The Hamiltonian ring is not a complex algebra

The Hamiltonian algebra (http://planetmath.org/QuaternionAlgebra2) $ℍ$ contains isomorphic copies of the real $ℝ$ and complex $ℂ$ numbers. However, the reals are a central subalgebra of $ℍ$ which makes $ℍ$ into a real algebra. This makes identifying $ℝ$ in $ℍ$ canonical: $1\in ℍ$ determines a unique embedding $ℝ\to ℍ:r\mapsto r1$. Yet $ℍ$ is not a complex algebra. The goal presently is to outline some of the incongruities of $ℂ=⟨1,i⟩$ and $ℍ=⟨1,\widehat{ı},\widehat{ȷ},\widehat{k}⟩$ 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 $ℍ$ is not a complex algebra.

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

• It is also evident that the usual, notationally driven, embedding of $ℂ$ into $ℍ$ is non-central. That is, $ℂ$ embeds as $a+bi\mapsto a+b\widehat{i}$, into $ℍ=⟨1,\widehat{ı},\widehat{ȷ},\widehat{k}⟩$. This is not central:

  $$(1+\widehat{ı})\widehat{ȷ}=\widehat{ȷ}+\widehat{k}\ne \widehat{ȷ}(1+\widehat{ı})=\widehat{ȷ}-\widehat{k}.$$

• Further evidence of the incompatiblity of $ℍ$ and $ℂ$ comes from considering polynomials. If $x^2+1$ is considered as a polynomial over $ℂ[x]$ then it has exactly two roots $i,-i$ as expected. However, if it is considered as a polynomial over $ℍ[x]$ we arrive at 6 obvious roots: $\{\widehat{ı},-\widehat{ı},\widehat{ȷ},-\widehat{ȷ},\widehat{k},-\widehat{k}\}$. But indeed, given any $q\in ℍ$, $q\ne 0$, then $q\widehat{ı}{q}^{-1}$ is also a root. Thus there are an infinite number of roots to $x^2+1$. Therefore declaring $\widehat{ı}=\sqrt{-1}$ can be greatly misleading. Such a conflict does not arise for polynomials with real roots since $ℝ$ is a central subalgebra.