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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894: 現代数学の系譜 雑談 ◆yH25M02vWFhP  [] 2020/08/29(土) 13:29:38.29 ID:T0GrcKp2

>>893
つづき

Basic Properties 8.1
Let M be an R-module.

(b) Every linearly independent subset S of M is contained in a maximal linearly
independent subset of M.
(c) M has a maximal linearly independent subset.
(d) If M is a free R-module, then every basis for M is a maximal linearly indepen
dent subset of M.
PROOF: 略

P217
Proposition 8.2
Let D be a division ring. Then the following statements are equivalent for a subset
B of a D-module M:
(a) B is a basis for M.
(b) B is a maximal linearly independent subset of M. Since every module has a
maximal linearly independent subset, every module over a division ring D has
a basis and is therefore a free D-module.
PROOF: Because the reader has already shown that every basis of a module is
a maximal linearly independent subset of the module, we only have to show that
(b) implies (a).
略
P218
This finishes the proof that a maximal linearly independent subset B of a
D-module M is a basis for M because it generates M.

Having established that all modules over division rings are free, we will have
a complete description of all nonzero rings R with the property that all R -modules
are free if we show that any nonzero ring with this property must be a division
ring. Because we are trying to describe when a ring is a division ring in terms of its
module theory, it is reasonable to expect that a module-theoretic description of
when a ring is a division ring would be helpful. We do this now in terms of the
properties of the R-module R.

つづく

http://rio2016.5ch.net/test/read.cgi/math/1595166668/894

895: 現代数学の系譜 雑談 ◆yH25M02vWFhP  [] 2020/08/29(土) 13:30:50.73 ID:T0GrcKp2

>>894
つづき

Suppose R is a division ring. We claim that the R -module R has the following
properties: (a) R ± (0) and (b) (0) and R are the only submodules of R. By
definition, a division ring R is not zero so (a) is trivially satisfied. Suppose now
that M is a nonzero submodule of a division ring R. Then there is a nonzero x in
M. Because R is a division ring there is a y in R such that yx = 1. Because yx is in
M, it follows that 1 is in M and so r=r\ is in M for all r in R, which means that
M = R. So we see that a division ring R also satisfies (b), that is, it has the
property that (0) and R are the only submodules of R.
On the other hand, it is not difficult to see that a nonzero ring R which has the
property that (0) and R are its only submodules, is a division ring. To show this we
first show that if x is a nonzero element of R and yx = 0, then y = 0. The set M of
all y in R such that yx = 0 is a submodule of R, because it is the kernel of the
morphismof R -module R →R given by rl→ncfora!l rin R. Now M±R because 1
is not in R (remember R is not the zero ring). Therefore, M = (0) because (0) and R
are the only submodules of R. Hence, if yx = 0, then y = 0 because it is in M and
M = (0).

Next we observe that if x is a nonzero element of R, then there is a y in R
such that yx = 1. For the subset Rx is a submodule of R which is not the zero
submodule of R because it contains the nonzero element x. Hence, Rx = R be
cause (0) and R are the only submodules of R and Rx±0. This means that there is
a y in R such that yx = 1.

つづく

http://rio2016.5ch.net/test/read.cgi/math/1595166668/895

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