Question

a=根号2加根号3,证明,存在有理数域上的不可约多项式f（x）,使f（a）=0

Answer 1

证明：因为（√2+√3)(√2-√3)=-1, (√2+√3)+(√2-√3)=2√2故√2+√3是方程x^2-2√2x-1=0的根x^2-2√2x-1=0,乘以x^2+2√2x-1得：（x^2-1)^2-(2√2x)^2=0,即：x^4-10x^2+1=0取f(x)=x^4-10x^2+1,则f(x)为有理数域上...