Every $R$-module $M$ has a maximal linearly independent subset $A$. If submodule $M_A$ is generated by $A$,...
$begingroup$
I am stuck at proving
Every $R$-module $M$ has a maximal linearly independent subset $A$. If the submodule $M_A$ is generated by $A$, then $M_A$ is free on $A$, and for a non-zero submodule $M_1subseteq M$, $M_1 cap M_A neq 0$.
I know for the first part I should just apply Zorn's lemma, but I am not clear about the rest of the problem. Could anyone help me with that?
abstract-algebra modules free-modules
edited Dec 26 '18 at 21:39
user26857
39.4k124183
asked Nov 16 '18 at 5:46
Eric CurtisEric Curtis
214
$endgroup$
add a comment |
$begingroup$
I am stuck at proving
Every $R$-module $M$ has a maximal linearly independent subset $A$. If the submodule $M_A$ is generated by $A$, then $M_A$ is free on $A$, and for a non-zero submodule $M_1subseteq M$, $M_1 cap M_A neq 0$.
I know for the first part I should just apply Zorn's lemma, but I am not clear about the rest of the problem. Could anyone help me with that?
abstract-algebra modules free-modules
edited Dec 26 '18 at 21:39
user26857
39.4k124183
asked Nov 16 '18 at 5:46
Eric CurtisEric Curtis
214
$endgroup$
add a comment |
$begingroup$
I am stuck at proving
Every $R$-module $M$ has a maximal linearly independent subset $A$. If the submodule $M_A$ is generated by $A$, then $M_A$ is free on $A$, and for a non-zero submodule $M_1subseteq M$, $M_1 cap M_A neq 0$.
I know for the first part I should just apply Zorn's lemma, but I am not clear about the rest of the problem. Could anyone help me with that?
abstract-algebra modules free-modules
edited Dec 26 '18 at 21:39
user26857
39.4k124183
asked Nov 16 '18 at 5:46
Eric CurtisEric Curtis
214
$endgroup$
I am stuck at proving
Every $R$-module $M$ has a maximal linearly independent subset $A$. If the submodule $M_A$ is generated by $A$, then $M_A$ is free on $A$, and for a non-zero submodule $M_1subseteq M$, $M_1 cap M_A neq 0$.
I know for the first part I should just apply Zorn's lemma, but I am not clear about the rest of the problem. Could anyone help me with that?
abstract-algebra modules free-modules
abstract-algebra modules free-modules
edited Dec 26 '18 at 21:39
user26857
39.4k124183
asked Nov 16 '18 at 5:46
Eric CurtisEric Curtis
214
edited Dec 26 '18 at 21:39
user26857
39.4k124183
asked Nov 16 '18 at 5:46
Eric CurtisEric Curtis
214
edited Dec 26 '18 at 21:39
user26857
39.4k124183
edited Dec 26 '18 at 21:39
user26857
39.4k124183
edited Dec 26 '18 at 21:39
user26857
39.4k124183
39.4k124183
asked Nov 16 '18 at 5:46
Eric CurtisEric Curtis
214
asked Nov 16 '18 at 5:46
Eric CurtisEric Curtis
214
asked Nov 16 '18 at 5:46
Eric CurtisEric Curtis
214
214
add a comment |
add a comment |
1 Answer
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$begingroup$
As you suggested, the existence of a maximal linearly independent set is a fairly standard use of Zorn's lemma.
To show that $M_A$ is free, you need an isomorphism $M_Ato bigoplus_{lambdainLambda} Re_lambda$ for some index set $Lambda$ and where the $e_lambda$ are formal symbols. The natural choice is to let $Lambda=A$ and send $amapsto e_a$. Extend this by $R$-linearity to show it is a homomorphism; surjectivity is easy, and injectivity follows from linear independence.
The last statement isn't true, and I don't see an easy fix. As a counterexample, let $R=Bbb Z$ and $M=Bbb Z_2$. Then $M$ does not have any nonempty linearly independent subsets, so $M_A=0$. This is not just a problem with the empty set or the zero module; we can just as easily consider $M=Bbb Zoplus Bbb Z_2$, where a qualitatively similar issue arises with $A={(1,0)}$.
answered Dec 25 '18 at 4:31
aleph_twoaleph_two
24912
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1 Answer
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$begingroup$
As you suggested, the existence of a maximal linearly independent set is a fairly standard use of Zorn's lemma.
To show that $M_A$ is free, you need an isomorphism $M_Ato bigoplus_{lambdainLambda} Re_lambda$ for some index set $Lambda$ and where the $e_lambda$ are formal symbols. The natural choice is to let $Lambda=A$ and send $amapsto e_a$. Extend this by $R$-linearity to show it is a homomorphism; surjectivity is easy, and injectivity follows from linear independence.
The last statement isn't true, and I don't see an easy fix. As a counterexample, let $R=Bbb Z$ and $M=Bbb Z_2$. Then $M$ does not have any nonempty linearly independent subsets, so $M_A=0$. This is not just a problem with the empty set or the zero module; we can just as easily consider $M=Bbb Zoplus Bbb Z_2$, where a qualitatively similar issue arises with $A={(1,0)}$.
answered Dec 25 '18 at 4:31
aleph_twoaleph_two
24912
$endgroup$
add a comment |
$begingroup$
As you suggested, the existence of a maximal linearly independent set is a fairly standard use of Zorn's lemma.
To show that $M_A$ is free, you need an isomorphism $M_Ato bigoplus_{lambdainLambda} Re_lambda$ for some index set $Lambda$ and where the $e_lambda$ are formal symbols. The natural choice is to let $Lambda=A$ and send $amapsto e_a$. Extend this by $R$-linearity to show it is a homomorphism; surjectivity is easy, and injectivity follows from linear independence.
The last statement isn't true, and I don't see an easy fix. As a counterexample, let $R=Bbb Z$ and $M=Bbb Z_2$. Then $M$ does not have any nonempty linearly independent subsets, so $M_A=0$. This is not just a problem with the empty set or the zero module; we can just as easily consider $M=Bbb Zoplus Bbb Z_2$, where a qualitatively similar issue arises with $A={(1,0)}$.
answered Dec 25 '18 at 4:31
aleph_twoaleph_two
24912
$endgroup$
add a comment |
$begingroup$
As you suggested, the existence of a maximal linearly independent set is a fairly standard use of Zorn's lemma.
To show that $M_A$ is free, you need an isomorphism $M_Ato bigoplus_{lambdainLambda} Re_lambda$ for some index set $Lambda$ and where the $e_lambda$ are formal symbols. The natural choice is to let $Lambda=A$ and send $amapsto e_a$. Extend this by $R$-linearity to show it is a homomorphism; surjectivity is easy, and injectivity follows from linear independence.
The last statement isn't true, and I don't see an easy fix. As a counterexample, let $R=Bbb Z$ and $M=Bbb Z_2$. Then $M$ does not have any nonempty linearly independent subsets, so $M_A=0$. This is not just a problem with the empty set or the zero module; we can just as easily consider $M=Bbb Zoplus Bbb Z_2$, where a qualitatively similar issue arises with $A={(1,0)}$.
answered Dec 25 '18 at 4:31
aleph_twoaleph_two
24912
$endgroup$
As you suggested, the existence of a maximal linearly independent set is a fairly standard use of Zorn's lemma.
To show that $M_A$ is free, you need an isomorphism $M_Ato bigoplus_{lambdainLambda} Re_lambda$ for some index set $Lambda$ and where the $e_lambda$ are formal symbols. The natural choice is to let $Lambda=A$ and send $amapsto e_a$. Extend this by $R$-linearity to show it is a homomorphism; surjectivity is easy, and injectivity follows from linear independence.
The last statement isn't true, and I don't see an easy fix. As a counterexample, let $R=Bbb Z$ and $M=Bbb Z_2$. Then $M$ does not have any nonempty linearly independent subsets, so $M_A=0$. This is not just a problem with the empty set or the zero module; we can just as easily consider $M=Bbb Zoplus Bbb Z_2$, where a qualitatively similar issue arises with $A={(1,0)}$.
answered Dec 25 '18 at 4:31
aleph_twoaleph_two
24912
answered Dec 25 '18 at 4:31
aleph_twoaleph_two
24912
answered Dec 25 '18 at 4:31
aleph_twoaleph_two
24912
answered Dec 25 '18 at 4:31
aleph_twoaleph_two
24912
24912
add a comment |
add a comment |
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