User:J D Bowen/Math735 HW10

13.5.5 For any prime $ p \ $ and any nonzero $ a \in F_p \ $ prove that $ x^p - x + a \ $ is irreducible and separable over $ F_p \ $.

Note that by Proposition 37 from the text, it suffices to show that f(x) = $ x^p - x + a \ $ is irreducible over $ F_p \ $. Then let $ \alpha \ $ be a root of f(x). Then

f($ \alpha \ $ + 1) = $( \alpha + 1)^p - ( \alpha + 1) + a \ $

= $ \alpha^2 + 1 - \alpha -1 + a \ $

= $ \alpha^2 - \alpha + a \ $

Then, for any $ \alpha \ $ the root of $ f \ $, $ \alpha + 1 \ $ is also a root. Thus by induction, each $ \alpha' \in F_p \ $ is a root of $ f \ $.

Now consider, f(0) = $ 0^p + 0 + a \ $ = 0 $ \implies a \ $ = 0, which is a contradiction. And therefore $ f \ $ has no roots.

So suppose that $ f \ $ is reducible, where $ f = g_1, g_2, \dots, g_n \ $. Then $ \exists \ $ some extension of $ F_p \ $, which contains the root $ \beta \ $ of $ f \ $. But, we proved that each $ \beta + k \ $ is also a factor for $ k \in F_p \ $, and thus our extension field is a splitting field.

Now since our $ \beta \ $ is arbitrary, the deg ($ g_i \ $) = [$ F_p(\beta) : F_p \ $] for any i. Since $ f \ $ has no roots

$ \prod_{1 \le i \le n}^{} \ $ deg ($ g_i \ $) = $ p \ $

for $ p \ $ prime $ f \ $ must be irreducible.