[過去ﾛｸﾞ] ガロア第一論文と乗数イデアル他関連資料スレ2 (1002ﾚｽ)
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358(3): １３２人目の素数さん [] 2023/03/13(月)21:11 ID:UeELXD7y(5/14)
>>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
359(3): １３２人目の素数さん [] 2023/03/13(月)21:12 ID:UeELXD7y(6/14)
>>358
つづき
The finish
For a, b in V define B(a, b) = (?ab ? ba)/2. Because of the identity (a + b)2 ? a2 ? b2 = ab + ba, it follows that B(a, b) is real. Furthermore, since a2 <= 0, we have: B(a, a) > 0 for a ≠ 0. Thus B is a positive definite symmetric bilinear form, in other words, an inner product on V.
Let W be a subspace of V that generates D as an algebra and which is minimal with respect to this property. Let e1, ..., en be an orthonormal basis of W with respect to B. Then orthonormality implies that:
e_{i}^{2}=-1,\quad e_{i}e_{j}=-e_{j}e_{i}.
If n = 0, then D is isomorphic to R.
If n = 1, then D is generated by 1 and e1 subject to the relation e2
1 = ?1. Hence it is isomorphic to C.
If n = 2, it has been shown above that D is generated by 1, e1, e2 subject to the relations
e_{1}^{2}=e_{2}^{2}=-1,\quad e_{1}e_{2}=-e_{2}e_{1},\quad (e_{1}e_{2})(e_{1}e_{2})=-1.
These are precisely the relations for H.
つづく
368: １３２人目の素数さん [sage] 2023/03/14(火)07:40 ID:bQV51cAg(1/8)
>>367
１、検索結果を読んでも全く理解できず全コピペ
さすが大学１年の４月で落ちこぼれた真正●●
Q, >>357-360を読んで肝心な部分をまとめて
2048バイト以内（すなわち１コメント）で書け
１には絶対できないと予言する
勝った！（完全勝利宣言！！！）
381(8): １３２人目の素数さん [sage] 2023/03/14(火)20:13 ID:bQV51cAg(3/8)
>>377-379
東京●●大と大阪●●大
落ちこぼれ同士の共鳴
> 数学科以外で自分より上がいると、
> 落ちこぼれた自分がみじめで許せないんだ
誰が上？貴様が？
正則行列も知らず
任意の正方行列に逆行列があると
大嘘ぶっこいた馬鹿野郎の貴様が？
悪いが貴様より下なんかいねえよｗ
で、>>357-360のコピペの要約もできんのか？
こんなもんハードル下げまくってるぞ
それでも答えられんのか？
じゃ解答で二匹の落ちこぼれのゴキブリを焼き尽くすかｗ
まず358はR上の多元体で1以外の基底は
みな2乗すると-1になるといってる
この証明には代数学の基本定理とケイリー・ハミルトンの定理を使ってる
ま、どっちの定理の証明も1には生涯理解できまいから全部省略するｗ
次に359は多元体をR上の線形空間とみなした場合の生成元の基底を取ったとき
生成元の数が1つなら複素数C (e1^2=-1）
生成元の数が2つなら四元数H （e1^2=e2^2=-1、e1e2=-e2e1 ゆえに(e1e2)^2=-e1^2e2^2=-1）
最後に360は生成元の数が3以上だとe1e2en=1となるから、
358に述べた定理によって多元体にならないと言ってる
たったこんだけだぞ、なんで書けないんだ？
正真正銘のパクチー野郎か？１と乙は？（嘲）
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