Suppose $\mathbb{F}$ is a field of characteristic $p$. Show that if $a, b \in$ $\mathbb{F}$ and $a^{p}=b^{p}$, then $a=b$

Question

I'm trying to do Exercise 2.6.12 from textbook Groups, Matrices, and Vector Spaces - A Group Theoretic Approach to Linear Algebra by James B. Carrell. Could you please confirm if my attempt is fine or contains logical mistakes?

Suppose $\mathbb{F}$ is a field of characteristic $p .$ Show that if $a, b \in$ $\mathbb{F}$ and $a^{p}=b^{p}$, then $a=b$.

My attempt:

We have $(a + b)^p = a^p + b^p$. Substitute $c$ for $a+b$, we get $c^p = a^p + (c-a)^p$ and thus $c^p-a^p = (c-a)^p$. Apply this identity to $a,b$, we get $0 = (b-a)^p$ and thus $b-a = 0$. Finally, we have $a=b$.

Answer 1

Your proof is fine. Here is a slightly simpler one: $$ (a-b)^p = a^p-b^p = 0 \implies a-b=0 \implies a=b $$ This is clear when $p$ is odd but also works when $p=2$ because $-x=x$.