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NinjaSharkSet5Problem1

First notice $ (x-c^\frac{1}{p})^p = x^p-c $.

So $ F(c^\frac{1}{p}) $ is the splitting field of $ x^p-c $.

Now suppose that the polynomial is reducible in some field K.



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Abstract algebra continues the conceptual developments of linear algebra, on an even grander scale.

Dr. Paul Garrett