Maximal Ideals
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From: William Elliot on 3 Aug 2010 04:48 On Mon, 2 Aug 2010, Hagen wrote: >>> On Aug 1, 4:29 am, William Elliot <ma.. >> Let R be the set of functions of (Z_2)^N with finite support and >> pointwise addition and multiplication. R is a Boolean ring without any >> maximal ideals and for any a in R, there is no maximal ideal within >> ideals not containing a. >> Does R have any proper prime ideals? I doubt it. > > You mean R consists of those elements of (Z_2)^N having finite support? Yes. > Fix some k \in N and let P be the set of all r \in R such that r(k)=0. > Isn't P a proper prime ideal then? > It's a proper ideal. Ok, it's a prime ideal. In general, if R is a subring of a product of rings, prod{ Rj | j in I }, then is I_j = { r | r(j) = 0 } a proper prime ideal? >> Let R1 be the extension of R with an (multiplicative) identity 1, >> added. R1 is the set of functions of (Z_2)^N with finite or cofinite >> support and pointwise addition and multiplication. > > 'the extension' ? Is it unique in some sense? I would hope so; as the smallest subring of (Z_2)^N containing R and 1. > So R1 equals (Z_2)^N. That I dispute. >> R1 does have maximal ideals and as maximal ideals in Boolean rings, >> are prime ideals, R1 has prime ideals. > > Maximal ideals are always prime - not just in Boolean rings. > If I maximal ideal, ab in I, a not in I, then b in (a,I) .. . = { ra + na + i | r in ring, n in Z, i in I }. Ok, that's easy to see for communative rings with identity and Boolean rings with or without idenity. What if the ring doesn't have identity or isn't communative? >> The intersection of all maximal ideals is {0}. For a given point a, >> there's a maximal ideal within the ideals excluding a, which is a >> maximal ideal, hence a prime ideal. Is R1 a PID? > > Every FINITELY GENERATED ideal in a Boolean ring is principal. Let R be a Boolean ring and a in R. Is there a prime ideal excluding a?
From: William Elliot on 3 Aug 2010 05:57 On Mon, 2 Aug 2010, Arturo Magidin wrote: >> If a is not in the intersection of all maximal ideals >> and I is an ideal maximal within ideals not containing a, >> then I is a maximal ideal. >> >> Steps for a proof. >> There's some maximal ideal J with a not in J. >> J subset I; I = J is maximal. > > There is no reason whatsoever why two ideals that are maximal among > those that do not contain I need to be related to one another by > inclusion. > Ok, I see the problem, confusing maximal with maximum. How does one approach this problem?
From: Hagen on 3 Aug 2010 02:27 > On Mon, 2 Aug 2010, Hagen wrote: > >>> On Aug 1, 4:29 am, William Elliot <ma.. > > >> Let R be the set of functions of (Z_2)^N with > finite support and > >> pointwise addition and multiplication. R is a > Boolean ring without any > >> maximal ideals and for any a in R, there is no > maximal ideal within > >> ideals not containing a. > > >> Does R have any proper prime ideals? I doubt it. > > > > You mean R consists of those elements of (Z_2)^N > having finite support? > > Yes. > > > Fix some k \in N and let P be the set of all r \in > R such that r(k)=0. > > Isn't P a proper prime ideal then? > > > It's a proper ideal. Ok, it's a prime ideal. > > In general, if R is a subring of a product of rings, > prod{ Rj | j in I }, then is I_j = { r | r(j) = 0 } > a proper prime ideal? No. You have to assume that the ring Rj is an integral domain. In this case I_j either is a proper prime ideal or equals R. > >> Let R1 be the extension of R with an > (multiplicative) identity 1, > >> added. R1 is the set of functions of (Z_2)^N with > finite or cofinite > >> support and pointwise addition and multiplication. > > > > 'the extension' ? Is it unique in some sense? > > I would hope so; as the smallest subring of (Z_2)^N > containing R and 1. > > > So R1 equals (Z_2)^N. > > That I dispute. You say that R1 contains all elements of cofinite support. The empty set is finite, thus R1 contains all elements of (Z_2)^N. Ok, lets forget about this and stick to the definition of R1 as the smallest subring of (Z_2)^N that contains R and 1. Then R1 = {1+r . r \in R} \cup R Hence for an element of R1 either has finite support or has infinitely many components equal to 1. Thus the element r with r(2k)=0 and r(2k+1)=1 is not in R1. > >> R1 does have maximal ideals and as maximal ideals > in Boolean rings, > >> are prime ideals, R1 has prime ideals. > > > > Maximal ideals are always prime - not just in > Boolean rings. > > > If I maximal ideal, ab in I, a not in I, then b in > (a,I) > . . = { ra + na + i | r in ring, n in Z, i in I }. > > Ok, that's easy to see for communative rings with > identity > and Boolean rings with or without idenity. > What if the ring doesn't have identity or isn't > communative? We are talking about a commutative ring with unity here - namely about R1. H > >> The intersection of all maximal ideals is {0}. > For a given point a, > >> there's a maximal ideal within the ideals > excluding a, which is a > >> maximal ideal, hence a prime ideal. Is R1 a PID? > > > > Every FINITELY GENERATED ideal in a Boolean ring is > principal. > > Let R be a Boolean ring and a in R. > Is there a prime ideal excluding a? > > >
From: W. Dale Hall on 3 Aug 2010 06:37 William Elliot wrote: > On Mon, 2 Aug 2010, Arturo Magidin wrote: > >>> If a is not in the intersection of all maximal ideals >>> and I is an ideal maximal within ideals not containing a, >>> then I is a maximal ideal. >>> >>> Steps for a proof. >>> There's some maximal ideal J with a not in J. >>> J subset I; I = J is maximal. >> >> There is no reason whatsoever why two ideals that are maximal among >> those that do not contain I need to be related to one another by >> inclusion. >> > Ok, I see the problem, confusing maximal with maximum. > How does one approach this problem? How about by recognizing that the property "maximal with respect to property P" depends essentially on the defining property "P" ? That'd do for a start. Second, realizing that partial orderings aren't always total orderings. I don't suspect you don't already know these things, only that you've neglected to realize you already know them. Just my two bits' worth. Dale
From: William Elliot on 3 Aug 2010 07:09 On Tue, 3 Aug 2010, Hagen wrote: >>>>> On Aug 1, 4:29 am, William Elliot <ma.. >> >>> You mean R consists of those elements of (Z_2)^N >> having finite support? >> Yes. >> >> In general, if R is a subring of a product of rings, >> prod{ Rj | j in I }, then is I_j = { r | r(j) = 0 } >> a proper prime ideal? > > No. You have to assume that the ring Rj is an > integral domain. In this case I_j either is a proper > prime ideal or equals R. > All that's needed is for Rj to not have any zero divisors. >>>> Let R1 be the extension of R with an (multiplicative) identity 1, >>>> added. R1 is the set of functions of (Z_2)^N with finite or cofinite >>>> support and pointwise addition and multiplication. >>> >>> 'the extension' ? Is it unique in some sense? >> >> I would hope so; as the smallest subring of (Z_2)^N >> containing R and 1. >> > Then R1 = {1+r . r \in R} \cup R > Hence for an element of R1 either has finite support > or has infinitely many components equal to 1. Or equivalently, cofinite support. > Thus the element r with r(2k)=0 and r(2k+1)=1 is > not in R1. > Agreed, R1 /= (Z_2)^N. >>> Maximal ideals are always prime - not just in Boolean rings. >>> >> If I maximal ideal, ab in I, a not in I, then b in >> (a,I) >> . . = { ra + na + i | r in ring, n in Z, i in I }. >> >> Ok, that's easy to see for communative rings with >> identity and Boolean rings with or without idenity. >> What if the ring doesn't have identity or isn't communative? > > We are talking about a commutative ring with unity > here - namely about R1. In other other than Boolean rings as you mention, are there any classes of rings beyond communative rings with identity, for which maximal rings are prime? >> Let R be a Boolean ring and a in R. >> Is there a prime ideal excluding a? I'm still trying to fix a flawed proof for which a hopefully yes answer to that question would complete the proof.
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