Appendix C: Zorn's Lemma
1. Let K ⊆ M be modules, M finitely generated, say
If 
we must show that 
contains maximal members. Suppose {Xi 
i ∊ I} is a chain from 
and put U = ∪ i∊IXi . It is clear that U is a submodule and that K ⊆ U, and we claim that U ≠ M . For if U = M then each xi ∊ U, and so each xi ∊ Xk for some k . Since the Xi form a chain, this means that {x1, x2, . . ., xn} ⊆ Xm for some m . Since the xi generate M this means that M ⊆ Xm, contradicting the fact that 
This shows that 
and so U is an upper bound for the Xi . Hence 
has maximal members by Zorn's lemma, as required.
we must show that contains maximal members. Suppose {Xi i ∊ I} is a chain from and put U = ∪ i∊IXi . It is clear that U is a submodule and that K ⊆ U, and we claim that U ≠ M . For if U = M then each xi ∊ U, and so each xi ∊ Xk for some k . Since the Xi form a chain, this means that {x1, x2, . . ., xn} ⊆ Xm for some m . Since the xi generate M this means that M ⊆ Xm, contradicting the fact that This shows that and so U is an upper bound for the Xi . Hence has maximal members by Zorn's lemma, as required.
2. Let K ⊆ M be modules.
a. Let 
is a submodule and K ∩ X = 0} . Then 
is nonempty because 
so let {Xi i ∊ I} be a chain from 
and put U = ∪ i∊IXi . It is clear that U is a submodule, and K ∩ U = 0 because K ∩ U ⊆ K ∩ Xi = 0 for each i . Hence U is an upper bound for the chain {Xi i ∊ I}, so 
contains maximal members by Zorn's lemma.
is a submodule and K ∩ X = 0} . Then is nonempty because so let {Xi i ∊ I} be a chain from and put U = ∪ i∊IXi . It is clear that U is a submodule, and K ∩ U = 0 because K ∩ U ⊆ K ∩ Xi = 0 for each i . Hence U is an upper bound for the chain {Xi i ∊ I}, so contains maximal members by Zorn's lemma.
3. Let 
be the set of all prime ideals of R, and partially order 
downward: Let P ≤ Q mean P ⊇ Q . We must find a maximal element in 
Let {Pi i ∊ I} be a chain from 
and put Q = ∩ i∊IPi . We claim that Q is an upper bound on {Pi i ∊ I} . Clearly Q ⊆ Pi for each i, so it remains to show that Q is a prime ideal. If rs ∊ Q where r, s∊ R ; we must show that either r ∊ Q or s ∊ Q . Suppose on the contrary that r ∉ Q and s ∉ Q . Then r ∉ Pi for some i, and s ∉ Pj for some j . Since the Pi form a chain, one of Pi ⊆ Pj or Pj ⊆ Pi must hold; assume Pi ⊆ Pj . Then r ∉ Pj and s ∉ Pj but rs ∊ Pj (because rs ∊ Q ⊆ Pj) . This contradicts the fact that Pj is a prime ideal. Since we also obtain a contradiction if Pj ⊆ Pi, this proves that Q is a prime ideal.
be the set of all prime ideals of R, and partially order downward: Let P ≤ Q mean P ⊇ Q . We must find a maximal element in Let {Pi i ∊ I} be a chain from and put Q = ∩ i∊IPi . We claim that Q is an upper bound on {Pi i ∊ I} . Clearly Q ⊆ Pi for each i, so it remains to show that Q is a prime ideal. If rs ∊ Q where r, s∊ R ; we must show that either r ∊ Q or s ∊ Q . Suppose on the contrary that r ∉ Q and s ∉ Q . Then r ∉ Pi for some i, and s ∉ Pj for some j . Since the Pi form a chain, one of Pi ⊆ Pj or Pj ⊆ Pi must hold; assume Pi ⊆ Pj . Then r ∉ Pj and s ∉ Pj but rs ∊ Pj (because rs ∊ Q ⊆ Pj) . This contradicts the fact that Pj is a prime ideal. Since we also obtain a contradiction if Pj ⊆ Pi, this proves that Q is a prime ideal.
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