Maximal Ideals
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From: Arturo Magidin on 1 Aug 2010 10:41 On Aug 1, 4:29 am, William Elliot <ma...(a)rdrop.remove.com> wrote: > On Sat, 31 Jul 2010, Arturo Magidin wrote: > > On Jul 31, 11:05 pm, William Elliot > >> By Zorn's lemma, if x /= 0, there's a maximal ideal > >> I with x not in I. Is I a maximal ideal? > > > Not necessarily. > > > But why restrict to x? You have much larger notions of ideals that > > avoid specific subsets. > > Just x is all that's needed for the current problem. Perhaps (you gave no context); but you claimed that the terminology was ambiguous and then justified it by giving a single issue that arises, and insinuated that the only other notion of "maximal" is the one you are contemplating. As usual, you are cryptic and high handed, and as often, also wrong. > > >> Are the two notions the same for rings? > > > There are many notions; each x gives you a different notion; each > > multiplicative subset may even give you yet another one. > > >> If different, how are they distinguished? > > > "Maximal ideal" refers to ideals that are maximal, in the lattice of > > *ALL* proper ideals. If you want to consider some subset of ideals > > (such as the collection of all ideals that do not contain a given x), > > then you talk about "ideals maximal among those that ..." or some such > > expression. > > How would you express my question at the top of this post? "Ideals that are maximal among those that do not contain x". I thought that was obvious from the syntax of my sentence. > >> Are there cases when they are the same? > > > Yes: when x is a unit; or more generally, when x is not in the > > intersection of all maximal ideals. > > If a not in intersection of all maximal ideals and > I is a maximal ideal without a, then I is a maximal ideal. > > Proof. Some maximal J with a not in J; J subset I; > I = J is maximal. QED. This is not a valid proof of the statement. You have shown that if a is not in the intersection of all maximal ideals, then there exists an ideal that is both maximal and does not contain a. You did *not* prove, however, that if I is an ideal that is maximal among those that do not contain a, then I is maximal among all ideals, which is the statement you claimed to be proving. Yours is a basic misreading of the statement. > Let (R, disjoint union, intersection, nulset) be the > Boolean ring of finite subsets of an infinite set. > > Then R doesn't have any maximal ideals nor is there > any (relative) maximal ideal I with x not in I. Ouch! > > In a Boolean ring R without units, if a /= 0 > is there a prime ideal I with a not in I? What if a is nilpotent? -- Arturo Magidin
From: William Elliot on 2 Aug 2010 04:38 On Sun, 1 Aug 2010, Arturo Magidin wrote: > On Aug 1, 4:29�am, William Elliot <ma.. 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. > Yours is a basic misreading of the statement. > Have I made ample corrections? >> Let (R, disjoint union, intersection, nulset) be the >> Boolean ring of finite subsets of an infinite set. >> >> Then R doesn't have any maximal ideals nor is there >> any (relative) maximal ideal I with x not in I. �Ouch! >> >> In a Boolean ring R without units, if a /= 0 >> is there a prime ideal I with a not in I? > > What if a is nilpotent? > Can there be any nilpotents in a Boolean ring other than 0? 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. 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. R1 does have maximal ideals and as maximal ideals in Boolean rings, are prime ideals, R1 has prime ideals. 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? If R is a Boolean ring with (multiplicative) identity, does it have the same ideal properties as listed above that R1 has?
From: Hagen on 2 Aug 2010 04:36 > On Sun, 1 Aug 2010, Arturo Magidin wrote: > > On Aug 1, 4:29 am, William Elliot <ma.. > > 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. > > > Yours is a basic misreading of the statement. > > > Have I made ample corrections? > > >> Let (R, disjoint union, intersection, nulset) be > the > >> Boolean ring of finite subsets of an infinite set. > >> > >> Then R doesn't have any maximal ideals nor is > there > >> any (relative) maximal ideal I with x not in I. > Ouch! > >> > >> In a Boolean ring R without units, if a /= 0 > >> is there a prime ideal I with a not in I? > > > > What if a is nilpotent? > > > Can there be any nilpotents in a Boolean ring other > than 0? > > 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? 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? > 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? So R1 equals (Z_2)^N. > R1 does have maximal ideals and as maximal ideals in > Boolean > rings, are prime ideals, R1 has prime ideals. The Maximal ideals are always prime - not just in Boolean rings. > 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. H > If R is a Boolean ring with (multiplicative) > identity, does > it have the same ideal properties as listed above > that R1 has? >
From: Arturo Magidin on 2 Aug 2010 12:53 On Aug 2, 3:38 am, William Elliot <ma...(a)rdrop.remove.com> wrote: > On Sun, 1 Aug 2010, Arturo Magidin wrote: > > On Aug 1, 4:29 am, William Elliot <ma.. > > 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. THIS IS NOT A PROOF OF THAT STATEMENT. 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. You have no warrant whatsoever to assume that J will be contained in I. You have shown that if J is a maximal ideal that does not contain a, then J is maximal among the ideals that do not contain a. But what you claimed to be proving was that if J is an ideal that is maximal among the ideals that do not contain a, then J is a maximal ideal. YOU DID NOT PROVE THAT. That's *TWICE* you've failed spectacularly at this very simple logical issue. For someone who is so high-and-mighty when complaining about basic writing skills and thinking skills of others, it is always rather ironic to see just how hard-headed and silly you can be and not notice it. > > Yours is a basic misreading of the statement. > > Have I made ample corrections? Hardly. You merely repeated the exact same stupid error. -- Arturo Magidin
From: Chip Eastham on 2 Aug 2010 14:44 On Aug 1, 1:55 am, Arturo Magidin <magi...(a)member.ams.org> wrote: > On Aug 1, 12:34 am, "W. Dale Hall" > > > > <wdunderscorehallatpacbelldotnet(a)last> wrote: > > William Elliot wrote: > > > The expression maximal ideal is ambiguous. > > > Here are two differing notions of maximal ideal. > > > > By Zorn's lemma, if x /= 0, there's a maximal ideal > > > I with x not in I. Is I a maximal ideal? > > > > Are the two notions the same for rings? > > > If different, how are they distinguished? > > > Are there cases when they are the same? > > > > ---- > > > I'd word the notion you've identified a bit > > more carefully, namely, as being "maximal > > with respect to not containing x as a member". > > The unrestricted "maximal ideal" notion is > > "maximal with respect to being a proper ideal > > of the ring". > > > I could imagine that they may not be identical, > > but haven't tried to come up with an example. > > Consider the ring R= Z/4Z; and let x = 2 + 4Z. The only ideal of R > that does not contain x is (0); and the only maximal ideal of R is > (2). > > -- > Arturo Magidin An even more familiar example, anticipating William's confusion below, take the ideals of Z that are maximal with respect to not containing 2. These are the prime ideals other than 2Z and the (principal but not prime) ideal 4Z. This is essentially just "reading down" Arturo's example into Z, but it illustrates that 4Z can be "maximal with respect to x not in I" (here 2 not in 4Z) without being maximal among all ideals, and without any resort to discussing nilpotent elements. William may be recalling a topic of ring theory in which one looks at ideals maximal with respect to not intersecting a multi- plicative set S. [Or maybe not.] regards, chip
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