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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.

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    $\begingroup$ 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. $\endgroup$ – Crostul Apr 5 at 15:28
  • $\begingroup$ @Crostul I see, thank you. I've shown that it's trivial by definition. $\endgroup$ – Ilan Aizelman WS Apr 5 at 15:50
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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.

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

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