# Thread: Relations between the set consisting 0 and zero divisors and prime ideals.

1. ## Relations between the set consisting 0 and zero divisors and prime ideals.

One exercise from my textbook:

The set consisting of 0 and all zero divisors in a commutative ring with identity contains at least one prime ideal.

Having thought about this problem for days, I only come up with the solution to the simplest case in which there's no zero divisors in such a ring.
But what about the general case when zero divisors do exist in such a ring?
One more stupid question... does this set consisting 0 and zero divisors form a ring? Even an ideal?

2. ## Re: Relations between the set consisting 0 and zero divisors and prime ideals.

the set of zero divisors need not be closed under addition:

to see this, note that 2 and 3 are zero divisors in Z6, since 2*3 = 0 (mod 6).

but 2+3 = 5, which is NOT a zero-divisor (in fact it is a unit, since 5*5 = 1 (mod 6)).

just of the top of my head, the thought i have is:

suppose a is a zero-divisor. then for any r in R, ra is also a zero-divisor.

if there exists b ≠ 0 with ab = 0 (which there does, since a is a zero divisor), then (ra)b = r(ab) = r0 = 0.

thus (a), the ideal generated by a, is a subset of the set of zero divisors.

so now ask yourself: does (a) always contain a prime ideal? well, obviously it does if a is a prime element, because then (a) is a prime ideal.

but what if a isn't a prime element of R?

this means that (a) is not a prime ideal, so there exists a product xy in R, with xy in (a), but x,y not in (a).

hmm....suppose we form the following set: Jx = {y in R: xy is in (a)}. is THIS an ideal of R?

suppose y,y' are in Jx. then xy and xy' are in (a), whence xy - xy' is in (a), so that y - y' is in Jx.

and clearly, for any r in R, x(ry) = r(xy) is in (a), so ry is in Jx.

thus Jx is an ideal of R. moreover, it is clear (since 1 is in R), that (a) ⊂ Jx, and that further, this inclusion is strict (since x is not in (a)).

i claim that Jx consists of all zero-divisors. to see this let y be in Jx.

then xy = ra, for some r in R, hence (bx)y = b(ra) = r(ab) = r0 = 0 (using the b we know exists with ab = 0, since a is a zero-divisor).

if Jx is prime, we are done.

but, maybe not, so rinse and repeat.

we get a chain of ideals: (a) ⊂ Jx1 ⊂ Jx2 ⊂.......

by Zorn's lemma, this chain has a maximal element (which is a maximal ideal), let's call it M.

(1) every element of M is a zero-divisor. this is elementary, once you think about it, but i will explain, anyway. to apply Zorn's lemma, we need an upper bound for our chain of ideals. i claim this is:

I = U{Jxk: k in N} of course, we need to show I is an ideal.

but suppose y,y' are in I. then x is in Jxk for some k, and y' is in some Jxk' for some k'. since our chain is linearly ordered, one of these contains the other, so y - y' is in the larger one, and thus in I.

similarly, for any r in R, since any y in I lies in some Jxk, ry is also in Jxk, and thus in I. note 1 is not in I, so M is a proper ideal of R.

since every element of I lies in some Jxk, and all of these are zero divisors, and M is contained in I, M consists entirely of zero divisors.

(2) M is a prime ideal. this is easy: suppose rs is in M, with r in R-M. since (r) + M is an ideal containing M, (r) + M = R, by the maximality of M. so 1 is in (r) + M, that is 1 = ar + m, for some a in R, and m in M.

hence s = 1s = (ar + m)s = a(rs) + ms, which is in M.

in actual practice, we often find M in a finite number of steps (this will happen for many "example rings" you try).