Boolean Boogie
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From: William Elliot on 10 Aug 2010 23:49 On Tue, 10 Aug 2010, quasi wrote: >>> 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? > > Ok, here's a counterexample (I think) ... > > Let A be the boolean ring with identity defined as the quotient > > A = Z_2[x,y] / (x^2-x, y^2-y) > > and let B = (x,y), the ideal of A generated by x,y. > > Then B is a Boolean ring without identity, and in the ring B, the zero > ideal is prime but not maximal. > 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. (If 1 in B, every element except 0 and 1, is a zero divisor.) Let a /= 0. Thus some x with ax /= x, a(ax + x) = 0; a(ax + x) in (0) = {0}; a = 0 or ax + x = 0, huh? That's a no, no. (0) isn't prime unless B = {0}. (0) is a prime ideal iff the ring has no zero divisors. Proposition. A prime ideal of a Boolean ring is maximal even if the ring is without identity. Conjecture. If R is a communative ring and for all a /= 0 (and /= 1, if 1 in R) is a zero divisor, is R a Boolean ring? ----
From: William Elliot on 11 Aug 2010 00:52 On Tue, 10 Aug 2010, quasi wrote: > On Tue, 10 Aug 2010 01:02:12 -0700, William Elliot > > Then the prime ideal P may or may not be maximal. My favorite Boolean ring without identity is B = { r in (Z_2)^I | support r finite }. It's equivalent to the Boolean algebra of finite subsets of I. In that ring, { r in B | r(k) = 0 } is a prime ideal and it's also maximal. Since every Boolean ring embeds into R = (Z_2)^I for some set I, does that mean prime ideals of Boolean rings are maximal? >>>>> 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. Does he get into commutative topological rings?
From: quasi on 11 Aug 2010 02:51 On Tue, 10 Aug 2010 20:49:23 -0700, William Elliot <marsh(a)rdrop.remove.com> wrote: >On Tue, 10 Aug 2010, quasi wrote: > >>>> 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? >> >> Ok, here's a counterexample (I think) ... >> >> Let A be the boolean ring with identity defined as the quotient >> >> A = Z_2[x,y] / (x^2-x, y^2-y) >> >> and let B = (x,y), the ideal of A generated by x,y. >> >> Then B is a Boolean ring without identity, and in the ring B, the zero >> ideal is prime but not maximal. Ok, I see my error. The ring B that I constructed has zero divisors, so the zero ideal is not a prime ideal. In fact, as you observe below, in a Boolean ring without identity, every nonzero element is a zero divisor. >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. >(If 1 in B, every element except 0 and 1, is a zero divisor.) > >Let a /= 0. Thus some x with ax /= x, a(ax + x) = 0; >a(ax + x) in (0) = {0}; a = 0 or ax + x = 0, huh? >That's a no, no. (0) isn't prime unless B = {0}. > >(0) is a prime ideal iff the ring has no zero divisors. > >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. >Conjecture. If R is a communative ring and for all >a /= 0 (and /= 1, if 1 in R) is a zero divisor, is >R a Boolean ring? Didn't you ask the same question several years ago, and shortly after that, didn't I post a proof that such a ring need not be Boolean? quasi
From: quasi on 11 Aug 2010 02:55 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. >Does he get into commutative topological rings? No. quasi
From: William Elliot on 11 Aug 2010 05:16
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 ---- |