Mathematical Reflections 2014, Issue 4 - Problem U312

Problem:
Let $p$ be a prime and let $R$ be a commutative ring with characteristic $p$. Prove that the
sets $S_k = \{x \in R \ | \ x^p=k\}$, where $k \in \{1,\ldots,p\}$, have the same number of elements.

Proposed by Corneliu Manescu-Avram.

Solution:
If for any $x \in R$ and $k \in \{1,\ldots,p\}$ it holds $x^p \neq k$, then there is nothing to prove. Assume that there is some $x \in R$ such that $x^p=k$ for some $k \in \{1,\ldots,p\}$. Hence, $k \in R$. Consider the set $S=\{x+k \ | \ x \in R\}$. Observe that the mapping $\varphi:R \longrightarrow S$ defined by $\varphi(x)=x+k$ is bijective. Indeed, if $\varphi(x)=\varphi(y)$, then $x+k=y+k$, i.e. $x=y$ and this proves that $\varphi$ is injective. If $y \in S$, then $y=x+k$ for some $x \in R$, so it's enough to take $x=y-k \in R$ in order to have $y=\varphi(x)$, and this proves that $\varphi$ is surjective. Let $n \in \{1,\ldots,p\}$ and consider the set $S_n = \{x \in R \ | \ x^p=n\}$. Since $(x+k)^p=x^p+k^p=x^p+k$, then $$\varphi(S_n)=\begin{cases} S_{n-k} & \textrm{if } n>k \\ S_{p-k+n} & \textrm{if } k \leq n. \end{cases}$$ It follows that $|S_n|=|S_{n-k}|$ or $|S_n|=|S_{p-k+n}|$. By arbitrarily of $n \in \{1,\ldots,p\}$, we conclude that $|S_1|=|S_2|=\ldots=|S_p|$.