Submodule of free module over a p.i.d. is free even when the module is not finitely generated?
$begingroup$
I have heard that any submodule of a free module over a p.i.d. is free.
I can prove this for finitely generated modules over a p.i.d. But the proof involves induction on the number of generators, so it does not apply to modules that are not finitely generated.
Does the result still hold? What's the argument?
abstract-algebra ring-theory modules principal-ideal-domains
edited Jun 18 '14 at 21:49
user26857
39.3k124183
asked Jun 25 '12 at 18:21
Ben Blum-SmithBen Blum-Smith
10.2k23086
$endgroup$
add a comment |
$begingroup$
I have heard that any submodule of a free module over a p.i.d. is free.
I can prove this for finitely generated modules over a p.i.d. But the proof involves induction on the number of generators, so it does not apply to modules that are not finitely generated.
Does the result still hold? What's the argument?
abstract-algebra ring-theory modules principal-ideal-domains
edited Jun 18 '14 at 21:49
user26857
39.3k124183
asked Jun 25 '12 at 18:21
Ben Blum-SmithBen Blum-Smith
10.2k23086
$endgroup$
2
$begingroup$
Rotman has it in Advanced Modern Algebra, page 651!
$endgroup$
– Dedalus
Jun 25 '12 at 18:29
$begingroup$
(The result still holds)
$endgroup$
– Dedalus
Jun 25 '12 at 18:29
add a comment |
$begingroup$
I have heard that any submodule of a free module over a p.i.d. is free.
I can prove this for finitely generated modules over a p.i.d. But the proof involves induction on the number of generators, so it does not apply to modules that are not finitely generated.
Does the result still hold? What's the argument?
abstract-algebra ring-theory modules principal-ideal-domains
edited Jun 18 '14 at 21:49
user26857
39.3k124183
asked Jun 25 '12 at 18:21
Ben Blum-SmithBen Blum-Smith
10.2k23086
$endgroup$
I have heard that any submodule of a free module over a p.i.d. is free.
I can prove this for finitely generated modules over a p.i.d. But the proof involves induction on the number of generators, so it does not apply to modules that are not finitely generated.
Does the result still hold? What's the argument?
abstract-algebra ring-theory modules principal-ideal-domains
abstract-algebra ring-theory modules principal-ideal-domains
edited Jun 18 '14 at 21:49
user26857
39.3k124183
asked Jun 25 '12 at 18:21
Ben Blum-SmithBen Blum-Smith
10.2k23086
edited Jun 18 '14 at 21:49
user26857
39.3k124183
asked Jun 25 '12 at 18:21
Ben Blum-SmithBen Blum-Smith
10.2k23086
edited Jun 18 '14 at 21:49
user26857
39.3k124183
edited Jun 18 '14 at 21:49
user26857
39.3k124183
edited Jun 18 '14 at 21:49
user26857
39.3k124183
39.3k124183
asked Jun 25 '12 at 18:21
Ben Blum-SmithBen Blum-Smith
10.2k23086
asked Jun 25 '12 at 18:21
Ben Blum-SmithBen Blum-Smith
10.2k23086
asked Jun 25 '12 at 18:21
Ben Blum-SmithBen Blum-Smith
10.2k23086
10.2k23086
2
$begingroup$
Rotman has it in Advanced Modern Algebra, page 651!
$endgroup$
– Dedalus
Jun 25 '12 at 18:29
$begingroup$
(The result still holds)
$endgroup$
– Dedalus
Jun 25 '12 at 18:29
add a comment |
2
$begingroup$
Rotman has it in Advanced Modern Algebra, page 651!
$endgroup$
– Dedalus
Jun 25 '12 at 18:29
$begingroup$
(The result still holds)
$endgroup$
– Dedalus
Jun 25 '12 at 18:29
2
2
$begingroup$
Rotman has it in Advanced Modern Algebra, page 651!
$endgroup$
– Dedalus
Jun 25 '12 at 18:29
$begingroup$
Rotman has it in Advanced Modern Algebra, page 651!
$endgroup$
– Dedalus
Jun 25 '12 at 18:29
$begingroup$
(The result still holds)
$endgroup$
– Dedalus
Jun 25 '12 at 18:29
$begingroup$
(The result still holds)
$endgroup$
– Dedalus
Jun 25 '12 at 18:29
add a comment |
1 Answer
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$begingroup$
Let $F$ be a free $R$-module, where $R$ is a PID, and $U$ be a submodule. Then $U$ is also free (and the rank is at most the rank of $F$). Here is a hint for the proof.
Let ${e_i}_{i in I}$ be a basis of $F$. Choose a well-ordering $leq$ on $I$ (this requires the Axiom of Choice). Let $p_i : F to R$ be the projection on the $i$th coordinate. Let $F_i$ be the submodule of $F$ generated by the $e_j$ with $j leq i$. Let $U_i = U cap F_i$. Then $p_i(U_i)$ is a submodule of $R$, i.e. has the form $R a_i$. Choose some $u_i in U_i$ with $p_i(u_i)=a_i$. If $a_i=0$, we may also choose $u_i=0$.
Now show that the $u_i neq 0$ constitute a basis of $U$. Hint: Transfinite induction.
The same proof shows the more general result: If $R$ is a hereditary ring (every ideal of $R$ is projective over $R$), then any submodule of a free $R$-module is a direct sum of ideals of $R$.
answered Jun 25 '12 at 18:59
Martin BrandenburgMartin Brandenburg
108k13163334
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add a comment |
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1 Answer
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1 Answer
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oldest
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$begingroup$
Let $F$ be a free $R$-module, where $R$ is a PID, and $U$ be a submodule. Then $U$ is also free (and the rank is at most the rank of $F$). Here is a hint for the proof.
Let ${e_i}_{i in I}$ be a basis of $F$. Choose a well-ordering $leq$ on $I$ (this requires the Axiom of Choice). Let $p_i : F to R$ be the projection on the $i$th coordinate. Let $F_i$ be the submodule of $F$ generated by the $e_j$ with $j leq i$. Let $U_i = U cap F_i$. Then $p_i(U_i)$ is a submodule of $R$, i.e. has the form $R a_i$. Choose some $u_i in U_i$ with $p_i(u_i)=a_i$. If $a_i=0$, we may also choose $u_i=0$.
Now show that the $u_i neq 0$ constitute a basis of $U$. Hint: Transfinite induction.
The same proof shows the more general result: If $R$ is a hereditary ring (every ideal of $R$ is projective over $R$), then any submodule of a free $R$-module is a direct sum of ideals of $R$.
answered Jun 25 '12 at 18:59
Martin BrandenburgMartin Brandenburg
108k13163334
$endgroup$
add a comment |
$begingroup$
Let $F$ be a free $R$-module, where $R$ is a PID, and $U$ be a submodule. Then $U$ is also free (and the rank is at most the rank of $F$). Here is a hint for the proof.
Let ${e_i}_{i in I}$ be a basis of $F$. Choose a well-ordering $leq$ on $I$ (this requires the Axiom of Choice). Let $p_i : F to R$ be the projection on the $i$th coordinate. Let $F_i$ be the submodule of $F$ generated by the $e_j$ with $j leq i$. Let $U_i = U cap F_i$. Then $p_i(U_i)$ is a submodule of $R$, i.e. has the form $R a_i$. Choose some $u_i in U_i$ with $p_i(u_i)=a_i$. If $a_i=0$, we may also choose $u_i=0$.
Now show that the $u_i neq 0$ constitute a basis of $U$. Hint: Transfinite induction.
The same proof shows the more general result: If $R$ is a hereditary ring (every ideal of $R$ is projective over $R$), then any submodule of a free $R$-module is a direct sum of ideals of $R$.
answered Jun 25 '12 at 18:59
Martin BrandenburgMartin Brandenburg
108k13163334
$endgroup$
add a comment |
$begingroup$
Let $F$ be a free $R$-module, where $R$ is a PID, and $U$ be a submodule. Then $U$ is also free (and the rank is at most the rank of $F$). Here is a hint for the proof.
Let ${e_i}_{i in I}$ be a basis of $F$. Choose a well-ordering $leq$ on $I$ (this requires the Axiom of Choice). Let $p_i : F to R$ be the projection on the $i$th coordinate. Let $F_i$ be the submodule of $F$ generated by the $e_j$ with $j leq i$. Let $U_i = U cap F_i$. Then $p_i(U_i)$ is a submodule of $R$, i.e. has the form $R a_i$. Choose some $u_i in U_i$ with $p_i(u_i)=a_i$. If $a_i=0$, we may also choose $u_i=0$.
Now show that the $u_i neq 0$ constitute a basis of $U$. Hint: Transfinite induction.
The same proof shows the more general result: If $R$ is a hereditary ring (every ideal of $R$ is projective over $R$), then any submodule of a free $R$-module is a direct sum of ideals of $R$.
answered Jun 25 '12 at 18:59
Martin BrandenburgMartin Brandenburg
108k13163334
$endgroup$
Let $F$ be a free $R$-module, where $R$ is a PID, and $U$ be a submodule. Then $U$ is also free (and the rank is at most the rank of $F$). Here is a hint for the proof.
Let ${e_i}_{i in I}$ be a basis of $F$. Choose a well-ordering $leq$ on $I$ (this requires the Axiom of Choice). Let $p_i : F to R$ be the projection on the $i$th coordinate. Let $F_i$ be the submodule of $F$ generated by the $e_j$ with $j leq i$. Let $U_i = U cap F_i$. Then $p_i(U_i)$ is a submodule of $R$, i.e. has the form $R a_i$. Choose some $u_i in U_i$ with $p_i(u_i)=a_i$. If $a_i=0$, we may also choose $u_i=0$.
Now show that the $u_i neq 0$ constitute a basis of $U$. Hint: Transfinite induction.
The same proof shows the more general result: If $R$ is a hereditary ring (every ideal of $R$ is projective over $R$), then any submodule of a free $R$-module is a direct sum of ideals of $R$.
answered Jun 25 '12 at 18:59
Martin BrandenburgMartin Brandenburg
108k13163334
answered Jun 25 '12 at 18:59
Martin BrandenburgMartin Brandenburg
108k13163334
answered Jun 25 '12 at 18:59
Martin BrandenburgMartin Brandenburg
108k13163334
answered Jun 25 '12 at 18:59
Martin BrandenburgMartin Brandenburg
108k13163334
108k13163334
add a comment |
add a comment |
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$begingroup$
Rotman has it in Advanced Modern Algebra, page 651!
$endgroup$
– Dedalus
Jun 25 '12 at 18:29
$begingroup$
(The result still holds)
$endgroup$
– Dedalus
Jun 25 '12 at 18:29