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From: quasi on 9 Aug 2010 13:12 On Mon, 09 Aug 2010 12:03:53 -0500, quasi <quasi(a)null.set> wrote: >On Mon, 09 Aug 2010 12:01:02 -0500, quasi <quasi(a)null.set> wrote: > >>On Mon, 9 Aug 2010 03:56:56 -0700, William Elliot >><marsh(a)rdrop.remove.com> wrote: >> >>>Let B be a Boolean ring and a, a point in B. >>>Is there a prime ideal that exclude a? >> >>Yes, provided the element a is nonzero. >> >>By the Boolean property, the set {a} is multiplicatively closed. Also, >>since a is nonzero, the ideal 0 is disjoint from S. > >Correction: disjoint from {a} > >>By Zorn's lemma, >>we can extend the ideal 0 to an ideal which is maximal (with respect >>to inclusion) in the set ideals not containing a. Such an ideal is >>necessarily prime. In fact, such an deal, P say, is actually maximal in B. Since P is prime , a^2 = a and a not in P implies 1-a in P, but then any ideal J strictly containing P must contain a (by the maximality condition on P). It follows that J = B, hence P is a maximal ideal of B. >< A clear discussion of these ideas can be found in >>the first few pages of Kaplansky's wonderful text "Commutative Rings". quasi
From: quasi on 9 Aug 2010 13:35 On Mon, 09 Aug 2010 12:12:31 -0500, quasi <quasi(a)null.set> wrote: >On Mon, 09 Aug 2010 12:03:53 -0500, quasi <quasi(a)null.set> wrote: > >>On Mon, 09 Aug 2010 12:01:02 -0500, quasi <quasi(a)null.set> wrote: >> >>>On Mon, 9 Aug 2010 03:56:56 -0700, William Elliot >>><marsh(a)rdrop.remove.com> wrote: >>> >>>>Let B be a Boolean ring and a, a point in B. >>>>Is there a prime ideal that exclude a? >>> >>>Yes, provided the element a is nonzero. >>> >>>By the Boolean property, the set {a} is multiplicatively closed. Also, >>>since a is nonzero, the ideal 0 is disjoint from S. >> >>Correction: disjoint from {a} >> >>>By Zorn's lemma, >>>we can extend the ideal 0 to an ideal which is maximal (with respect >>>to inclusion) in the set ideals not containing a. Such an ideal is Should be: "in the set *of* ideals" >>>necessarily prime. > >In fact, such an deal, P say, is actually maximal in B. Since P is Such a deal -- supersize P to maximal -- today only! Of course that should have been "In fact, such an ideal" >prime , a^2 = a and a not in P implies 1-a in P, but then any ideal J >strictly containing P must contain a (by the maximality condition on >P). It follows that J = B, hence P is a maximal ideal of B. > >>< A clear discussion of these ideas can be found in >>>the first few pages of Kaplansky's wonderful text "Commutative Rings". Hopefully all typos corrected. quasi
From: William Elliot on 10 Aug 2010 02:41 On Mon, 9 Aug 2010, David C. Ullrich wrote: > <marsh(a)rdrop.remove.com> wrote: > >> Let B be a Boolean ring and a, a point in B. >> Is there a prime ideal that exclude a? > > Clearly not if a = 0. Yes if a <> 0: Also not if a = 1, which may or may not be in B. What if 1 not in B? > Let b = 1 - a. We need only show that b is not invertible; > then b is contained in a maximal (proper) ideal, which > as Neil points out cannot also contain a. > > But suppose that bc = 1. Then > > bc^2 = c > bc = c > c = 1 1 = bc = bcc = 1c = c > b = 1. 1 = bc = b1 = b >> Please don't use the Boolean ring representation theorem, >> as the question is a key step in a proof of the theorem. > >
From: William Elliot on 10 Aug 2010 04:02 On Mon, 9 Aug 2010, quasi wrote: >>> <marsh(a)rdrop.remove.com> wrote: >>> >>>> Let B be a Boolean ring and a, a point in B. >>>> Is there a prime ideal that exclude a? >>> >>> Yes, provided the element a is nonzero. Nor the identity. >>> By the Boolean property, the set {a} is multiplicatively closed. >>> Also, since a is nonzero, the ideal 0 is disjoint from {a}. >> >>> By Zorn's lemma, we can extend the ideal 0 to an ideal which is >>> maximal (with respect to inclusion) in the set ideals not containing >>> a. Such an ideal is necessarily prime. Why? Is this correct? If rs in I, the ideal, r not in I, s not in I, then a in (r,I) = Br + Zr + I and a in (s,I) = Bs + Zs + I; some b,c in B, i,j in I, n,m in {0,1} with a = br + nr + i = cs + ms + j Thus a = aa = bcrs + ncrs + csi + mbrs + nmrs + msi + brj + nrj + ij in I, which is a no-no. > In fact, such an deal, P say, is actually maximal in B. Since P is > prime , a^2 = a and a not in P implies 1-a in P, but then any ideal J > strictly containing P must contain a (by the maximality condition on > P). It follows that J = B, hence P is a maximal ideal of B. If, which the premise did not affirm, 1 in B. What if 1 not in B? >>> 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. ----
From: William Elliot on 10 Aug 2010 05:48 On Tue, 10 Aug 2010, William Elliot wrote: >>>> >>>>> Let B be a Boolean ring and a, a point in B. >>>>> Is there a prime ideal that exclude a? >>>> >>>> Yes, provided the element a is nonzero. > > Nor the identity. > >>>> By the Boolean property, the set {a} is multiplicatively closed. >>>> Also, since a is nonzero, the ideal 0 is disjoint from {a}. >>> >>>> By Zorn's lemma, we can extend the ideal 0 to an ideal which is maximal >>>> (with respect to inclusion) in the set ideals not containing a. Such an >>>> ideal is necessarily prime. > > Why? Is this correct? > If rs in I, the ideal, r not in I, s not in I, then > a in (r,I) = Br + Zr + I and a in (s,I) = Bs + Zs + I; > some b,c in B, i,j in I, n,m in {0,1} > with a = br + nr + i = cs + ms + j > > Thus a = aa > = bcrs + ncrs + csi + mbrs + nmrs + msi + brj + nrj + ij in I, > which is a no-no. > >> In fact, such an deal, P say, is actually maximal in B. Since P is >> prime , a^2 = a and a not in P implies 1-a in P, but then any ideal J >> strictly containing P must contain a (by the maximality condition on >> P). It follows that J = B, hence P is a maximal ideal of B. > > If, which the premise did not affirm, 1 in B. What if 1 not in B? > A more general proposition can be similarly proved. If B is a Boolean ring with identity and I a proper prime ideal, then I is maximal ideal. If J ideal, I subset J and a in J\I, then a not in I; a(a + 1) = 0 in I; a + 1 in I; a + 1 in J; 1 = a + a + 1 in J; J = B. Can the same be said of Boolean rings without identity?
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