Ideals
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From: William Elliot on 4 Aug 2010 05:45 Back to basics. Let R be a communative 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?
From: G. A. Edgar on 4 Aug 2010 07:22 In article <20100804024142.B7612(a)agora.rdrop.com>, William Elliot <marsh(a)rdrop.remove.com> wrote: > Back to basics. Let R be a communative 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? Zorn's Lemma -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/
From: Gerry on 4 Aug 2010 09:02 On Aug 4, 9:22 pm, "G. A. Edgar" <ed...(a)math.ohio-state.edu.invalid> wrote: > In article <20100804024142.B7...(a)agora.rdrop.com>, William Elliot > > <ma...(a)rdrop.remove.com> wrote: > > Back to basics. Let R be a communative 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? > > Zorn's Lemma And to show that R bears maximal fruit, use Zorn's Lemon. -- GM
From: Arturo Magidin on 4 Aug 2010 12:14 On Aug 4, 4:45 am, William Elliot <ma...(a)rdrop.remove.com> wrote: > Back to basics. Let R be a communative ring with unity. If you don't mind, I'll assume R to be a commutative ring with unity instead. > 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. 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. Hence, J_0 is in P, and is an upper bound for C in P. Thus, P satisfies the hypotheses of Zorn's Lemma. By Zorn's Lemma, P has maximal elements. Let M be a maximal element of P. Then M is a proper ideal of R that contains I. If N is an ideal of R such that M<=N<=R, then either N=R, or N=/=R; in the latter case, N is in P, and since M<=N and M is maximal, we conclude that M=N. Thus, M is a maximal ideal of R. QED -- Arturo Magidin
From: William Elliot on 5 Aug 2010 03:34 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. Let Rf = { r in (Z_2)^N | support r is finite }. 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 } Ouch. I was thinking \/C = { r in (Z_2)^N | r(1) = 0 }, 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. 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? > 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. 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. > Hence, J_0 is in P, and is an upper bound for C in P. Thus, P > satisfies the hypotheses of Zorn's Lemma. > > By Zorn's Lemma, P has maximal elements. Let M be a maximal element of > P. Then M is a proper ideal of R that contains I. If N is an ideal of > R such that M<=N<=R, then either N=R, or N=/=R; in the latter case, N > is in P, and since M<=N and M is maximal, we conclude that M=N. Thus, > M is a maximal ideal of R. QED
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