[Math] Does a finite commutative ring necessarily have a unity

abstract-algebrafinite-ringsring-theoryrngs

Does a finite commutative ring necessarily have a unity?

I ask because of the following theorem given in my lecture notes:

Theorem. In a finite commutative ring every non-zero-divisor is a unit.

If it had said "finite commutative ring with unity…" there would be no question to ask, I understand that part. What I'm asking about is whether or not we can omit explicitly stating it because it follows from the finiteness of our commutative ring.

[Clarification] The way I'm learning ring theory now, a "ring" is defined as an additive Abelian group further equipped (I hope I'm using the right terminology) with an associative multiplication operation which distributes over addition. In this definition we do not require the existence of 1.

In other words, when I say "ring" I mean a rng.

Best Answer

Let $A$ be a finite commutative ring (not assumed to contain an identity). Suppose that $a \in A$ is not a zero-divisor. Then multiplication by $a$ induces an injection from $A$ to itself, which is necessarily a bijection, since $A$ is finite. Thus multiplication by $a$ is a permutation of the finite set $A$, and hence multiplication by some power of $a$ (which by associativity is the same as some power of multiplication by $a$) is the identity permutation of $A$. That is, some power of $a$ acts as the identity under multiplication, which is to say, it is a (and hence the) multiplicative identity of $A$.

In short, if a finite commutative ring $A$ contains a non-zero divisor, then it necessarily contains an identity, and every non-zero divisor in $A$ is a unit.

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