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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 $d(x, y) = \sqrt{x^2 + y^2}$, takes values in $K$.

Certainly this is not true for $\mathbb Q$: we have $d(1, 1) = \sqrt{2} \notin \mathbb Q$. 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 $d$?

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    $\begingroup$ Just so you know, there are other distances besides euclidean distance. $\endgroup$ – PyRulez Mar 21 '19 at 23:47
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Consider the tower of fields

$K_0:=\mathbb{Q}$,

$K_{i+1}:=K_i(\sqrt{x^2+y^2}| x,y\in K_i)$,

$K:=\bigcup_i K_i$.

Then $K$ is closed under $d$ and contains $1+\sqrt 5$ but not $\sqrt{1+\sqrt 5}$, as I have found by following the Pythagorean fields Wikipedia link given by @Dirk in his answer: If $\sqrt{1+\sqrt 5}$ 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 $\mathbb{Q}(\sqrt 5)$, which implies that it is a sum of squares in $\mathbb{Q}(\sqrt 5)$, which is impossible because that would entail that $1-\sqrt 5$, which is negative, is also a sum of squares in $\mathbb{Q}(\sqrt 5)$.

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, $\sqrt 2\in K$ but $\sqrt[4]2\not\in K$, 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.

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  • $\begingroup$ "which implies that it is a sum of squares in $\mathbb Q(\sqrt{5})$" -- 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 '19 at 13:04
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    $\begingroup$ @MeesdeVries Not obvious, it is a consequence of Theorems 5.10-5.13 in Martin's book. $\endgroup$ – Jose Brox Mar 21 '19 at 13:06
  • $\begingroup$ $K$ is countable, right? $\endgroup$ – PyRulez Mar 21 '19 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 '19 at 21:52
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    $\begingroup$ @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 '19 at 22:57
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edit: Look what I found: Wiki

The field $$\mathbb{Q}(\sqrt{p} \mid p \in \mathbb{P})$$ might be a good candidate.
At least, all fields closed under $d$ must contain this field.

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  • $\begingroup$ Why must a field closed under $d$ contain $\sqrt{3}$? $\endgroup$ – FredH Mar 21 '19 at 10:51
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    $\begingroup$ @FredH, it must contain $\sqrt{2} = d(1, 1)$, and thus it must contain $\sqrt{3} = d(1, \sqrt{2})$. $\endgroup$ – Mees de Vries Mar 21 '19 at 10:53
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    $\begingroup$ See en.wikipedia.org/wiki/Spiral_of_Theodorus $\endgroup$ – lhf Mar 21 '19 at 11:16
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    $\begingroup$ I don't think this works: $d(\sqrt 2 + 1, 1) = \sqrt{2\sqrt2 + 4}$, but that doesn't look like a sum of square roots of rational numbers. $\endgroup$ – Arthur Mar 21 '19 at 12:05

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