《Galois theory》

H.E. p. 58 (S43)

* * * 11: 51

第五段

Thus, if the given equation f(x) = 0 is solvable by radicals, there is a sequence of fields K ⊂ K' ⊂ K'' ⊂ ... ⊂ K^(μ),...

---- 这样，如果给定的方程 f(x) = 0 根式可解，则存在域的序列 K ⊂ K' ⊂ K'' ⊂ ... ⊂ K^(μ), ...

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... which starts with the known quantities K, adjoins at the ith stage a pi-1th root of an element ki-1 of K^(i-1) to K^(i-1) to obtain K^(i), ...

---- 始于已知量 K, 在第 i 阶段添加第 i-1个根 (出自 K^(i-1) 的元素 ki-1) 到 K^(i-1) 以获得 K^(i), ...

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... and ends with a field K^(μ) which contains n roots of a, b, c, ... of the given equation f(x) = 0.

---- 止于域 K^(μ) 包含给定方程 f(x) = 0 的 n 个根 a, b, c, ... 。

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Moreover, it can be assumed that K^(i-1) contains primitive pi-1th roots of unity.

---- 另外，可以假定 K^(i-1) 包含原始的第 pi-1 个单位跟。

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评论：递增的域序列及形成方式。

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第四段

The four arithmetic operations can be performed on known quantities without leaving K -- that is the definition of a field.

---- 四则运算可以在已知量上进行而不离开 K -- 这是域的定义。

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However, if the solution calls for taking a pth root, say of a quantity k of K, then although it is possible that there is a pth root of k in K, ...

---- 然而，如果解要求取第 p 个根，比如 K 中的量 k 的根，则尽管 K 中存在 k 的第 p 个根是可能的，...

... most likely it will not be possible to do this within K and the realm of known quantities will have to be extended to include the needed pth root of k.

---- 更可能的是这不会在 K 中进行，而是已知量的范围将不得不扩张以便包含所需要的 k 的第 p 个根。

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Galois called such an extension of the known quantities the adjunction of a pth root, and this name has been used ever since.

---- 伽罗瓦称已知量的这个扩张为第 p 个根的 添加，此名称得以沿用。

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The result of an adjunction is a new, larger field K' ⊃ K whose elements can all be expressed rationally in terms of (p)sqrt(k) and quantities in K.

---- 添加的结果是一个新的、更大的域 它的元素都可以有理地表达为 (p)sqrt(k) 和 K 中的量。

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(The theorem on simple algebraic extensions can be applied to obtain such a field K' provided x^p - k is shown to be irreducible. ...

---- ( 简单代数扩张定理可用于获得这样的域 K' 只要 x^p - k 是不可约的。...

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It was noted in S42 that this is the case when K contains primitive pth roots of unity. ...

---- 在 S42 中指出过，当 K 包含原始的第 p 个单位根时就是这种情况。...

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It is even true -- see Exercise 6 of the Eighth Set -- without this assumption. ...

---- 甚至没有此假定时这也是对的，见第八套练习 6。...

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However, Galois ignored the technicality of constructing K', and consideration of it here will be postponed to S62.)

---- 然而，伽罗瓦忽略了构造 K' 的技术，关于此事的考虑将推迟到 S62。)

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If the solution involves the extraction of another root not already contained in K', ...

---- 如果解涉及到提取另一个根而它并未包含在 K' 中，...

... then it will be necessary to perform another adjunction K'' ⊃ K', say of a p1th root of an element k1 of K', and so on, ...

---- 则有必要执行另一个添加 K'' ⊃ K'，比如 K' 中的元素 k1 的第 p1 个根，依此类推，...

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... until one arrives at a field K^(μ) which contains all the quantities that are involved in the solution of the equation.

---- 直到域 K^(μ) 包含求解方程所涉及到的所有的量。

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评论：第五段是第四段的概括；或者说，后者是前者的展开。

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