純粋・応用数学（含むガロア理論）3

[過去ﾛｸﾞ] 純粋・応用数学（含むガロア理論）3 (1002ﾚｽ)
上下前次1-新
通常表示 512ﾊﾞｲﾄ分割ﾚｽ栞
抽出解除ﾚｽ栞

このｽﾚｯﾄﾞは過去ﾛｸﾞ倉庫に格納されています｡
次ｽﾚ検索歴削→次ｽﾚ栞削→次ｽﾚ過去ﾛｸﾞﾒﾆｭｰ

ﾘﾛｰﾄﾞ規制です｡10分ほどで解除するので､他のﾌﾞﾗｳｻﾞへ避難してください｡

898: 現代数学の系譜 雑談 ◆yH25M02vWFhP  [] 2020/08/29(土) 13:37:45.02 ID:T0GrcKp2

>>897
つづき

Hence, R is a division ring because we have already seen that a ring R is a division
ring if and only if the R-module R is a simple R-module. Thus, our problem of
showing that a nonzero ring R is a division ring if every R-module is free is solved
once we establish that every nonzero ring has a simple module. To this end it is
convenient to have the following.

P220
Definition
Let M be a nonzero R-module. A submodule M' of M is said to be a maximal
submodule of M if and only if M'±M and M' and M are the only submodules of
M containing M'.
The following characterization of maximal submodules of a module is an
almost immediate consequence of the definition.
Basic Property 8.6
A submodule M' of the R-module M is a maximal submodule of M if and only if
M/M' is a simple R-module.
PROOF: This is a direct consequence of the isomorphism established by the
canonical surjective morphism kM,M-:M-≫MIM' between the set of submodules
of M containing M' and the set of submodules of MIM'.
Hence, in order to show that a nonzero ring R has simple modules, it suffices
to show that the R- module R has a maximal submodule M because in that case
R/M is a simple R-module.
Proposition 8.7
Let R be a nonzero ring. Then every submodule M' of R, different from R, is
contained in a maximal submodule of R. Consequently, the ring R has at least one
maximal submodule M which means that R also has the simple R-module R/M.
PROOF:

つづく

http://rio2016.5ch.net/test/read.cgi/math/1595166668/898

899: 現代数学の系譜 雑談 ◆yH25M02vWFhP  [] 2020/08/29(土) 13:38:07.43 ID:T0GrcKp2

>>898
つづき

P221
Because F is an inductive set, it must have a maximal element M by Zorn's
lemma. We leave it to the reader to verify that M is a maximal submodule of R.
Because M obviously contains M', the first part of the proposition is proven.
In the light of this result, to see that R contains at least one maximal submodule,
all we have to do is find some submodule M' of R different from R.
Because R is not the zero ring, the zero submodule of R will do.
The rest of the proposition now follows trivially from our previous characterization of maximal submodules.

In the light of this discussion, we have also established the following.
Theorem 8.8
For a nonzero ring R, the following statements are equivalent:
(a) R is a division ring.
(b) Every R-module is a free R-module.
(c) Every nonzero R-module generated by a single element is a free R-module.
(引用終り)
以上

http://rio2016.5ch.net/test/read.cgi/math/1595166668/899

上下前次1-新書関写板覧索設栞歴

ｽﾚ情報赤ﾚｽ抽出画像ﾚｽ抽出歴の未読ｽﾚ AAｻﾑﾈｲﾙ

ぬこの手ぬこTOP 0.033s