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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

1 Answers1

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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 $j$.

Gerry Myerson
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  • +1. It's implicitly used that since the $x_i$ are algebraic over $\bf Q$, it holds that ${\bf Q} (x_i) = {\bf Q}[x_i]$ and then $h$ can exist. – leo Nov 23 '13 at 16:02