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.