If a is algebraic over F of degree at least 2, then no power of x-a is in F[x]

Let F be a field and let \alpha be algebraic over F of degree at least 2. Prove that (x-\alpha)^n is not in F[x] for any n.


Suppose to the contrary that (x-\alpha)^n \in F[x]. Then the minimal polynomial p(x) of \alpha over F divides (x-\alpha)^n. Since the roots of (x-\alpha)^n are all the same and those of p(x) are all distinct, in fact p(x) = x - \alpha; but then \alpha has degree 1 over F, a contradiction.

So no power of x-\alpha can be in F[x].

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