# The Hamiltonian ring is not a complex algebra

The Hamiltonian algebra (http://planetmath.org/QuaternionAlgebra2) $\mathbb{H}$ contains isomorphic   copies of the real $\mathbb{R}$ and complex $\mathbb{C}$ numbers. However, the reals are a central subalgebra of $\mathbb{H}$ which makes $\mathbb{H}$ into a real algebra. This makes identifying $\mathbb{R}$ in $\mathbb{H}$ canonical: $1\in\mathbb{H}$ determines a unique embedding   $\mathbb{R}\rightarrow\mathbb{H}:r\mapsto r1$. Yet $\mathbb{H}$ is not a complex algebra. The goal presently is to outline some of the incongruities of $\mathbb{C}=\langle 1,i\rangle$ and $\mathbb{H}=\langle 1,\hat{\imath},\hat{\jmath},\hat{k}\rangle$ which may be obscured by the notational overlap of the letter $i$.

###### Proposition 1.

There are no proper finite dimensional division rings over algebraically closed fields.

###### Proof.

Let $D$ be a finite dimensional division ring over an algebraically closed field $K$. This means that $K$ is a central subalgebra of $D$. Let $a\in D$ and consider $K(a)$. Since $K$ is central in $D$, $K(a)$ is commutative   , and so $K(a)$ is a field extension of $K$. But as $D$ is a finite dimensional $K$ space, so is $K(a)$. As any finite dimensional extension   of $K$ is algebraic, $K(a)$ is an algebraic extension  . Yet $K$ is algebraically closed  so $K(a)=K$. Thus $a\in K$ so in fact $D=K$. ∎

Title The Hamiltonian ring is not a complex algebra TheHamiltonianRingIsNotAComplexAlgebra 2013-03-22 16:01:57 2013-03-22 16:01:57 Algeboy (12884) Algeboy (12884) 10 Algeboy (12884) Result msc 16W99