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$(f,g)=1$,$(f,h)=1$,求$(f,gh)=$?

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其妙 Posted at 2014-5-20 17:35:37 |Read mode
设多项式函数$f(x)$、$g(x)$、$h(x)$,满足$(f,g)=1$,$(f,h)=1$,求$(f,gh)=$?
注:$(f,g)$表示多项式函数$f(x)$、$g(x)$首项系数(最高次项系数)为1的最大公因式.
妙不可言,不明其妙,不着一字,各释其妙!

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 Author| 其妙 Posted at 2014-5-20 23:56:43
回复 1# 其妙
例如$f(x)=(3x-2)(x-1),g(x)=(3x-2)(2x+1)$,则$(f,g)=x-\dfrac23$

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 Author| 其妙 Posted at 2014-6-10 16:48:36
符号[f,g]叫最小公因式?首项系数为1

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 Author| 其妙 Posted at 2014-6-22 21:45:11
这道题是不是有点难?

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tommywong Posted at 2014-6-23 06:13:42
概念性问题被果断略过了......

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realnumber Posted at 2014-6-23 11:00:20
难道答案不是1?

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realnumber Posted at 2014-6-23 11:07:12
(f,g)=1,即存在多项式u,v,有uf+vg=1,vg=1-uf
(f,h)=1,即存在多项式s,t,有sf+th=1,th=1-sf
所以tvgh=(1-uf)(1-sf)=1-f(u+s-suf)
即tvgh+f(u+s-suf)=1
所以(gh,f)=1

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 Author| 其妙 Posted at 2014-6-23 15:25:56
回复 7# realnumber
非常棒!

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 Author| 其妙 Posted at 2014-7-1 20:45:14
我也来做做:
设$uf+vg=1$,则$(uh)f+v(gh)=h$,令$(f,gh)=d$,则$d|f,d|gh$,于是$d|h$,故$d|(f,h)=1$,所以$d=1$,即:$(f,gh)=1$,证毕.

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