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R和C之间没有集合对加减乘除封闭

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hbghlyj posted 2023-3-19 02:45 |Read mode
不存在$F$对加减乘除封闭且满足$\mathbb{R} \subsetneq F \subsetneq \mathbb{C}$.

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original poster hbghlyj posted 2023-3-19 02:48
$F$ 包含一个元素 $a+bi$,其中 $b\neq 0$。
由于$F$包含$\mathbb R$,它包含$b^{-1}$和$-a$,所以它包含$b^{-1}\left((a+bi)-a\right)=i $。因此它包含所有复数,即 $F=\mathbb{C}$。

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original poster hbghlyj posted 2023-3-19 02:51
Is there a field extension over the real numbers that is not the same as the field of complex numbers?
扩张 $[ℂ:ℝ]$ 的度数为 2,因此不存在中间的真扩张。
假设这样的扩域 $F$ 存在。根据塔定理,我们有 $[ℂ:F][F:ℝ]=2$。要使其成为真扩张,每个因数都必须大于 1。矛盾。

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