Does a finite commutative ring necessarily have a unity? 
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.
 A: 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.
A: 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."]
A: 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.
A: If all the elements of the ring are zero divisors, it is false.
A: No. Consider the ideal generated by $2$ in $\mathbb{Z}/4\mathbb{Z}.$
A: 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?
