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 $\u2102$ 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^{} $\mathbb{R}\to \mathbb{H}:r\mapsto r1$. Yet $\mathbb{H}$ is not a complex algebra. The goal presently is to outline some of the incongruities of $\u2102=\u27e81,i\u27e9$ and $\mathbb{H}=\u27e81,\widehat{\u0131},\widehat{\u0237},\widehat{k}\u27e9$ 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 $D=K$. ∎

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In particular, this proposition^{} proves $\mathbb{H}$ is not a complex algebra.

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Alternatively, from the WedderburnArtin theorem we know the only semisimple^{} complex algebra of dimension^{} 2 is $\u2102\oplus \u2102$. This has proper ideals^{} and so it cannot be the division ring $\mathbb{H}$.

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It is also evident that the usual, notationally driven, embedding of $\u2102$ into $\mathbb{H}$ is noncentral. That is, $\u2102$ embeds as $a+bi\mapsto a+b\widehat{i}$, into $\mathbb{H}=\u27e81,\widehat{\u0131},\widehat{\u0237},\widehat{k}\u27e9$. This is not central:
$$(1+\widehat{\u0131})\widehat{\u0237}=\widehat{\u0237}+\widehat{k}\ne \widehat{\u0237}(1+\widehat{\u0131})=\widehat{\u0237}\widehat{k}.$$ 
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Further evidence of the incompatiblity of $\mathbb{H}$ and $\u2102$ comes from considering polynomials^{}. If ${x}^{2}+1$ is considered as a polynomial over $\u2102[x]$ then it has exactly two roots $i,i$ as expected. However, if it is considered as a polynomial over $\mathbb{H}[x]$ we arrive at 6 obvious roots: $\{\widehat{\u0131},\widehat{\u0131},\widehat{\u0237},\widehat{\u0237},\widehat{k},\widehat{k}\}$. But indeed, given any $q\in \mathbb{H}$, $q\ne 0$, then $q\widehat{\u0131}{q}^{1}$ is also a root. Thus there are an infinite^{} number of roots to ${x}^{2}+1$. Therefore declaring $\widehat{\u0131}=\sqrt{1}$ can be greatly misleading. Such a conflict does not arise for polynomials with real roots since $\mathbb{R}$ is a central subalgebra.
Title  The Hamiltonian ring is not a complex algebra 

Canonical name  TheHamiltonianRingIsNotAComplexAlgebra 
Date of creation  20130322 16:01:57 
Last modified on  20130322 16:01:57 
Owner  Algeboy (12884) 
Last modified by  Algeboy (12884) 
Numerical id  10 
Author  Algeboy (12884) 
Entry type  Result 
Classification  msc 16W99 