Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.4 Exercise 7.4.21
Solution:
[We will assume Wedderburn’s Theorem.]
Let $R$ be a finite ring (not necessarily commutative) with $1 \neq 0$ having no zero divisors. Let $a \in R$ be nonzero, and define $\varphi_a : R \rightarrow R$ by $\varphi_a(r) = ar$. Note that if $\varphi_a(r) = \varphi_a(s)$, we have $a(r-s) = 0$, and since $a \neq 0$, $r = s$. Thus $\varphi_a$ is injective. Since $R$ is finite, $\varphi_a$ is surjective. Thus for some $b \in R$, we have $\varphi_a(b) = ab = 1$. Similarly, with $\psi_a(r) = ra$, there exists $c \in R$ such that $ca = 1$. By Exercise 7.1.28, $a$ is a unit in $R$. Thus $R$ is a division ring.
By Wedderburn’s Theorem, $R$ is commutative; thus $R$ is a field.
