# A problem about local rings.

• Jan 13th 2013, 01:00 AM
Rita
One problem:
A commutative ring R with identity is local iff for all r,s in R, r+s=1R implies r or s is a unit.

I've proved the easy part of the proof (=>), but for the converse one? So far, I only have a result that if K is a set consisting of all nonunits of R, then for any u, v in K, u+v won't be 1R. Does it help? Any suggestion?
• Jan 13th 2013, 03:24 AM
Deveno
Re: A problem about local rings.
note that your condition can be rephrased as:

either r or 1R - r is a unit, for any r in R.

you don't say what definition of local you are using, there are several equivalent formulations. another statement equivalent to the one you gave is:

if I,J are co-prime ideals, then either I or J (or both) is not proper (that is: one (or both) of them is the entire ring).

still another way to say the same thing is:

if r is a non-unit, and s is a non-unit, so is r+s (for if not: we have r+s = u, so u-1r + u-1s = 1R, so that one of u-1r, u-1s is a unit.

let's say it is u-1r. then x(u-1r) = 1R, for some x in R, whence (xu-1)r = 1R, so that r is a unit).

in fact, using the set K you give, we have: K is closed under subtraction (taking it as obvious -k is not a unit if k is a non-unit). suppose k is in K, and r is any element of R. clearly rk is in K as well, for rk a unit implies k a unit.

so K is, in fact, an ideal of R. but this ideal must therefore be maximal, since if we include anything else, it will be a unit, and thus will include the entire ring.

suppose J is any other maximal ideal. since J does not include any units (since maximal ideals are proper), J ⊆ K.since J is maximal and K is proper, J = K. this shows R has a UNIQUE maximal ideal, that is: R is local.

the main reason why rings fail to be local is that the sum of two non-units may actually be a unit. for example, with Z, we have -2 + 3 = 1. this is the same as saying (2) and (3) are both proper co-prime ideals of Z, something that cannot happen in a local ring.

we can "force locality" of a (commutative) ring R, given a prime ideal P like so:

let S = R - P. then S is multiplicatively closed (since P is a prime ideal), so we can form S-1R in an analogous fashion to the field of fractions of an integral domain:

S-1R = RxS/~ where (r,s) ~ (r',s') iff t(rs') = t(r's), for some t in S.

we have a ring-homomorphism R-->S-1R given by r--> [(rs,s)] = [(r,1)]. in this new ring, the ideal generated by P is the unique maximal ideal.

(if S contains zero-divisors, this will not be an injection, unfortunately).
• Jan 14th 2013, 06:37 PM
LoidaWard
Re: A problem about local rings.
suppose J is any other maximal ideal. since J does not include any units (since maximal ideals are proper), J ⊆ K.since J is maximal and K is proper, J = K. this shows R has a UNIQUE maximal ideal, that is: R is local.

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