Prove that an irreducible polynomial in $\displaystyle Q[x]$ has no multiples roots
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Let $\displaystyle f(x)\in \mathbb{Q}[x]$ be irreducible. Then consider the derivative. $\displaystyle f'(x)$ is a polynomial of degree n-1 and since $\displaystyle char(\mathbb{Q})=0$, so it is not identically 0. Up to constant factors, the only factors of $\displaystyle f(x)$ are $\displaystyle f(x)$ and 1, so $\displaystyle f(x)$ and $\displaystyle f'(x)$ are relatively prime, thus $\displaystyle f(x)$ is separable, ie has no multiple roots.