TITLE: How prove this polynomial such $\dfrac{1}{f(x_{i})}=h(x_{i}),i=1,2,\cdots,s$.
QUESTION [5 upvotes]: Question:  let  polynomial  $\phi{(x)}$ 
is an irreducible polynomials on the rational number field.and assume that $x_{1},x_{2},\cdots,x_{s}$ is $\phi{(x)}$ complex roots, for any $f(x)$  is rational coefficients polynomial,such
$f(x)\nmid\phi{(x)}$;
show that:

There exist polynomial (where the coefficients is rational) $h(x)$,such
  $\dfrac{1}{f(x_{i})}=h(x_{i}),i=1,2,\cdots,s$.

my try:
since 
$\phi{(x)}$ is irreducible polymials on the rational number field,and have complex roots,then
$$\phi{(x)}=A(x-x_{1})(x-x_{2})\cdots (x-x_{s}),x_{i}\in C$$
then I can't.Thank you

REPLY [2 votes]: Let $\bf Q$ be the rationals, let $K={\bf Q}(x_1,\dots,x_s)$ be the splitting field of $\phi(x)$ over $\bf Q$, let $E={\bf Q}(x_1)$. Then $f(x_1)$ is in $E$, so $1/f(x_1)$ is in $E$, so $1/f(x_1)=h(x_1)$ for some polynomial $h$ with rational coefficients. Applying elements of the Galois group of $K/{\bf Q}$, we get $1/f(x_j)=h(x_j)$ for all $j$.