We say a field $F$ has the property $*$ if the equation $x^2 + y^2=-1$ has no solution in $F$. For an example if $F$ is a subfield of real numbers then $F$ satisfies $*$. On the other hand if $ F $ is a finite field then by Chevalley-Warning theorem the equation $ x^2 + y^2 =-1 $ always has a solution. I want to know which fields that satisfy property $ * $. Does this type of field belong to a special class of field?
5
-
1$\begingroup$ For what it's worth, these are precisely the fields over which the quaternion group $Q_8$ has a four dimensional irreducible representation. You might also want to look up the theory of Severi-Brauer varieties. $\endgroup$– Dave BensonJul 10 at 6:51
-
$\begingroup$ "Additionally, it is assumed": you don't need this assumption, which is an immediate consequence of the definition. $\endgroup$– YCorJul 10 at 7:02
-
$\begingroup$ Yes we don't need this assumption. I have edited. $\endgroup$– SkyJul 10 at 7:14
-
2$\begingroup$ There are many such fields with rather disparate properties, and I don’t think there is any kind of classification. FWIW, these are called the fields of Stufe 2. $\endgroup$– Emil JeřábekJul 10 at 7:37
-
1$\begingroup$ I mean, of Stufe more than $2$. $\endgroup$– Emil JeřábekJul 10 at 7:46
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
2
In the notation of Lam's Quadratic forms over fields, the Stufe (a German word) or level (its English translation) $s(F)$ of a field is the minimal $n$ such that $-1$ is the sum of $n$ squares.
A theorem of Pfister says that $s(F)$ is always a power of $2$ (or $\infty$).
So you are taking of fields $F$ such that $s(F) > 2$ (or $s(F) \geq 4$, if you are willing to use Pfister's theorem).
-
2$\begingroup$ Why would you be not willing to use Pfister's theorem? In this special case, it is an immediate consequence of the Brahmagupta–Fibonacci identity, hence it is has a short, completely elementary proof. $\endgroup$ Jul 10 at 8:26
-