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Is computing distance a lesser capability than computing square roots?Why algebraic closures?Is an algebra the smallest one generated by a certain subset of it?What is the automorphism group of the field of all constructible numbers?Elementary proof for $sqrtp_n+1 notin mathbbQ(sqrtp_1, sqrtp_2, ldots, sqrtp_n)$ where $p_i$ are different prime numbers.Why is $mathbbQ(sqrt2)subseteqmathbbQ(sqrt2+sqrt[3]5)subseteqmathbbQ(sqrt2,sqrt[3]5)$What is the known form of this transcendental number theorem suggesting $etimespi$ is very unlikely to be algebraic?Show that $mathbbQ(sqrtn,sqrt-n)subseteqmathbbQ(sqrtn+sqrt-n)$ by evaluating $alpha^3+2nalpha$ and $alpha^3-2nalpha$Question about field extensionDoes every equation involving $+,-,times,div,sqrt,mathbb Q$ only have solutions in the algebraic numbers

25

5

$begingroup$

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

Jose Brox

3,36211129

asked Mar 21 at 10:20

Mees de VriesMees de Vries

17.6k13059

$endgroup$

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  $begingroup$
  Just so you know, there are other distances besides euclidean distance.
  $endgroup$
  – PyRulez
  Mar 21 at 23:47
  

  
  
  
  

add a comment | 

25

5

$begingroup$

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

Jose Brox

3,36211129

asked Mar 21 at 10:20

Mees de VriesMees de Vries

17.6k13059

$endgroup$

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  $begingroup$
  Just so you know, there are other distances besides euclidean distance.
  $endgroup$
  – PyRulez
  Mar 21 at 23:47
  

  
  
  
  

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

Jose Brox

3,36211129

asked Mar 21 at 10:20

Mees de VriesMees de Vries

17.6k13059

$endgroup$

share|cite|improve this question

edited yesterday

Jose Brox

3,36211129

asked Mar 21 at 10:20

Mees de VriesMees de Vries

17.6k13059

share|cite|improve this question

edited yesterday

Jose Brox

3,36211129

asked Mar 21 at 10:20

Mees de VriesMees de Vries

17.6k13059

edited yesterday

Jose Brox

3,36211129

edited yesterday

Jose Brox

3,36211129

asked Mar 21 at 10:20

Mees de VriesMees de Vries

17.6k13059

asked Mar 21 at 10:20

Mees de VriesMees de Vries

17.6k13059

2 Answers
2

active

oldest

votes

19

$begingroup$

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 Mar 21 at 13:08

answered Mar 21 at 12:54

Jose BroxJose Brox

3,36211129

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  "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
  Mar 21 at 13:04
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @MeesdeVries Not obvious, it is a consequence of Theorems 5.10-5.13 in Martin's book.
  $endgroup$
  – Jose Brox
  Mar 21 at 13:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $K$ is countable, right?
  $endgroup$
  – PyRulez
  Mar 21 at 20:00
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @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
  Mar 21 at 21:52
  
  

  
  
  
  


• 
  
  
  

  
  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
  Mar 21 at 22:57
  
  

  
  
  
  

add a comment | 

8

$begingroup$

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 Mar 21 at 12:19

answered Mar 21 at 10:44

DirkDirk

4,438218

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why must a field closed under $d$ contain $sqrt3$?
  $endgroup$
  – FredH
  Mar 21 at 10:51
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @FredH, it must contain $sqrt2 = d(1, 1)$, and thus it must contain $sqrt3 = d(1, sqrt2)$.
  $endgroup$
  – Mees de Vries
  Mar 21 at 10:53
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  See en.wikipedia.org/wiki/Spiral_of_Theodorus
  $endgroup$
  – lhf
  Mar 21 at 11:16
  

  
  
  
  


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  4
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Mar 21 at 12:05
  
  

  
  
  
  

add a comment | 

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19

$begingroup$

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 Mar 21 at 13:08

answered Mar 21 at 12:54

Jose BroxJose Brox

3,36211129

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  "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
  Mar 21 at 13:04
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @MeesdeVries Not obvious, it is a consequence of Theorems 5.10-5.13 in Martin's book.
  $endgroup$
  – Jose Brox
  Mar 21 at 13:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $K$ is countable, right?
  $endgroup$
  – PyRulez
  Mar 21 at 20:00
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @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
  Mar 21 at 21:52
  
  

  
  
  
  


• 
  
  
  

  
  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
  Mar 21 at 22:57
  
  

  
  
  
  

add a comment | 

19

$begingroup$

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 Mar 21 at 13:08

answered Mar 21 at 12:54

Jose BroxJose Brox

3,36211129

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  "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
  Mar 21 at 13:04
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @MeesdeVries Not obvious, it is a consequence of Theorems 5.10-5.13 in Martin's book.
  $endgroup$
  – Jose Brox
  Mar 21 at 13:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $K$ is countable, right?
  $endgroup$
  – PyRulez
  Mar 21 at 20:00
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @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
  Mar 21 at 21:52
  
  

  
  
  
  


• 
  
  
  

  
  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
  Mar 21 at 22:57
  
  

  
  
  
  

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 Mar 21 at 13:08

answered Mar 21 at 12:54

Jose BroxJose Brox

3,36211129

$endgroup$

share|cite|improve this answer

edited Mar 21 at 13:08

answered Mar 21 at 12:54

Jose BroxJose Brox

3,36211129

answered Mar 21 at 12:54

Jose BroxJose Brox

3,36211129

answered Mar 21 at 12:54

Jose BroxJose Brox

3,36211129

8

$begingroup$

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 Mar 21 at 12:19

answered Mar 21 at 10:44

DirkDirk

4,438218

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why must a field closed under $d$ contain $sqrt3$?
  $endgroup$
  – FredH
  Mar 21 at 10:51
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @FredH, it must contain $sqrt2 = d(1, 1)$, and thus it must contain $sqrt3 = d(1, sqrt2)$.
  $endgroup$
  – Mees de Vries
  Mar 21 at 10:53
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  See en.wikipedia.org/wiki/Spiral_of_Theodorus
  $endgroup$
  – lhf
  Mar 21 at 11:16
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Mar 21 at 12:05
  
  

  
  
  
  

add a comment | 

8

$begingroup$

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 Mar 21 at 12:19

answered Mar 21 at 10:44

DirkDirk

4,438218

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why must a field closed under $d$ contain $sqrt3$?
  $endgroup$
  – FredH
  Mar 21 at 10:51
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @FredH, it must contain $sqrt2 = d(1, 1)$, and thus it must contain $sqrt3 = d(1, sqrt2)$.
  $endgroup$
  – Mees de Vries
  Mar 21 at 10:53
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  See en.wikipedia.org/wiki/Spiral_of_Theodorus
  $endgroup$
  – lhf
  Mar 21 at 11:16
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Mar 21 at 12:05
  
  

  
  
  
  

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 Mar 21 at 12:19

answered Mar 21 at 10:44

DirkDirk

4,438218

$endgroup$

share|cite|improve this answer

edited Mar 21 at 12:19

answered Mar 21 at 10:44

DirkDirk

4,438218

answered Mar 21 at 10:44

DirkDirk

4,438218

answered Mar 21 at 10:44

DirkDirk

4,438218