Problem 1

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 \supset F $ in which $ (x-c^\frac{1}{p})^m $ is one of the irreducible factors of $ x^p-c $ for some m < p.

Then $ c^\frac{m}{p} \in K $ but because p is prime, there is an n so that $ nm \equiv 1 $ (mod p) and so $ c^\frac{nm}{p} = c^r c^\frac{1}{p} $. Thus $ c^\frac{1}{p} \in K $ so $ K \supset F(c^\frac{1}{p}) $.

Thus, if $ x^p-c $ is not irreducible over F, $ F \supset F(c^\frac{1}{p}) $ and F contains a root of $ x^p-c $.



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

Dr. Paul Garrett