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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.

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  • $\begingroup$ I donot know what a commutative ring without 1 is, what about $2\mathbb{Z}/4\mathbb{Z}$? $\endgroup$
    – wxu
    Jun 14, 2011 at 0:07
  • $\begingroup$ Look in that book for the definition of ring. Sometimes it includes 1. A discussion in Mathoverflow: mathoverflow.net/questions/22579 $\endgroup$
    – GEdgar
    Jun 14, 2011 at 0:16
  • $\begingroup$ @GEdgar: Yes I am aware of the different definitions of a "ring". I added a clarification; in my case we do not require 1. $\endgroup$
    – Josh
    Jun 14, 2011 at 0:20
  • $\begingroup$ @wxu: Excuse my oversight, see the clarification :) $\endgroup$
    – Josh
    Jun 14, 2011 at 0:22
  • $\begingroup$ @Rasmus: Dear Rasmus, I'm not sure that your edits clarified the question. As originally written, the word "unity" was used to mean "multiplicative identity" (of which it is one synonym), while "unit" in the block-quote means "invertible element". Your edits have now removed this distinction --- you have the word unit meaning both multiplicative identity and invertible element, which is potentially confusing ... . Regards, $\endgroup$
    – Matt E
    Jun 14, 2011 at 9:21

6 Answers 6

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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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  • $\begingroup$ Thanks everyone for the answers, all of them helped me understand this. Matt's answer just finalised it. $\endgroup$
    – Josh
    Jun 14, 2011 at 1:06
  • $\begingroup$ Does this work for finite-dimensional algebras as well? I guess so. That is, let $A$ be a commutative ring and a finite dimensional $\mathbb{K}$-vector space s.t. the ring multiplication is bilinear. Then every element of $A$ that is not a zero divisor is a unity. The proof is exactly the same, just replace "an injection is necessarily a bijection since $A$ is finite" with "a linear injection is a linear bijection because $A$ is finite dimensional". What do you think? $\endgroup$
    – Giuseppe Negro
    Jun 14, 2011 at 2:24
  • $\begingroup$ @dissonance: Dear dissonance, I agree. Regards, $\endgroup$
    – Matt E
    Jun 14, 2011 at 6:36
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    $\begingroup$ for a finite ring (commutative or not) the following statements are equivalent: the ring has a unit element, the ring has a non-zero-divisor $\endgroup$
    – miracle173
    Jun 14, 2011 at 10:19
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    $\begingroup$ @Noah: Dear Noah, Any matrix which does not have zero as an eigenvalue has an inverse, which can be expressed as a polynomial in the given matrix. (Consider the minimal polyomial.) Regards, $\endgroup$
    – Matt E
    Jun 14, 2011 at 14:49
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No. Consider the ideal generated by $2$ in $\mathbb{Z}/4\mathbb{Z}.$

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  • $\begingroup$ Aha. Of course. Thanks for that. $\endgroup$
    – Josh
    Jun 14, 2011 at 0:52
  • $\begingroup$ I think $2$ is not a non-zero-divisor in this ring because of $2*2=0$ $\endgroup$
    – miracle173
    Jun 14, 2011 at 10:40
  • $\begingroup$ It is still a finite commutative ring, however. $\endgroup$
    – Joe Z.
    Jan 30, 2013 at 16:34
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The usual fashion nowadays is to build the existence of a multiplicative identity into the definition of commutative ring. However, the stated result is correct even if one does not.

This is not because the existence of a multiplicative identity follows from finiteness. As already posted examples show, if one does not build a "$1$" into the definition of ring, there are finite rings with no $1$.

However, if there is even one non-zero divisor, then it is easy to prove that a finite ring must have a $1$. So one can say that in a finite ring, either every object other than $0$ is a zero divisor, or there is a multiplicative identity.

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  • $\begingroup$ You mean "unity" right? A "unit" is an invertible element I think. $\endgroup$
    – Josh
    Jun 14, 2011 at 0:56
  • $\begingroup$ But thank you, this clarified things a lot. $\endgroup$
    – Josh
    Jun 14, 2011 at 0:56
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    $\begingroup$ @Josh Chen: Either way! If there is an element $\ne 0$ which is not a zero divisor then (i) there is a multiplicative identity and (ii) everything which is not zero, not a zero divisor is invertible, that is, a unit. But thank you for pointing out the ambiguity. I changed the "unit" to "multiplicative identity." But more is true. The proof is basically the same as the usual proof that a finite integral domain is a field. $\endgroup$
    – André Nicolas
    Jun 14, 2011 at 2:22
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On any additive abelian group one can define an identically zero multiplication operation. Taking the group to be nontrivial but finite gives an example of a finite rng without unity. (Note that jspecter's example is of this form.)

On the other hand any proper ideal in a ring gives an example of a rng, but one has to be a little careful here: some other element could act as an identity on the ideal. One can avoid this by choosing rings without nontrivial idempotent elements, a good example being any local ring. (This leads back again to jspecter's example.)

[Note that the "rng" above is not a typo: it is a rather standard term for the algebraic object which satisfies all the axioms for a ring except the existence of a multiplicative identity. The point is that the vast majority of mathematicians nowadays mean by "ring" an object having a multiplicative identity and by a "ring homomorphism" a map preserving that identity. I had to restrain myself from answering, "Yes, every ring has a unity, by definition."]

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  • $\begingroup$ Referring to your last line, thank you for your restraint. :) And for your helpful answer. $\endgroup$
    – Josh
    Jun 14, 2011 at 0:54
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    $\begingroup$ I've heard it quipped that "rng" is pronounced "(w)rong." $\endgroup$
    – Cheerful Parsnip
    Jun 14, 2011 at 1:33
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If all the elements of the ring are zero divisors, it is false.

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Think about $0 \,\,2\,\, 4\,\, 6\,\, 8\,\, 10$ in $\mathbb{Z}_{12}$ is a ring without unity! And the proof from Matt E is not suitable here because all the elements are zero divisor. Am I right?

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