The Hamiltonian ring is not a complex algebra
The Hamiltonian algebra (http://planetmath.org/QuaternionAlgebra2) contains isomorphic copies of the real and complex numbers. However, the reals are a central subalgebra of which makes into a real algebra. This makes identifying in canonical: determines a unique embedding . Yet is not a complex algebra. The goal presently is to outline some of the incongruities of and which may be obscured by the notational overlap of the letter .
There are no proper finite dimensional division rings over algebraically closed fields.
Let be a finite dimensional division ring over an algebraically closed field . This means that is a central subalgebra of . Let and consider . Since is central in , is commutative, and so is a field extension of . But as is a finite dimensional space, so is . As any finite dimensional extension of is algebraic, is an algebraic extension. Yet is algebraically closed so . Thus so in fact . ∎
In particular, this proposition proves is not a complex algebra.
It is also evident that the usual, notationally driven, embedding of into is non-central. That is, embeds as , into . This is not central:
Further evidence of the incompatiblity of and comes from considering polynomials. If is considered as a polynomial over then it has exactly two roots as expected. However, if it is considered as a polynomial over we arrive at 6 obvious roots: . But indeed, given any , , then is also a root. Thus there are an infinite number of roots to . Therefore declaring can be greatly misleading. Such a conflict does not arise for polynomials with real roots since is a central subalgebra.
|Title||The Hamiltonian ring is not a complex algebra|
|Date of creation||2013-03-22 16:01:57|
|Last modified on||2013-03-22 16:01:57|
|Last modified by||Algeboy (12884)|