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[数论] 分解$x^{5^2}-x\bmod5$

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hbghlyj posted 2024-5-7 18:54 |Read mode
Last edited by hbghlyj 2024-11-6 16:23wolframalpha.com/input?i=factor x^{5^2}-x modulo 5
\begin{aligned}
&x^{5^2}-x\bmod5\\
&=x (x + 1) (x + 2) (x + 3) (x + 4) (x^2 + 2) (x^2 + 3) (x^2 + x + 1) (x^2 + x + 2) (x^2 + 2 x + 3) (x^2 + 2 x + 4) (x^2 + 3 x + 3) (x^2 + 3 x + 4) (x^2 + 4 x + 1) (x^2 + 4 x + 2)\bmod 5
\end{aligned}前面$x ( x + 1 ) ( x + 2 ) ( x + 3 ) ( x + 4 )$我知道是由于$x=0,1,2,3,4\bmod5$是$x^{5^2}-x$的根,

后面一堆都是不可约二次多项式,怎么得到的

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original poster hbghlyj posted 2024-11-7 00:22
一般地,所有$d|n$次不可约多项式之积$$
  x^{p^n}-x = \prod_{d|n} \prod_{\stackrel{f(x) \text{ monic, irred.}}{\text{deg}(f)=d}} f(x).
$$

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