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From: William Elliot on 11 Aug 2010 05:26 On Wed, 11 Aug 2010, quasi wrote: > On Tue, 10 Aug 2010 21:52:55 -0700, William Elliot > <marsh(a)rdrop.remove.com> wrote: > >>>>>>> A clear discussion of these ideas can be found in the first >>>>>>> few pages of Kaplansky's wonderful text "Commutative Rings". >> >>>> Currently beyond my reach. >> >>> Well, when your reach extends further, reach for it -- I think you >>> would really appreciate his style of writing, as well as his cool >>> choice of exercises. >>> >> What are the page numbers to which you're referring. I'll see >> if I can fetch them through local library reference service. >> >> Edition number and date of publication may also be useful. > > 2nd Edition, 1974. > > The early pages. > Without the exact pages, I couldn't make a local library reference request.
From: William Elliot on 11 Aug 2010 06:14 On Wed, 11 Aug 2010, William Elliot wrote: > On Wed, 11 Aug 2010, quasi wrote: >> On Tue, 10 Aug 2010 20:49:23 -0700, William Elliot > >>> If B is a Boolean ring without identity, then >>> not some a with for all x, ax = x; >>> for all a, some x with ax /= x, a(ax + x) = 0. >>> >>> ie, every non-zero element is a zero divisor. >> Right. >> >>> Proposition. A prime ideal of a Boolean ring is >>> maximal even if the ring is without identity. >> >> Right, otherwise you could mod out by a non-maximal prime ideal and >> get a Boolean ring (with or without identity), but lacking zero >> divisors. >> > Clearly if I is an ideal, then B/I is a Boolean ring. > If (a+I)(b+I) = 0+I, then ab+I = 0+I; ab in I; > a in I or b in I; a+I = 0+I or b+I = 0+I. Ok. > > How's that a contradiction? There are two Boolean > rings without zero divisors, {0} and {0,1}. > > If B/I = {0}, then I = B, a no, no. > If B/I = {0,1}, then some a not in I with B = I \/ a+I. > > If b not in I, then b in a+I; b+I = a+I; > I, b+I subset (b,I); B = I \/ a+I = I \/ b+I = (b,I). > > In conclusion. If I is an ideal of a Boolean ring B, then > I is prime ideal iff I is maximal ideal I is maximal ideal among ideals excluding a iff I is proper prime ideal iff I is maximal ideal.
From: quasi on 11 Aug 2010 07:16 On Wed, 11 Aug 2010 02:26:47 -0700, William Elliot <marsh(a)rdrop.remove.com> wrote: >On Wed, 11 Aug 2010, quasi wrote: >> On Tue, 10 Aug 2010 21:52:55 -0700, William Elliot >> <marsh(a)rdrop.remove.com> wrote: >> >>>>>>>> A clear discussion of these ideas can be found in the first >>>>>>>> few pages of Kaplansky's wonderful text "Commutative Rings". >>> >>>>> Currently beyond my reach. >>> >>>> Well, when your reach extends further, reach for it -- I think you >>>> would really appreciate his style of writing, as well as his cool >>>> choice of exercises. >>>> >>> What are the page numbers to which you're referring. I'll see >>> if I can fetch them through local library reference service. >>> >>> Edition number and date of publication may also be useful. >> >> 2nd Edition, 1974. >> >> The early pages. >> >Without the exact pages, I couldn't make >a local library reference request. I don't have a copy at present. But the book really should be read from the beginning. How far is up to you. quasi
From: quasi on 11 Aug 2010 07:21
On Wed, 11 Aug 2010 03:14:56 -0700, William Elliot <marsh(a)rdrop.remove.com> wrote: >On Wed, 11 Aug 2010, William Elliot wrote: >> On Wed, 11 Aug 2010, quasi wrote: >>> On Tue, 10 Aug 2010 20:49:23 -0700, William Elliot >> >>>> If B is a Boolean ring without identity, then >>>> not some a with for all x, ax = x; >>>> for all a, some x with ax /= x, a(ax + x) = 0. >>>> >>>> ie, every non-zero element is a zero divisor. >>> Right. >>> >>>> Proposition. A prime ideal of a Boolean ring is >>>> maximal even if the ring is without identity. >>> >>> Right, otherwise you could mod out by a non-maximal prime ideal and >>> get a Boolean ring (with or without identity), but lacking zero >>> divisors. >>> >> Clearly if I is an ideal, then B/I is a Boolean ring. >> If (a+I)(b+I) = 0+I, then ab+I = 0+I; ab in I; >> a in I or b in I; a+I = 0+I or b+I = 0+I. Ok. >> >> How's that a contradiction? There are two Boolean >> rings without zero divisors, {0} and {0,1}. >> >> If B/I = {0}, then I = B, a no, no. >> If B/I = {0,1}, then some a not in I with B = I \/ a+I. >> >> If b not in I, then b in a+I; b+I = a+I; >> I, b+I subset (b,I); B = I \/ a+I = I \/ b+I = (b,I). >> >> In conclusion. If I is an ideal of a Boolean ring B, then >> I is prime ideal iff I is maximal ideal Right. >I is maximal ideal among ideals excluding a >iff I is proper prime ideal iff I is maximal ideal. Those iff's should be forward implications. quasi |