Ideals
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From: William Elliot on 6 Aug 2010 04:55 On Thu, 5 Aug 2010, Arturo Magidin wrote: >> >> 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 }, > > 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. For example. Define r_n:N -> Z_2 with r_n(k) = 1 if 1 <= k <= n, = 0 otherwise. C = { (r_n) | n in N\1 } is a chain of ideals of Rf. \/C = 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? It was a set theory exercise. >>> 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 mutates mutandi for right ideals. > No... Really? Yes. Recall that this is in addition to conclusion above from the premise that R has an identity. In fact, a left or right identity should suffice depending upon the handedness of the ideal. Are definitions regarding left and right ideals and left and right identities the same throughout all texts? Are these correct or crossed? A subring I, of a ring R is a left ideal when RI subset I a right ideal when IR subset I. An idempotent d of a ring R is a left identity when for all x, dx = x and a right identity when for all x, xd = x, ----
From: William Elliot on 6 Aug 2010 04:57 On Thu, 5 Aug 2010, Arturo Magidin 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. It's correct in the context of rings with identities. > 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 "multiple left" or "multiple right" identities of > the ring (G,+,*). { }
From: William Elliot on 6 Aug 2010 05:00 On Thu, 5 Aug 2010, Butch Malahide wrote: > 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): > Buzzards. What about left and right identities? How are they defined? > [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] >
From: José Carlos Santos on 6 Aug 2010 06:44 On 06-08-2010 9:57, William Elliot 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. > > It's correct in the context of rings with identities. Trivial. *Any* assertion which begins with "If there is no identity" is trivially true in the context of rings with identities. Best regards, Jose Carlos Santos
From: Achava Nakhash, the Loving Snake on 6 Aug 2010 13:27 On Aug 6, 1:57 am, William Elliot <ma...(a)rdrop.remove.com> wrote: > On Thu, 5 Aug 2010, Arturo Magidin 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. l, > > written additively, and define a multiplication * on G by a*b = 0 for > > all a and b in G. > > > Kindly exhibit the "multiple left" or "multiple right" identities of > > the ring (G,+,*). > Intesrestingly, while your assertion was false, it is true that a ring with more than one left identity has infinitely many of them. This is a problem in Jacoboson's 3-part tome on algebra, and I did this a very long time ago. It was a fun problem, and I recommend it to everyone as an interesting problem to work on. Regards, Achava
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