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[数论] 证明环同构

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hbghlyj Posted 2023-1-22 22:18 |Read mode
Last edited by hbghlyj 2023-2-26 00:25定义\[ℚ(\sqrt{-5})=\set{r+s\sqrt{-5}:r,s∈ℚ}\]
和\[ℚ[x]/\left<x^2-2x+6\right>=\set{f(x)\bmod x^2-2x+6:f(x)\text{是有理系数多项式}}\]
求证:存在一个双射$f : ℚ(\sqrt{-5})→ℚ[x]/\left<x^2-2x+6\right>$,使得:
  • $f(a + b) = f(a) + f(b)$,对于$ℚ(\sqrt{-5})$内的所有$a$和$b$;
  • $f(ab) = f(a) f(b)$,对于$ℚ(\sqrt{-5})$内的所有$a$和$b$;
  • $f(1) = 1$。

来源RingHomomorphismsAndIsomorphisms.pdf Exercise 12

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 Author| hbghlyj Posted 2023-1-22 22:31
注意到$(x-1)^2+5=x^2-2x+6$
令$f(r+s\sqrt{-5})=r+s(x-1)\bmod (x-1)^2+5$

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