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)
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897: 現代数学の系譜 雑談 ◆yH25M02vWFhP  [] 2020/08/29(土) 13:37:06.72 ID:T0GrcKp2

>>896
つづき

Basic Properties 8.4
Let R be an arbitrary ring and M a nonzero R-module. The following conditions
are equivalent:
(a) M is generated by each nonzero element in M.
(b) For every R-module X, every morphism f:X→M is either zero or an
epimorphism.
(c) For every R-module X, every morphism f:M→X is either zero or a
monomorphism.
As an immediate consequence of these basic properties, we have the
following.

Corollary 8.5
Let M be a simple R-module. Then every endomorphism of M is either zero or an
automorphism. Hence, EndR( M ), the ring of endomorphisms of M, is a division
ring.
The main point to establish about simple modules in connection with our
problem of showing that a nonzero ring R is a division ring if every R-module is
free is that every nonzero ring R has at least one simple R-module.
Suppose we know that our nonzero ring R, which has the property that every
R-module is free, also has a simple R-module M. Then the simple R-module M
must have a basis B since M is a free R-module. Because M ≠ (0), we know that
B is not empty. We now show that B consists of exactly one element. Let b be an
element of B. Then by one of our characterizations of simple modules (Basic
Property 8.4), we know that the element b generates M since b ± 0. By Basic
Property 7.6, it follows that {b} = B. Hence, B consists of a single element.
But we have already shown that a free module over a ring R has a basis
consisting of one element if and only if it is isomorphic to R. Hence, the simple
R-module M is isomorphic to R which means that R is a simple R-module.

つづく

http://rio2016.5ch.net/test/read.cgi/math/1595166668/897

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

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