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357: １３２人目の素数さん [] 2023/03/13(月) 21:11:19.15 ID:UeELXD7y

>>356
つづき

（参考）英語版に詳しい証明がある、ただし文字化けなおさず。本文参照ください
https://en.wikipedia.org/wiki/Frobenius_theorem_(real_division_algebras)
Frobenius theorem (real division algebras)
In mathematics, more specifically in abstract algebra, the Frobenius theorem, proved by Ferdinand Georg Frobenius in 1877, characterizes the finite-dimensional associative division algebras over the real numbers. According to the theorem, every such algebra is isomorphic to one of the following:
R (the real numbers)
C (the complex numbers)
H (the quaternions).
These algebras have real dimension 1, 2, and 4, respectively. Of these three algebras, R and C are commutative, but H is not.

Proof
The main ingredients for the following proof are the Cayley?Hamilton theorem and the fundamental theorem of algebra.

Introducing some notation
Let D be the division algebra in question.
Let n be the dimension of D.
We identify the real multiples of 1 with R.
When we write a <= 0 for an element a of D, we tacitly assume that a is contained in R.
We can consider D as a finite-dimensional R-vector space. Any element d of D defines an endomorphism of D by left-multiplication, we identify d with that endomorphism. Therefore, we can speak about the trace of d, and its characteristic and minimal polynomials.
For any z in C define the following real quadratic polynomial:

Q(z;x)=x^{2}-2\operatorname {Re} (z)x+|z|^{2}=(x-z)(x-{\overline {z}})\in \mathbf {R} [x].
Note that if z ∈ C ? R then Q(z; x) is irreducible over R.

つづく

http://rio2016.5ch.net/test/read.cgi/math/1677671318/357

358: １３２人目の素数さん [] 2023/03/13(月) 21:11:52.56 ID:UeELXD7y

>>357
つづき

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.

つづく

http://rio2016.5ch.net/test/read.cgi/math/1677671318/358

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