# 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 $D=K$. ∎

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

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

• It is also evident that the usual, notationally driven, embedding of $\mathbb{C}$ into $\mathbb{H}$ is non-central. That is, $\mathbb{C}$ embeds as $a+bi\mapsto a+b\hat{i}$, into $\mathbb{H}=\langle 1,\hat{\imath},\hat{\jmath},\hat{k}\rangle$. This is not central:

 $(1+\hat{\imath})\hat{\jmath}=\hat{\jmath}+\hat{k}\neq\hat{\jmath}(1+\hat{% \imath})=\hat{\jmath}-\hat{k}.$
• Further evidence of the incompatiblity of $\mathbb{H}$ and $\mathbb{C}$ comes from considering polynomials. If $x^{2}+1$ is considered as a polynomial over $\mathbb{C}[x]$ 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: $\{\hat{\imath},-\hat{\imath},\hat{\jmath},-\hat{\jmath},\hat{k},-\hat{k}\}$. But indeed, given any $q\in\mathbb{H}$, $q\neq 0$, then $q\hat{\imath}q^{-1}$ is also a root. Thus there are an infinite number of roots to $x^{2}+1$. Therefore declaring $\hat{\imath}=\sqrt{-1}$ 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 TheHamiltonianRingIsNotAComplexAlgebra 2013-03-22 16:01:57 2013-03-22 16:01:57 Algeboy (12884) Algeboy (12884) 10 Algeboy (12884) Result msc 16W99