday

Mees de Vries

asked yesterday

Mees de VriesMees de Vries

17.5k12958

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Just so you know, there are other distances besides euclidean distance.
  $endgroup$
  – PyRulez
  22 hours ago
  

  
  
  
  

add a comment | 

20

4

$begingroup$

Consider a field $K$, let's say $K subseteq mathbb R$. We can consider the 'plane' $K times K$. I am wondering in which cases the distance function $d: K times K to mathbb R$, defined as is normal by $d(x, y) = sqrtx^2 + y^2$, takes values in $K$.

Certainly this is not true for $mathbb Q$: we have $d(1, 1) = sqrt2 notin mathbb Q$. If we take any $K$ which is closed under taking square roots of non-negative numbers, then certainly $d$ will take values in $K$.

However, a priori it might still be true that $a in K$ positive has no square root, yet this does not provide an obstruction because there is no way to write $a = x^2 + y^2$. Thus I am wondering:

Are there fields $K subseteq mathbb R$ which do not have all square roots of positive numbers, yet are closed under $d$?

abstract-algebra field-theory

share|cite|improve this question

edited yesterday

Mees de Vries

asked yesterday

Mees de VriesMees de Vries

17.5k12958

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Just so you know, there are other distances besides euclidean distance.
  $endgroup$
  – PyRulez
  22 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Consider a field $K$, let's say $K subseteq mathbb R$. We can consider the 'plane' $K times K$. I am wondering in which cases the distance function $d: K times K to mathbb R$, defined as is normal by $d(x, y) = sqrtx^2 + y^2$, takes values in $K$.

Certainly this is not true for $mathbb Q$: we have $d(1, 1) = sqrt2 notin mathbb Q$. If we take any $K$ which is closed under taking square roots of non-negative numbers, then certainly $d$ will take values in $K$.

However, a priori it might still be true that $a in K$ positive has no square root, yet this does not provide an obstruction because there is no way to write $a = x^2 + y^2$. Thus I am wondering:

Are there fields $K subseteq mathbb R$ which do not have all square roots of positive numbers, yet are closed under $d$?

abstract-algebra field-theory

share|cite|improve this question

edited yesterday

Mees de Vries

asked yesterday

Mees de VriesMees de Vries

17.5k12958

$endgroup$

share|cite|improve this question

edited yesterday

Mees de Vries

asked yesterday

Mees de VriesMees de Vries

17.5k12958

share|cite|improve this question

edited yesterday

Mees de Vries

asked yesterday

Mees de VriesMees de Vries

17.5k12958

asked yesterday

Mees de VriesMees de Vries

17.5k12958

asked yesterday

Mees de VriesMees de Vries

17.5k12958

2 Answers
2

active

oldest

votes

16

$begingroup$

Consider the tower of fields

$K_0:=mathbbQ$,

$K_i+1:=K_i(sqrtx^2+y^2| x,yin K_i)$,

$K:=bigcup_i K_i$.

Then $K$ is closed under $d$ and contains $1+sqrt 5$ but not $sqrt1+sqrt 5$, as I have found by following the Pythagorean fields Wikipedia link given by @Dirk in his answer: If $sqrt1+sqrt 5$ were in $K$ then $1+sqrt 5$ would be a sum of two squares in some extension $K_i$, and then it would be so in an extension of $mathbbQ(sqrt 5)$, which implies that it is a sum of squares in $mathbbQ(sqrt 5)$, which is impossible because that would entail that $1-sqrt 5$, which is negative, is also a sum of squares in $mathbbQ(sqrt 5)$.

The details can be found in Chapter 5 of the book Geometric constructions by Martin. The relevant results are Theorems 5.10-5.15.

Similarly, $sqrt 2in K$ but $sqrt[4]2notin K$, and more in general, this is true for any positive number which is not a sum of squares in the first extension in which it appears.

Geometrically, numbers in $K$ correspond to constructible points by ruler and dividers. Hence $sqrt[4]2$ is constructible by rule and compass but not by rule and dividers.

share|cite|improve this answer

edited yesterday

answered yesterday

Jose BroxJose Brox

3,33211129

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  "which implies that it is a sum of squares in $mathbb Q(sqrt5)$" -- I do not see directly how this follows from the previous sentence, but I could very well be missing something obvious.
  $endgroup$
  – Mees de Vries
  yesterday
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @MeesdeVries Not obvious, it is a consequence of Theorems 5.10-5.13 in Martin's book.
  $endgroup$
  – Jose Brox
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $K$ is countable, right?
  $endgroup$
  – PyRulez
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @PyRulez yeah the same construction which makes $mathbb Q$ countable from $mathbb N^2$ plus skipping over duplicates should work to imply that $K_i+1$ is countable given that $K_i$ is countable, skipping over duplicates; by induction therefore all $K_i$ are countable; then we should be able to repeat the same construction again with $K_m(n)$, again skipping over duplicates, to find that $K$ is countable.
  $endgroup$
  – CR Drost
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @PyRulez Yes: clearly, all elements of $K$ are algebraic, and algebraic numbers are countable (there is a countable number of rational polynomials, with a finite number of roots each)
  $endgroup$
  – Jose Brox
  23 hours ago
  
  

  
  
  
  

add a comment | 

5

$begingroup$

edit: Look what I found:
Wiki

---

The field
$$mathbbQ(sqrtp mid p in mathbbP)$$
might be a good candidate.

At least, all fields closed under $d$ must contain this field.

share|cite|improve this answer

edited yesterday

answered yesterday

DirkDirk

4,408218

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why must a field closed under $d$ contain $sqrt3$?
  $endgroup$
  – FredH
  yesterday
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @FredH, it must contain $sqrt2 = d(1, 1)$, and thus it must contain $sqrt3 = d(1, sqrt2)$.
  $endgroup$
  – Mees de Vries
  yesterday
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  See en.wikipedia.org/wiki/Spiral_of_Theodorus
  $endgroup$
  – lhf
  yesterday
  

  
  
  
  


• 
  
  
  

  
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  $begingroup$
  I don't think this works: $d(sqrt 2 + 1, 1) = sqrt2sqrt2 + 4$, but that doesn't look like a sum of square roots of rational numbers.
  $endgroup$
  – Arthur
  yesterday
  
  

  
  
  
  

add a comment | 

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16

$begingroup$

Consider the tower of fields

$K_0:=mathbbQ$,

$K_i+1:=K_i(sqrtx^2+y^2| x,yin K_i)$,

$K:=bigcup_i K_i$.

Then $K$ is closed under $d$ and contains $1+sqrt 5$ but not $sqrt1+sqrt 5$, as I have found by following the Pythagorean fields Wikipedia link given by @Dirk in his answer: If $sqrt1+sqrt 5$ were in $K$ then $1+sqrt 5$ would be a sum of two squares in some extension $K_i$, and then it would be so in an extension of $mathbbQ(sqrt 5)$, which implies that it is a sum of squares in $mathbbQ(sqrt 5)$, which is impossible because that would entail that $1-sqrt 5$, which is negative, is also a sum of squares in $mathbbQ(sqrt 5)$.

The details can be found in Chapter 5 of the book Geometric constructions by Martin. The relevant results are Theorems 5.10-5.15.

Similarly, $sqrt 2in K$ but $sqrt[4]2notin K$, and more in general, this is true for any positive number which is not a sum of squares in the first extension in which it appears.

Geometrically, numbers in $K$ correspond to constructible points by ruler and dividers. Hence $sqrt[4]2$ is constructible by rule and compass but not by rule and dividers.

share|cite|improve this answer

edited yesterday

answered yesterday

Jose BroxJose Brox

3,33211129

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  "which implies that it is a sum of squares in $mathbb Q(sqrt5)$" -- I do not see directly how this follows from the previous sentence, but I could very well be missing something obvious.
  $endgroup$
  – Mees de Vries
  yesterday
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @MeesdeVries Not obvious, it is a consequence of Theorems 5.10-5.13 in Martin's book.
  $endgroup$
  – Jose Brox
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $K$ is countable, right?
  $endgroup$
  – PyRulez
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @PyRulez yeah the same construction which makes $mathbb Q$ countable from $mathbb N^2$ plus skipping over duplicates should work to imply that $K_i+1$ is countable given that $K_i$ is countable, skipping over duplicates; by induction therefore all $K_i$ are countable; then we should be able to repeat the same construction again with $K_m(n)$, again skipping over duplicates, to find that $K$ is countable.
  $endgroup$
  – CR Drost
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @PyRulez Yes: clearly, all elements of $K$ are algebraic, and algebraic numbers are countable (there is a countable number of rational polynomials, with a finite number of roots each)
  $endgroup$
  – Jose Brox
  23 hours ago
  
  

  
  
  
  

add a comment | 

16

$begingroup$

Consider the tower of fields

$K_0:=mathbbQ$,

$K_i+1:=K_i(sqrtx^2+y^2| x,yin K_i)$,

$K:=bigcup_i K_i$.

Then $K$ is closed under $d$ and contains $1+sqrt 5$ but not $sqrt1+sqrt 5$, as I have found by following the Pythagorean fields Wikipedia link given by @Dirk in his answer: If $sqrt1+sqrt 5$ were in $K$ then $1+sqrt 5$ would be a sum of two squares in some extension $K_i$, and then it would be so in an extension of $mathbbQ(sqrt 5)$, which implies that it is a sum of squares in $mathbbQ(sqrt 5)$, which is impossible because that would entail that $1-sqrt 5$, which is negative, is also a sum of squares in $mathbbQ(sqrt 5)$.

The details can be found in Chapter 5 of the book Geometric constructions by Martin. The relevant results are Theorems 5.10-5.15.

Similarly, $sqrt 2in K$ but $sqrt[4]2notin K$, and more in general, this is true for any positive number which is not a sum of squares in the first extension in which it appears.

Geometrically, numbers in $K$ correspond to constructible points by ruler and dividers. Hence $sqrt[4]2$ is constructible by rule and compass but not by rule and dividers.

share|cite|improve this answer

edited yesterday

answered yesterday

Jose BroxJose Brox

3,33211129

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  "which implies that it is a sum of squares in $mathbb Q(sqrt5)$" -- I do not see directly how this follows from the previous sentence, but I could very well be missing something obvious.
  $endgroup$
  – Mees de Vries
  yesterday
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @MeesdeVries Not obvious, it is a consequence of Theorems 5.10-5.13 in Martin's book.
  $endgroup$
  – Jose Brox
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $K$ is countable, right?
  $endgroup$
  – PyRulez
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @PyRulez yeah the same construction which makes $mathbb Q$ countable from $mathbb N^2$ plus skipping over duplicates should work to imply that $K_i+1$ is countable given that $K_i$ is countable, skipping over duplicates; by induction therefore all $K_i$ are countable; then we should be able to repeat the same construction again with $K_m(n)$, again skipping over duplicates, to find that $K$ is countable.
  $endgroup$
  – CR Drost
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @PyRulez Yes: clearly, all elements of $K$ are algebraic, and algebraic numbers are countable (there is a countable number of rational polynomials, with a finite number of roots each)
  $endgroup$
  – Jose Brox
  23 hours ago
  
  

  
  
  
  

add a comment | 

$begingroup$

Consider the tower of fields

$K_0:=mathbbQ$,

$K_i+1:=K_i(sqrtx^2+y^2| x,yin K_i)$,

$K:=bigcup_i K_i$.

Then $K$ is closed under $d$ and contains $1+sqrt 5$ but not $sqrt1+sqrt 5$, as I have found by following the Pythagorean fields Wikipedia link given by @Dirk in his answer: If $sqrt1+sqrt 5$ were in $K$ then $1+sqrt 5$ would be a sum of two squares in some extension $K_i$, and then it would be so in an extension of $mathbbQ(sqrt 5)$, which implies that it is a sum of squares in $mathbbQ(sqrt 5)$, which is impossible because that would entail that $1-sqrt 5$, which is negative, is also a sum of squares in $mathbbQ(sqrt 5)$.

The details can be found in Chapter 5 of the book Geometric constructions by Martin. The relevant results are Theorems 5.10-5.15.

Similarly, $sqrt 2in K$ but $sqrt[4]2notin K$, and more in general, this is true for any positive number which is not a sum of squares in the first extension in which it appears.

Geometrically, numbers in $K$ correspond to constructible points by ruler and dividers. Hence $sqrt[4]2$ is constructible by rule and compass but not by rule and dividers.

share|cite|improve this answer

edited yesterday

answered yesterday

Jose BroxJose Brox

3,33211129

$endgroup$

share|cite|improve this answer

edited yesterday

answered yesterday

Jose BroxJose Brox

3,33211129

answered yesterday

Jose BroxJose Brox

3,33211129

answered yesterday

Jose BroxJose Brox

3,33211129

5

$begingroup$

edit: Look what I found:
Wiki

---

The field
$$mathbbQ(sqrtp mid p in mathbbP)$$
might be a good candidate.

At least, all fields closed under $d$ must contain this field.

share|cite|improve this answer

edited yesterday

answered yesterday

DirkDirk

4,408218

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why must a field closed under $d$ contain $sqrt3$?
  $endgroup$
  – FredH
  yesterday
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @FredH, it must contain $sqrt2 = d(1, 1)$, and thus it must contain $sqrt3 = d(1, sqrt2)$.
  $endgroup$
  – Mees de Vries
  yesterday
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  See en.wikipedia.org/wiki/Spiral_of_Theodorus
  $endgroup$
  – lhf
  yesterday
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  I don't think this works: $d(sqrt 2 + 1, 1) = sqrt2sqrt2 + 4$, but that doesn't look like a sum of square roots of rational numbers.
  $endgroup$
  – Arthur
  yesterday
  
  

  
  
  
  

add a comment | 

5

$begingroup$

edit: Look what I found:
Wiki

---

The field
$$mathbbQ(sqrtp mid p in mathbbP)$$
might be a good candidate.

At least, all fields closed under $d$ must contain this field.

share|cite|improve this answer

edited yesterday

answered yesterday

DirkDirk

4,408218

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why must a field closed under $d$ contain $sqrt3$?
  $endgroup$
  – FredH
  yesterday
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @FredH, it must contain $sqrt2 = d(1, 1)$, and thus it must contain $sqrt3 = d(1, sqrt2)$.
  $endgroup$
  – Mees de Vries
  yesterday
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  See en.wikipedia.org/wiki/Spiral_of_Theodorus
  $endgroup$
  – lhf
  yesterday
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  I don't think this works: $d(sqrt 2 + 1, 1) = sqrt2sqrt2 + 4$, but that doesn't look like a sum of square roots of rational numbers.
  $endgroup$
  – Arthur
  yesterday
  
  

  
  
  
  

add a comment | 

$begingroup$

edit: Look what I found:
Wiki

---

The field
$$mathbbQ(sqrtp mid p in mathbbP)$$
might be a good candidate.

At least, all fields closed under $d$ must contain this field.

share|cite|improve this answer

edited yesterday

answered yesterday

DirkDirk

4,408218

$endgroup$

share|cite|improve this answer

edited yesterday

answered yesterday

DirkDirk

4,408218

answered yesterday

DirkDirk

4,408218

answered yesterday

DirkDirk

4,408218