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[parent] Hamiltonians versus the complex numbers (Result)

The Hamiltonian algebra $ \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$. $ \qedsymbol$
  • 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:
    $\displaystyle (1+\hat{\imath})\hat{\jmath}=\hat{\jmath}+\hat{k}\neq \hat{\jmath}(1+\hat{\imath})=\hat{\jmath}-\hat{k}.$
  • 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.



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Cross-references: infinite, obvious, roots, polynomials, proper ideals, dimension, semisimple, Wedderburn-Artin theorem, proposition, algebraic extension, algebraic, extension, field extension, commutative, fields, algebraically closed, division rings, finite dimensional, embedding, canonical, algebra, subalgebra, complex, real, isomorphic, contains

This is version 6 of Hamiltonians versus the complex numbers, born on 2006-06-23, modified 2006-06-25.
Object id is 8076, canonical name is HamiltoniansVersesTheComplexNumbers.
Accessed 628 times total.

Classification:
AMS MSC16W99 (Associative rings and algebras :: Rings and algebras with additional structure :: Miscellaneous)

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