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[注:下文是群邮件的内容。文末收集了 S43 的笔记链接。]
《Galois theory》 H.E. p. 57 (S43) * * * ??? 第三段 If the expression of a root involves taking a jth root and j is not prime, then j = j1j2 where j1 and j2 are smaller than j, ... ---- 如果根的表达式涉及取 第 j 个根,而 j 不是素数,则 j = j1j2 其中 j1 和 j2 比 j 小,... . ... and instead of taking a jth root one can take a j1th root of a j2th root. ---- 代替第 j 个根的是,取第 j2 个根的第 j1 个根。 . 评论:“jth root” 前面出现不定冠词,或有别的含义(?)。 . If j1 or j2 is not prime, it can be further decomposed, until in the end the expressions of the roots of f(x) = 0 involve taking roots of prime order only. ---- 如果 j1 或 j2 不是素数,则可以继续分解,直到 f(x) = 0 的根的表达式最终只涉及取素数次序的根。 . Moreover, since, as Gauss showed,* the pth roots of unity for any prime p can be expressed in terms of radicals, ... ---- 而且,既然 (如高斯证明的那样) 任何素数 p 的第 p 个单位根都可以表达为根式,... . ... use can be made of pth roots of unity in a solution by radicals. ---- 那么在根式解中就可以用到第 p 个单位根。 . 注:这个后半句的 “use can be made of” 猛一看会让人会不明白。 . This means that, whenever a pth root of a known quantity is taken, all pth roots of the quantity are known, ... ---- 这意味着,只要取到某个已知量的一个 第p个根,则该量的所有 第p个根 就都知道了... . ... because all others are obtained from any one by multiplying by pth roots of unity. ---- 因为所有其它的可以通过给任何一个乘以 第 p 个单位根来得到。 . 小结:S43 是个概论(共一页半),包括七段话:1, 2, 3, 4, 5, 6, 7。 * * *16:19 符号大全、上下标.|| 常用:↑↓ π ΓΔΛΘΩμφΣ∈ ∉ ∪ ∩ ⊆ ⊇ ⊂ ⊃ Ø ∀ ∃ ≤ ≥ ⌊ ⌋ ⌈ ⌉ ≠ ≡ ⁻⁰ 1 2 3 ᵈ ⁺ ₊ ₀ ₁ ₂ ₃ ᵢ . |
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