Showing $F$ of Characteristic $p$ is Separable Provided $\phi(a) = a^p$ is Surjective on $F$

Question

Let $F$ be a field of characteristic prime $p$. Let $\phi: F \rightarrow F$ be defined as $\phi(a) = a^p$ for all $a \in F$.

Goals:

(i) Show that $\phi$ is an injective homomorphism of $F$.

(ii) Show that $F$ is separable provided $\phi$ is surjective.

Attempt at (i):

1. We have that $\phi$ is a homomorphism since $\phi(a_1 a_2) = (a_1 a_2)^p = a_1^p a_2^p = \phi(a_1) \phi(a_2)$, making use of the abelianness of $F$ to justify this step. Furthermore we have $\phi(a_1 + a_2) = (a_1 + a_2)^p = a_1^2 + p a_1 a_2 + a_2 ^p = a_1^2 + a_2^p$ since $p (a_1 a_2) = 0$ by the fact that $char(F) = p$.

2. For injectivity, let $a_1 \ne a_2$ in $F$.

3. Then $\phi(a_1) = a_1^p$ and $\phi(a_2) = a_2^p$.

Question: Why is $a_1^p = a_2^p$ absurd? Or else how can I show the injectivity of $\phi$?

Attempt at (ii):

1. Suppose $\phi$ is surjective, so that $b \in F \implies a^p = b$ for some $a \in F$.

2. Let $p(x) \in F[x]$ be irreducible s.t. $p(x) = a_0 + a_1 x + \ldots + a_n x^n$ for $a_i \in F$ and $n \ge 1$.

3. Then $p(x)$ is separable iff $gcd(p(x), p'(x)) = 1$.

4. Now $p'(x) = a_1 + 2 a_2 x + \ldots + n a_n x^{n-1}$.

5. Since $deg(p'(x)) < deg(p(x))$ and $p(x)$ is irreducible, the only way that $gcd(p(x), p'(x)) \ne 1$ is if $p'(x) = 0$.

6. Since $\phi$ is surjective, we can take

   $$ p(x) = a_0 + a_1 x + \ldots + a_n x^n = b_0^p + b_1^p x + \ldots + b_n^p x^n $$

   for the $b_i \in F$ s.t. $b_i^p = \phi(b_i) = a_i$

Question: Now how do I proceed from here to show that $p'(x) \ne 0$? Or is this not the right approach either?

Answer 1

i) Your algebra is a little confused in part 1) here. You need to show that $(x+y)^p = x^p + y^p$, and this requires looking at the binomial theorem seriously. You seem to be pretending that $p=2$, which is a good example, but not all that's needed.

Fortunately, once you have fixed this issue, you will have the tools necessary to show that $a\mapsto a^p$ is injective. Hint: if you think hard, you'll find that you already know how to factor $x^p +y^p$. Now apply the same reasoning to $x^p-y^p$.

ii) If $f'(x)$ is identically zero, then every nonzero term of $f(x)$ must be of the form $ax^{pk}$. As you've observed, we can also write this as $b^p x^{pk}$. Now use the very same idea from i) to show that $f(x) = \sum_i b_i^p x^{pk_i}$ factors. (Here's a thought: Write all p-th powers in terms of $\phi$.)