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[数论] 如果 $f(x)$ 不可约,$f(x^k)$ 不可约?

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hbghlyj posted 2024-12-10 19:51 |Read mode
Problems from the Book by Titu Andreescu (Example 9, page 494)
设 $f(x)$ 是一个具有整数系数的首一多项式,设 $p$ 是一个素数。如果 $f(x)$ 在 $\mathbb{Z}[x]$ 中是不可约的,并且 $\sqrt[p]{(-1)^{\deg f}f(0)}$ 是无理数,那么 $f(x^p)$ 在 $\mathbb{Z}[x]$ 中也是不可约的。

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