[過去ﾛｸﾞ] 純粋・応用数学（含むガロア理論）3 (1002ﾚｽ)

純粋・応用数学（含むガロア理論）3 http://rio2016.5ch.net/test/read.cgi/math/1595166668/

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737: 現代数学の系譜 雑談 ◆yH25M02vWFhP [sage] 2020/08/27(木) 11:49:40.74 ID:NVBIr97s >>736 追加 https://planetmath.org/ringswhoseeverymoduleisfree rings whose every module is free Author joking (16130) Last modified on 2013-03-22 この前半の証明も良いね〜 ”Recall that if R is a (nontrivial) ring and M is a R-module, then (nonempty) subset S⊆M is called linearly independent if for any m1,…,mn∈M and any r1,?,rn∈R the equality r1・m1+…+rn・mn=0 implies that r1=…=rn=0. If S⊆M is a linearly independent subset of generators of M, then S is called a basis of M. Of course not every module has a basis (it even doesn’t have to have linearly independent subsets). R-module is called free, if it has basis. In particular if R is a field, then it is well known that every R-module is free. What about the converse? Proposition. Let R be a unital ring. Then R is a division ring if and only if every left R-module is free. つづく http://rio2016.5ch.net/test/read.cgi/math/1595166668/737

738: 現代数学の系譜 雑談 ◆yH25M02vWFhP [sage] 2020/08/27(木) 11:50:05.99 ID:NVBIr97s >>737 つづき Proof. ,,⇒” First assume that R is a divison ring. Then obviously R has only two (left) ideals, namely 0 and R (because every nontrivial ideal contains invertible element and thus it contains 1, so it contains every element of R). Let M be a R-module and m∈M such that m≠0. Then we have homomorphism of R-modules f:R→M such that f(r)=r・m. Note that ker(f)≠R (because f(1)≠0) and thus ker(f)=0 (because ker(f) is a left ideal). It is clear that this implies that {m} is linearly independent subset of M. Now letΛ={P⊆M??P is linearly independent}. Therefore we proved that Λ≠Φ. Note that (Λ,⊆) is a poset (where ,,⊆” denotes the inclusion) in which every chain is bounded. Thus we may apply Zorn’s lemma. Let P0∈Λ be a maximal element in Λ. We will show that P0 is a basis (i.e. P0 generates M). Assume that m∈M is such that m not∈P0. Then P0∪{m} is linearly dependent (because P0 is maximal) and thus there exist m1,?,mn∈M and λ,λ1,?,λn∈R such that λ≠0 and λ・m+λ1・m1+?λn・mn=0. Since λ≠0, then λ is invertible in R (because R is a divison ring) and therefore m=(?λ?1λ1)・m1+?+(?λ?1λn)・mn. Thus P0 generates M, so every R-module is free. This completes this implication.” (引用終り) なるほどね この証明は、味わい深いですね〜 つづく http://rio2016.5ch.net/test/read.cgi/math/1595166668/738

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