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A nonzero finite commutative ring with no zero divisors is a field

Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.4 Exercise 7.4.20


Let $R$ be a finite commutative ring with no zero divisors.

Now let $I \subseteq R$ be a nonzero ideal, with $a \in I$ nonzero. Define $\varphi_a : R \rightarrow R$ by $\varphi_a(r) = ar$. If $\varphi_a(r) = \varphi_a(s)$, then $ar = as$, so that $a(r-s) = 0$. Since $R$ has no zero divisors and $a \neq 0$, $r-s=0$, so that $r=s$. Thus $\varphi_a$ is injective. Since $R$ is finite, $\varphi_a$ is also surjective, so that every element $r \in R$ has the form as for some $s \in R$. In particular, $R \subseteq I$. Thus $R$ has only the trivial ideals.

Now suppose that $a \in \mathfrak{N}(R)$ is nonzero, and let $m$ be minimal such that $a^m = 0$. Now $m \geq 2$. But then $aa^{m-1} = 0$, and since $a \neq 0$, $a^{m-1} = 0$, a contradiction. Thus the nilradical of $R$ is trivial.

Now let $a \in R$ be nonzero; then $(a) = R$. In particular, note that since $R$ is finite, there exists $m \geq 2$ such that $a^m = a^n$ for some $n < m$. Let $m$ be minimal with this property. Now $a^m-a^n = 0$. Suppose $n \geq 2$. Then $$a^{n-1}(a^{m-n+1} – a) = 0.$$ Since $a$ is not nilpotent, we have $a^{m-n+1} = a$, a contradiction. Thus $n=1$ and we have $a^m = a$.

Finally, let $x = ra \in R$. Now $$a^{m-1}x = a^{m-1}ra ra^m = ra = x.$$ Then $a^{m-1}$ satisfies the defining property of an identity element, and thus $R$ is a finite integral domain. We proved in a lemma to Exercise 7.1.18 that every finite integral domain is a field.


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