ガロア第一論文と乗数イデアル他関連資料スレ17

[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17 (1002ﾚｽ)
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938: 現代数学の系譜 雑談 ◆yH25M02vWFhP  [] 2025/05/27(火) 23:57:08.54 ID:mVXlvt9d

>>933
>（引用始）
>Theorem H (Galois).
>A polynomial f ∈ Q[x] is solvable by radicals if and only if its Galois group over Q is soluble.
>The proof, which we omit,…
>（引用終）

おお、君は だいぶ 数学イップスから 回復して
数学文献の ”大人読み”＝まずは どこが重要かを考えながら読むべし！
が出来るようになったね

えーと、もう少し引用すると P92より
(15.3). Recalling the definition of soluble group given in Section 10:
Theorem H (Galois). A polynomial f ∈ Q[x] is solvable by radicals if and only if its Galois group over Q is soluble.
The proof, which we omit, uses the full power of the Galois correspondence, with the sequence of extensions in a radical extension corresponding to the sequence of subgroups
{1} = H0 ⊳H1 ⊳···⊳Hn−1 ⊳Hn =G,
in a soluble group.
(15.4). As a small reality check of Theorem H, we saw in Section 11 that the Galois group over Qof a quadratic polynomial is either the trivial group {id} or the (Abelian) permutation group {id, (α, β)} where α,β ∈ C are the roots. Abelian groups are soluble– see (10.8)– and this syncs with quadratics being solvable by radicals via the quadratic formula.
Similarly, the possible Galois groups of cubic polynomials are shown in Figure 11.3. Apart from S3, these are also Abelian. But S3 is the symmetry group of an equilateral triangle lying in the plane– soluble by (10.9).
(15.5). Somewhat out of chronological order, we have:
Theorem 15.1 (Abels-Fubini). The polynomial f = x5 − 4x + 2 is not solvable by radicals. The roots of x5−4x+2 are algebraic numbers, yet there is no algebraic expression for them.
Proof. We show that the Galois group of f over Q is insoluble. Indeed, we show that the Galois
略
(引用終り)

フルの証明は略しているが、証明の概要は語っているよね
それから、この(15.3)節のTheorem H (Galois)に来るまでに、”solvable by radicals”についての説明はあったぞ
特に、3次と5次については、図解までして詳しくね
だから、君の論難は当たらないと思うよ ；ｐ）

http://rio2016.5ch.net/test/read.cgi/math/1746597368/938

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