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


(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?


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 $\phi$.)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.