I am confused by problem 10, section 3.2 from Herstein.
The stated problem is:
"Show that the commutative ring D is an integral domain iff for all
a,b,c in D and a ne 0, ab=ac implies b=c"

I read the logic of this proof as follows:

Let D be a commutative ring  and ( D is an integral domain iff for all
a,b,c in D and a ne 0, ab=ac implies b=c)

I am having trouble with the proof of the back implication of this
proposition.
In class, we "proved" this proposition as follows:
Suppose D is a commutative ring.
Suppose also that for all a,b,c in D, ab = ac implies b=c
Now, let x and y be elements in D such that x ne 0 and xy = 0
using the basic ring property that the zero element times any element
in D is zero, we can say:
xy = 0 = x0
which by our hypothesis implies y=0.
So we have proved that xy=0 and x ne 0 implies y = 0, which implies D
has no zero divisors (since it is logically equivalent to xy ne 0 or x
= 0 or y = 0, which implies the statement "xy=0 and x ne 0 and y ne 0"
is always false and can therefore never occur, hence we can have no
zero divisors).

PROBLEM: My problem rests with the fact that (in the proof of the back
implication) we have showed that D is a commutative ring (given by
hypothesis) and that D has no zero divisors, but we have not shown
that D has a unit element, which is the third condition required in
the definition of "integral domain"

I went to a few other sources, but they weren't of much help. In
Hungerford's book, "Abstract Algebra : An Introduction" on pg. 61, thm
3.10, he proves the forward implication of Herstein's proof but not
its converse (i.e. the back implication). In Birkhoff and MacLane,
they take a somewhat different approach and define integral domain
right off the bat w/o talking about rings. They simply seem to replace
the condition for no zero divisors by the cancellation law (see pages
1 & 2 of "A Survey of Modern Algebra"). Finally in the Scahum's
outline, "Theory and Problems of Modern Abstract Algebra" by Frank
Ayres, pg. 114, , where he says (in talking about integral domains):
" A word of caution is necessary here. The term integral domain is
used by some to denote any ring without divisors of zero and by others
to denote any commutative ring without divisors of zero"
but then he goes on and seems to say the same thing implied by
Birkhoff and MacLane
"As a result "having no zero divisors" (his quotes) in the definition
of integral domain may be replaced by "for which the Cancellation Law
for multiplication holds" (his quotes)"

So I don't see how, in problem 10 from Herstein, we can prove that the
ring also has a unit element when we are proving the back implication.
Am I missing something or is Herstein a bit off on this one?

Thanks,