| Line 1: | Line 1: | ||
[[Category:NinjaSharksSet5]] | [[Category:NinjaSharksSet5]] | ||
| − | = | + | =Problem 1= |
| Line 8: | Line 8: | ||
So <math> F(c^\frac{1}{p})</math> is the splitting field of <math> x^p-c </math>. | So <math> F(c^\frac{1}{p})</math> is the splitting field of <math> x^p-c </math>. | ||
| − | Now suppose that the polynomial is reducible in some field K. | + | Now suppose that the polynomial is reducible in some field <math>K \supset F </math> in which <math>(x-c^\frac{1}{p})^m </math> is one of the irreducible factors of <math>x^p-c </math> for some m < p. |
| + | |||
| + | Then <math>c^\frac{m}{p} \in K</math> but because p is prime, there is an n so that <math> nm \equiv 1</math> (mod p) and so <math> c^\frac{nm}{p} = c^r c^\frac{1}{p}</math>. Thus <math>c^\frac{1}{p} \in K</math> so <math>K \supset F(c^\frac{1}{p})</math>. | ||
| + | |||
| + | Thus, if <math> x^p-c </math> is not irreducible over F, <math>F \supset F(c^\frac{1}{p})</math> and F contains a root of <math> x^p-c </math>. | ||
Latest revision as of 08:06, 3 July 2013
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 $.
