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By Frobenius' Theorem we know that if $D$ be an algebraic non-commutative division algebra over $\mathbb{R}$ then ,as an $\mathbb{R}$-algebra, $D$ is isomorphic to $\mathbb{H}$.

We can also replace $\mathbb{R}$ by any real closed field. if we replace $\mathbb{R}$ by an Euclidean field can we say the same statement is true?

A Euclidean field is an ordered field $K$ for which every non-negative element is a square: that is, $0 \le x$ in $K$ implies that $x = y^2$ for some y in K.

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    $\begingroup$ If I were to try for a counterexample, I would start with a rank-$9$ division algebra over $\Bbb Q$, and see what happened when I extended the base from $\Bbb Q$ to $K$. $\endgroup$ – Lubin Jul 25 '18 at 19:04
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    $\begingroup$ Yes, I think I managed to do just what @Lubin foresaw. Barring a fumble, of course. I probably should have the real subfield of the seventh cyclotomic field as here, but for some reason I went with the ninth this time :-/ $\endgroup$ – Jyrki Lahtonen Jul 26 '18 at 8:09
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I don't think the conclusion holds for all such $K$.

Consider the field $K$ of (compass and straightedge) constructible real numbers. It is Euclidean for sure. It is ordered because it is a subfield of the reals. Because the regular 9-gon is not constructible, the number $z=z_1=2\cos(2\pi/9)\notin K$. Nor are its Galois conjugates $z_2=z^2-2=2\cos(4\pi/9)$ and $z_3=(z^2-2)^2-2=2\cos(8\pi/9)$. Their shared minimal polynomial over $\Bbb{Q}$ is the familiar $$ m(x):=(x-z_1)(x-z_2)(x-z_3)=x^3-3x+1. $$ This remains irreducible over $K$, so we see that $L=K(z)$ is a cubic extension. Furthermore, $L/K$ is Galois with a cyclic Galois group generated by the $K$-automorphism $\sigma:z_1\mapsto z_2\mapsto z_3\mapsto z_1$.

The idea is to construct a cyclic division algebra (locally see e.g. this answer) from the extension $L/K$. More specifically, I will apply a theorem of A. Albert. It calls for a so called non-norm element $a\in K$ such that $a\neq N_{L/K}(x)$ for all $x\in L$. Here $N_{L/K}$ is the (relative) norm $$N_{L/K}:L\to K, x\mapsto x\sigma(x)\sigma^2(x).$$

I claim that $2$ is a non-norm element. Assume contrariwise that $$2=N_{L/K}(x)$$ for some $x=a_0+a_1z+a_2z^2\in L, a_0,a_1,a_2\in K$. Let $F=\Bbb{Q}(a_0,a_1,a_2)$, and $E=F(z)$. Then $E/F$ is also a cubic cyclic Galois extension, and the assumption is also that $N_{E/F}(x)=2$. As the numbers $a_0,a_1,a_2$ are constructible we know that $[F:\Bbb{Q}]$ is a power of two, say $2^n$. Furthermore, we have a tower of quadratic extensions $$ \Bbb{Q}=F_0\subset F_1\subset \cdots\subset F_n=F. $$ Let $\mathfrak{p}$ be a prime ideal of $F$ lying above the rational prime $2$. Because inertia degree is multiplicative in a tower, it follows that the inertia degree $f(\mathfrak{p}|2)$ is also a power of two. This is because the inertia degree $f(\mathfrak{p}\cap F_i|\mathfrak{p}\cap F_{i-1})$ can only be $1$ or $2$ for all $i=1,2,\ldots,n$.

On the other hand, $m(x)$ is irreducible modulo two, so the prime $p=2$ is inert in the cubic extension $\Bbb{Q}(z)/\Bbb{Q}$. Therefore any prime ideal of $E$ lying above $p=2$ has inertia degree divisible by three. In particular, $\mathfrak{p}$ is inert in $E/F$.

Consider the fractional ideal $I$ of $E$ generated by $x$. Its ideal norm $N_{E/F}(I)$ was assumed to be the principal ideal $(2)$. But, the previous result implies any prime ideal $\mathfrak{p}|2$ appears as a factor in $N_{E/F}(I)$ with multiplicity divisible by three. On the other hand, trace it through the tower of intermediate fields again, in the prime ideal decomposition of $(2)$ as an ideal of $F$ all the primes appear with a multiplicity that is a power of two. This is contradiction.

Consequently Albert's theorem tells us that the cyclic algebra $$\mathcal{A}=L\oplus Lu\oplus Lu^2$$ with its $K$-linear multiplication defined by the rules

  • $u^3=2$, and
  • $ku=u\sigma(k)$ for all $k\in K$

is a $9$-dimensional division algebra over $K$.

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    $\begingroup$ There may be 1) easier ways of showing that $N_{L/K}$ is not surjective, 2) in particular that $2$ is not a norm of any element of $L$. Unfortunately I only see this kludgy route at the moment, and don't know of more powerful methods. $\endgroup$ – Jyrki Lahtonen Jul 26 '18 at 8:06
  • $\begingroup$ Thanks for answer. But can we say if we have a Euclidean field $E$ then the only quaternion division algebra over $E$ is the ordinary quaternion division algebra? $\endgroup$ – s.Bahari Jul 26 '18 at 8:16
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    $\begingroup$ @s.Bahari I think that is, indeed, the case. If $D$ is a 4-dimensional division algebra with center $E$, then $D$ is split by any quadratic extension $L$ of $E$ it contains. As $E$ is Euclidean, such an extension $L$ is of the form $L=E(\sqrt{-d})$ for some positive element $d\in E$. But $\sqrt{d}\in E$, so $L=E(\sqrt{-1})$. This should imply that $D$ is the usual quaternion algebra over $E$. $\endgroup$ – Jyrki Lahtonen Jul 26 '18 at 8:24
  • $\begingroup$ Thank again. This help me a lot. $\endgroup$ – s.Bahari Jul 26 '18 at 8:28
  • $\begingroup$ @s.Bahari I no longer trust the argument of my previous comment. We do get $E(i)$ as a subfield of $D$. By Skolem-Noether there is a corresponding $j\in D$ such that $ji=-ij$. $j$ is not in $E$, so $E(j)$ is another quadratic extension. But why would $j^2$ be in $E$? $\endgroup$ – Jyrki Lahtonen Jul 27 '18 at 5:14

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