A fact about polynomials over a perfect field

Let K be an extension of F, with F a perfect field. Suppose p(x) \in F[x] has no repeated irreducible factors; prove that p(x) has no repeated irreducible factors in K[x].

Suppose p has no repeated irreducible factors. Since F is perfect, the irreducible factors of p are separable (Prop. 37 on 549 in D&F), and so p is separable. (If not, then it would have a repeated irreducible factor.) That is, p has no repeated roots.

Now if p has a repeated irreducible factor over K, then it has repeated roots- a contradiction.

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