# Show that a finite ring (with identity) is a division ring if and only if it has no zero divisors.

Claim: Show that a finite ring (with identity) is a division ring if and only if it has no zero divisors.

Proof: Let $$R$$ be a finite nonzero ring with no zero divisors. Let $$x ∈ R$$ be a nonzero element. Since $$R$$ is finite, there are only finitely many distinct powers of $$x$$.

Suppose that $$x^m = x^n$$ for some $$m > n$$. Then $$0 = x^m − x^n = x^n(x^{m−n} − 1)$$.

Since $$R$$ has no zero divisors, one of $$x^n$$ and $$x^{m−n} − 1$$ must be zero. If $$x^n = 0$$, then $$x$$ is zero divisor, which is a contradiction. Therefore, $$x^{m−n} − 1 = 0$$, i.e. $$x ^{m−n} = x · x^{m−n−1} = 1$$.

Therefore, $$x$$ has an inverse, and since this holds for all nonzero $$x$$, $$R$$ is a division ring.

Question: does this proof shows both directions (if and only if), because I assume that it's a division ring with no zero divisors from the start, without implicitly saying that

If a finite ring is a division ring, then showing that there're no zero divisors, and the opposing direction,

If a finite ring has no zero divisors then it has to be a division ring.

• The proof you posted shows only one direction of the theorem. That's because the opposite direction is trivial. In any division ring there is no zero divisor. – Crostul Apr 5 at 15:28
• @Crostul I see, thank you. I've shown that it's trivial by definition. – Ilan Aizelman WS Apr 5 at 15:50

## 1 Answer

May one briefly recommend an alternative argument that is applicable to somewhat more general settings (in proving for instance that given a commutative field $$K$$ and a finite dimensional non-zero unital associative algebra $$A$$ with no zero-divisors, then likewise $$A$$ is a field with its internal ring structure).

Let $$A$$ be non-zero, finite and with no zero-divisors. For $$t \in A$$ define:

$$\gamma_t, \delta_t: A \to A \\ \gamma_t(x)=tx, \delta_t(x)=xt$$

to be the left, respectively right homothecy of factor $$t$$. From the axiom of distributivity it is obvious that $$\gamma_t, \delta_t \in \mathrm{End}_{\mathrm{Gr}}(A)$$ (both homothecies are endomorphisms of the additive group structure on $$A$$). The claim on the absence of zero-divisors entails that for $$t \neq 0_A$$ both homothecies will be injective, as their kernels are trivial. From elementary set theory we (should) know that any injective endomorphism of a finite set is automatically bijective (by endomorphism of a set $$M$$ I mean simply a map from $$M$$ to itself).

The surjectivity of the two homothecies means in particular that $$1_A \in \mathrm{Im}(\gamma_t)=tA$$ and similarly $$1_A \in \mathrm{Im}(\delta_t)=At$$; this means nothing else than the existence of both a left and a right inverse, hence $$t$$ is invertible and thus $$A^{\times} \subseteq \mathrm{U}(A)$$ (the term on the right-side is my notation for the group of units/invertible elements; the reverse inclusion will always be valid in arbitrary non-zero rings). Thus, $$A$$ is a field.

• Thank you for the alternative proof :) – Ilan Aizelman WS Apr 5 at 15:53