A finite commutative ring with at least two elements consisting of no zero divisors is a field

Question

My question was that finite ring of non divisors elements forms a field ... My approach to this was let all the elements be 0 , x1,x2 ,...xn then x1.x2.x3.....xn = xj for some j <=n then xj.( x1.x2....xj-1.xj+1....xn -1) = 0 implies x1.x2....xj-1.xj+1...xn= 1 hence 1 is in this ring ... but then we can say that its a finite integral domain which is a field ? Am I correct ? Anday any other proof ?

Answer 1

Here's how I would tackle this problem. Let $G$ be a finite commutative ring, since it's finite, we'll have $G = \{0; x_1; x_2; x_3; ...; x_n \}$.

• Prove that the sequence of $n$ elements: $x_1x_1$; $x_1x_2$; $x_1x_3$; ...; $x_1x_n$ are all distinct elements. Hint: Use the fact that $x_1$ is not a zero divisor.

• Hence every element $0 \neq g \in G$, there's exactly one index $j$, such that $g = x_1.x_j$.

• Consider the 'seem-to-be-infinite-but-actually-not' set $\{x_1; x_1^2; x_1^3; x_1^4; ... \}$, make a wise guess of an element that can be the multiplicative identity of $G$.

• Using the first dot, and the commutativity of this ring to actually prove that element is indeed the multiplicative identity.

• So now, your ring has 1. Now to prove that it's a field, you must prove that every $0 \neq g \in G$, $g$ has a multiplicative inverse. Consider this sequence $g; g^2; g^3; g^4; ...$, and use the fact that $g$ is a not a zero divisor; and the fact that $G$ has 1, to find the multiplicative inverse of $g$.

Cheers,