Algebraic division algebra over Euclidean field 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.
 A: 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 $K$.
