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[函数] 任何次数低于$n$的首1多项式$q(x)$都不能使$q(a)=0$

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abababa posted 2020-11-14 20:54 |Read mode
假设多项式$f(x),q(x)$的系数$c_i$都在数域$F$上并且都是首1多项式。给定一个数$a\not\in F$,如果$n$次首1多项式$f(x)$不能使$f(a)=0$,则任何次数低于$n$的首1多项式$q(x)$都不能使$q(a)=0$。
这个命题是对的吗?感觉上是正确的,设$q(x)$是$s$次多项式,如果把$q(x)$的每一项都乘上一个$x^{n-s}$,就能让最高次和$f(x)$一样,但是后面那些项要怎么补到跟$f(x)$一样,才能用上已知$f(a)\neq 0$?详细要怎么证明呢?

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zhcosin posted 2020-11-24 11:32
纯属乱弹琴啊,数域$F$定为实数域,取$a=i$,那么 $f(x)=x^3+x+1$不能使 $f(i)=0$,而次数更低的多项式 $g(x)=x^2+1$却使 $g(i)=0$

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original poster abababa posted 2020-11-24 12:43
Last edited by abababa 2020-11-24 16:26回复 2# zhcosin

$f(x)$是任意的,不是给定的。这个我已经想明白了,只要乘上$x^{n-s}$就能证明。所以如果验证$3$次多项式不是$x$的最小多项式,那不用再验证$2,1$次的了,只要向更高的次数验证。

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