定理(中国剩余定理) 设f1(x)……,fn(x)∈K(x]两两互素,则对任意给定 的a1(x)…,an(x)∈K(x,必存在8(x),q1(x)……,m(x)∈K[x 使得 g(x)=f(x)9(x)+a1(x),(记为g(x)=a1(x)mod(x),i=1,……,n 证明:f(x)与f1(x)…f-1(x)f+1(x)…·fn(x)互素,所以存 在u1(x),v1(x)使得 f(x)u1(x)+f1(x)…f-1(x)f+1(x)…fn(x)v1(x)=1 于是对任意的1≤k≤n 小(x)v1(x)I6()=∑n(x)01(x)I(x)+0(x)61(x)…-1(x) ∑41(x)2(x)(x)+a(x)-(x)(x)。 取8(x)=∑=141(x)21(x)(x)即可。可知,每个这样的解与给pê õ‘ª ŒúϪ ½n (¥I{½n)  f1 (x), . . . , fn (x) ∈ K [x] üüpƒ§Ké?¿‰½  a1 (x), . . . , an (x) ∈ K [x]§73 g (x), q1 (x), . . . , qn (x) ∈ K [x]§ ¦ g (x) = fi (x) qi (x)+ai (x), (Pg (x) ≡ ai (x) mod fi (x) §i = 1, . . . , n" y²µfi (x) † f1 (x)· · · fi−1 (x)fi+1 (x)· · · fn (x) pƒ§¤± 3 ui (x), vi (x) ¦ fi (x) ui (x) + f1 (x)· · · fi−1 (x)fi+1 (x)· · · fn (x) vi (x) = 1 u´é?¿ 1 ≤ k ≤ n n ∑ i=1 ai (x) vi (x)∏ j6=i fj (x) = n ∑ i6=k ai (x) vi (x)∏ j6=i fj (x) + ak (x)f1 (x)· · · fk−1 (x)fk+1 = n ∑ i6=k ai (x) vi (x)∏ j6=i fj (x) + ak (x) − fk (x) uk (x) "  g (x) = ∑ n i=1 ai (x) vi (x) ∏j6=i fj (x) =Œ"Œ§z‡ù)†‰ ½ g (x) ƒ ∏ n j=1 fj (x) ª"