Ideals
in [Math]
|
From: Butch Malahide on 5 Aug 2010 04:08 On Aug 5, 2:34 am, William Elliot <ma...(a)rdrop.remove.com> wrote: > > Thus, as commutativity hasn't been used, if R is a ring with unity and I > an ideal within R, then there's a maximal ideal J with I subset J. > > In addition, if I is a left ideal within R, then there's a maximal > left ideal J with I subset J and mutatus mutandi for right ideals. Offhand I'd guess that, if you want a maximal one-sided ideal, you can get by with just a one-sided multiplicative identity, instead of a 1. Um, right identities for left ideals and vice versa, if "left ideals" and "right identities" are defined the way I think they are.
From: William Elliot on 5 Aug 2010 05:40 On Thu, 5 Aug 2010, Butch Malahide wrote: > On Aug 5, 2:34�am, William Elliot <ma...(a)rdrop.remove.com> wrote: >> >> Thus, as commutativity hasn't been used, if R is a ring with unity and I >> an ideal within R, then there's a maximal ideal J with I subset J. >> >> In addition, if I is a left ideal within R, then there's a maximal >> left ideal J with I subset J and mutates mutandi for right ideals. > > Offhand I'd guess that, if you want a maximal one-sided ideal, you can > get by with just a one-sided multiplicative identity, instead of a 1. > Um, right identities for left ideals and vice versa, if "left ideals" > and "right identities" are defined the way I think they are. > If there is no identity, then there are either multiple left or multiple right identities. Without any further sleepy thought, I'll agree and perhaps sleep on it. I is a left ideal within a ring R when -I, I + I, IR subset I. How do you define left ideal? Since a ring with some right identities has no left identities Can there be a ring with a maximal left ideal and no maximal right ideal? Oh BTW, is 1x = x or x1 = x a left identity? Yawn and good night before I convince myself I can't tell the difference between left and right. -- Is the wizard who can turn left into right and back again into wrong a republican nyetnik? ----
From: Arturo Magidin on 5 Aug 2010 12:36 On Aug 5, 2:34 am, William Elliot <ma...(a)rdrop.remove.com> wrote: > On Wed, 4 Aug 2010, Arturo Magidin wrote: > > On Aug 4, 4:45 am, William Elliot <ma... > > Let R be a commutative ring with unity. > > > > >> How does one show that R has maximal ideals or even better, that > >> if I is an ideal, then there's a maximal ideal J with I subset J? > > > If I is a proper ideal (including (0)), let P be the set of all proper > > ideals of R that contain I, partially ordered by inclusion. We show > > that P satisfies the hypotheses of Zorn's Lemma. > > > Let C be a chain in in P. If C is empty, then I is an upper bound for > > C in P. If C is not empty, then consider J_0=\/C = \/_{J in C} J. > > > J_0 is nonempty, since C is nonempty. It contains I, since each J in C > > contains I and C is nonempty. It's contained in R because each J in C > > is contained in R. If a,b in J_0, then there exist J_a, J_b in C such > > that a in J_a and b in J_b. Since C is a chain, either J_a \subset J_b > > or J_b \subset J_a. Either way, there exists J in C such that a,b in > > J. Then a-b in J, so a-b in J_0. Thus, J_0 is a subgroup of R. If a in > > J_0 and r in R, then there exists J in C such that a in J, so ra in J, > > hence ra in J_0. Thus, J_0 is an ideal of R. > > Nowhere have you used 1 in R nor commutativity. One does not need commutativity (as I had noted elsewhere). The same argument shows that every left (resp. right, two-sided) proper ideal is contained in a left (resp. right, two-sided) ideal in a ring with unity. As for using "1", I use it below to show that J_0 is a *PROPER* ideal of R. Otherwise, you have no warrant for asserting that J_0 lies in P, which is needed for Zorn's Lemma. > Let Rf = { r in (Z_2)^N | support r is finite }. I.e., the direct sum of countably many copies of Z/2Z. Okay. This is a not a ring with unity. > Define r_n:N -> Z_2 with r_n(k) = 1 if 2 <= k <= n, = 0 otherwise.. > C = { (r_n) | n in N\1 } is a chain of ideals of Rf. > \/C = { r in Rf | r(1) = 0 } Correct. > Ouch. I was thinking > \/C = { r in (Z_2)^N | r(1) = 0 }, You mean you *weren't* thinking. How can a union of subsets of Rf result in a set that is not contained in Rf? > when actually > \/C = { r in (Z_2)^N | r(1) = 0 } /\ Rf. > > This is an example of an ideal that's not a principal ideal > for it's not finitely generated. > > Thus my unchallenged claim that Rf has no maximal > ideal is dismissed as hallucinated illusion, for > the usual Zorn's lemma proof holds. No; the Zorn's Lemma proof does not work for Rf, because you do not know whether the union of C will be a *PROPER* ideal. The fact that Rf has maximal ideals is established by direct construction: you can map onto a ring with maximal ideals, and then pull back (in your case, you are projecting onto the first coordinate, which is Z/2Z, and pulling back the maximal ideal (0) of Z/2Z). The Lattice Isomorphism Theorem gives you that the pullback must be a maximal ideal of Rf. > Now however, if instead of Rf, the ring is Ra = (Z_2)^N, > then still, doesn't > \/C = { r in (Z_2)^N | r(1) = 0 } /\ Rf? Yes, but this is no longer a maximal ideal of Ra. So what? > > Finally, we show that J_0 =/= R: since each J in C is proper, 1 is not > > in J for all J in C. Therefore, 1 is not in \/_{J in C} J = J_0. Thus, > > J_0 =/= R. > > Thus, as commutativity hasn't been used, if R is a ring with unity and I > an ideal within R, then there's a maximal ideal J with I subset J. More generally, if R is a ring (with or without 1), commutative or not, and a is a nonzero element, then there always exist ideals of R that are maximal with respect to the property of not containing the element a. When a=1, since "I is an ideal that does not contain 1" is equivalent to "I is a proper ideal", you get the existence of maximal ideals. > In addition, if I is a left ideal within R, then there's a maximal > left ideal J with I subset J and mutatus mutandi for right ideals. No... Really? -- Arturo Magidin
From: Arturo Magidin on 5 Aug 2010 12:38 On Aug 5, 4:40 am, William Elliot <ma...(a)rdrop.remove.com> wrote: > On Thu, 5 Aug 2010, Butch Malahide wrote: > > On Aug 5, 2:34 am, William Elliot <ma...(a)rdrop.remove.com> wrote: > > >> Thus, as commutativity hasn't been used, if R is a ring with unity and I > >> an ideal within R, then there's a maximal ideal J with I subset J. > > >> In addition, if I is a left ideal within R, then there's a maximal > >> left ideal J with I subset J and mutates mutandi for right ideals. > > > Offhand I'd guess that, if you want a maximal one-sided ideal, you can > > get by with just a one-sided multiplicative identity, instead of a 1. > > Um, right identities for left ideals and vice versa, if "left ideals" > > and "right identities" are defined the way I think they are. > > If there is no identity, then there are either multiple left or > multiple right identities. False. Pick your favorite abelian group G with more than two elements, written additively, and define a multiplication * on G by a*b = 0 for all a and b in G. Kindly exhibit the "mulitple left" or "multiple right" identities of the ring (G,+,*). -- ArturO Magidin
From: Butch Malahide on 5 Aug 2010 18:58 On Aug 5, 4:40 am, William Elliot <ma...(a)rdrop.remove.com> wrote: > > I is a left ideal within a ring R when -I, I + I, IR subset I. No, that's a *right* ideal. I quote from p. 49 (beginning of section 16 if you have a different edition) of B. L. van der Waerden, Modern Algebra, Volume 1, revised English edition (in the book o and m are small German letters): [BEGIN QUOTE] Let o be a ring. [. . .] A non-empty subset m of o is called an ideal, or better a right ideal, if 1. a in m and b in m imply a - b in m (module property), 2. a in m implies ar in m for an arbitrary r in o. In words: the module m shall contain all "right multiples" a.r for every a. Similarly, a module is called a left ideal if a in m implies ra in m for an arbitrary r in o. [END QUOTE]
First
|
Prev
|
Next
|
Last
Pages: 1 2 3 4 Prev: ===Christian Louboutin - www.vipchristianlouboutin.com Next: calculation of a norm |