23/03/13 21:11:52.56 UeELXD7y.net
>>357
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The claim
The key to the argument is the following
Claim. The set V of all elements a of D such that a2 <= 0 is a vector subspace of D of dimension n - 1. Moreover D = R 〇+ V as R-vector spaces, which implies that V generates D as an algebra.
Proof of Claim: Let m be the dimension of D as an R-vector space, and pick a in D with characteristic polynomial p(x). By the fundamental theorem of algebra, we can write
p(x)=(x-t_{1})\cdots (x-t_{r})(x-z_{1})(x-{\overline {z_{1}}})\cdots (x-z_{s})(x-{\overline {z_{s}}}),\qquad t_{i}\in \mathbf {R} ,\quad z_{j}\in \mathbf {C} \backslash \mathbf {R} .
We can rewrite p(x) in terms of the polynomials Q(z; x):
p(x)=(x-t_{1})\cdots (x-t_{r})Q(z_{1};x)\cdots Q(z_{s};x).
Since zj ∈ C\R, the polynomials Q(zj; x) are all irreducible over R. By the Cayley?Hamilton theorem, p(a) = 0 and because D is a division algebra, it follows that either a ? ti = 0 for some i or that Q(zj; a) = 0 for some j. The first case implies that a is real. In the second case, it follows that Q(zj; x) is the minimal polynomial of a. Because p(x) has the same complex roots as the minimal polynomial and because it is real it follows that
p(x)=Q(z_{j};x)^{k}=\left(x^{2}-2\operatorname {Re} (z_{j})x+|z_{j}|^{2}\right)^{k}
Since p(x) is the characteristic polynomial of a the coefficient of x2k?1 in p(x) is tr(a) up to a sign. Therefore, we read from the above equation we have: tr(a) = 0 if and only if Re(zj) = 0, in other words tr(a) = 0 if and only if a2 = ?|zj|2 < 0.
So V is the subset of all a with tr(a) = 0. In particular, it is a vector subspace. The rank?nullity theorem then implies that V has dimension n - 1 since it is the kernel of
{\displaystyle \operatorname {tr} :D\to \mathbf {R} }. Since R and V are disjoint (i.e. they satisfy
{\displaystyle \mathbf {R} \cap V=\{0\}}), and their dimensions sum to n, we have that D = R 〇+ V.
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