Understanding proof of: A submodule of a free module of finite rank is also free of finite rank?
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Sorry in advance for the long exposition; it is necessary to include it.
I am trying to understand the proof of this result from Dummit & Foote (so in particular, I can't use the result yet):
Let $R$ be a Principal Ideal Domain, let $M$ be a free $R$ - module of finite rank $n$ and let $N$ be a submodule of $M$. Then
$(1)$ $N$ is free of rank $m$, $m le n$ and
$(2)$ there exists a basis $y_1, y_2, ..., y_n$ of $M$ so that $a_1y_1, a_2y_2, ..., a_my_m$ is a basis of $N$ where $a_1, a_2, ..., a_m$ are nonzero elements of $R$ with the divisibility relations
$$a_1|a_2| cdots | a_m$$
Somewhere in the middle of the proof, it says:
We now verify that this element $y_1$ can be taken as one element in a basis for $M$ and that $a_1y_1$ can be taken as one element in a basis for $N$, namely that we have
$(a)$ $M=Ry_1 oplus ker v$, and
$(b)$ $N = Ra_1y_1 oplus (N cap ker v)$
This is the part I have a question about. It seems like there is a hidden theorem here which says
($pm$ some assumptions of $R$ being an integral domain or a P.I.D.)
If $M$ is a free $R$-module, $N$ is a submodule, and $M = Ry oplus N$, then there exists a basis of $M$ containing $y$.
However, I do not know how to prove this. If we already assumed the result, it would not be difficult; just take any basis for $N$ and adjoin it to $y$. However, without the result I am stuck.
Question: Did I identify the hidden theorem correctly? If so, how do we prove it?
abstract-algebra modules free-modules
asked Nov 17 at 15:26
Ovi
12.1k938108
add a comment |
up vote
0
down vote
favorite
Sorry in advance for the long exposition; it is necessary to include it.
I am trying to understand the proof of this result from Dummit & Foote (so in particular, I can't use the result yet):
Let $R$ be a Principal Ideal Domain, let $M$ be a free $R$ - module of finite rank $n$ and let $N$ be a submodule of $M$. Then
$(1)$ $N$ is free of rank $m$, $m le n$ and
$(2)$ there exists a basis $y_1, y_2, ..., y_n$ of $M$ so that $a_1y_1, a_2y_2, ..., a_my_m$ is a basis of $N$ where $a_1, a_2, ..., a_m$ are nonzero elements of $R$ with the divisibility relations
$$a_1|a_2| cdots | a_m$$
Somewhere in the middle of the proof, it says:
We now verify that this element $y_1$ can be taken as one element in a basis for $M$ and that $a_1y_1$ can be taken as one element in a basis for $N$, namely that we have
$(a)$ $M=Ry_1 oplus ker v$, and
$(b)$ $N = Ra_1y_1 oplus (N cap ker v)$
This is the part I have a question about. It seems like there is a hidden theorem here which says
($pm$ some assumptions of $R$ being an integral domain or a P.I.D.)
If $M$ is a free $R$-module, $N$ is a submodule, and $M = Ry oplus N$, then there exists a basis of $M$ containing $y$.
However, I do not know how to prove this. If we already assumed the result, it would not be difficult; just take any basis for $N$ and adjoin it to $y$. However, without the result I am stuck.
Question: Did I identify the hidden theorem correctly? If so, how do we prove it?
abstract-algebra modules free-modules
asked Nov 17 at 15:26
Ovi
12.1k938108
We can take any element of M and construct a basis. As an example, think of a vector space V. To construct a basis for V. One can start with any non-zero vector v$_1$ and then take another vector v$_2$ $notin$ Span(v$_1$). Then take v$_3$ $notin$ Span(v$_1$,v$_2$). Continue this process. A similar thing happens with free modules in general.
– Joel Pereira
Nov 17 at 15:59
@JoelPereira I understand why for later they might want to show that $M=Ry_1 oplus ker v$, but why mention explicitly that "we will show that $y_1$ can be taken as a basis for M"? Do you think they are just being redundant?
– Ovi
Nov 17 at 16:03
It's not true in general that $any$ y can be be used as part of the basis for BOTH M and N. It's true for M always, but not necessarily the case for a general submodule.
– Joel Pereira
Nov 17 at 16:07
@JoelPereira But as long as that $y$ is an element of both the module and the submodule, then it can be used for a basis, right?
– Ovi
Nov 18 at 14:04
Yes. But the proof is also saying that the a$_i$ have that divisibility condition.
– Joel Pereira
Nov 18 at 22:11
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Sorry in advance for the long exposition; it is necessary to include it.
I am trying to understand the proof of this result from Dummit & Foote (so in particular, I can't use the result yet):
Let $R$ be a Principal Ideal Domain, let $M$ be a free $R$ - module of finite rank $n$ and let $N$ be a submodule of $M$. Then
$(1)$ $N$ is free of rank $m$, $m le n$ and
$(2)$ there exists a basis $y_1, y_2, ..., y_n$ of $M$ so that $a_1y_1, a_2y_2, ..., a_my_m$ is a basis of $N$ where $a_1, a_2, ..., a_m$ are nonzero elements of $R$ with the divisibility relations
$$a_1|a_2| cdots | a_m$$
Somewhere in the middle of the proof, it says:
We now verify that this element $y_1$ can be taken as one element in a basis for $M$ and that $a_1y_1$ can be taken as one element in a basis for $N$, namely that we have
$(a)$ $M=Ry_1 oplus ker v$, and
$(b)$ $N = Ra_1y_1 oplus (N cap ker v)$
This is the part I have a question about. It seems like there is a hidden theorem here which says
($pm$ some assumptions of $R$ being an integral domain or a P.I.D.)
If $M$ is a free $R$-module, $N$ is a submodule, and $M = Ry oplus N$, then there exists a basis of $M$ containing $y$.
However, I do not know how to prove this. If we already assumed the result, it would not be difficult; just take any basis for $N$ and adjoin it to $y$. However, without the result I am stuck.
Question: Did I identify the hidden theorem correctly? If so, how do we prove it?
abstract-algebra modules free-modules
asked Nov 17 at 15:26
Ovi
12.1k938108
Sorry in advance for the long exposition; it is necessary to include it.
I am trying to understand the proof of this result from Dummit & Foote (so in particular, I can't use the result yet):
Let $R$ be a Principal Ideal Domain, let $M$ be a free $R$ - module of finite rank $n$ and let $N$ be a submodule of $M$. Then
$(1)$ $N$ is free of rank $m$, $m le n$ and
$(2)$ there exists a basis $y_1, y_2, ..., y_n$ of $M$ so that $a_1y_1, a_2y_2, ..., a_my_m$ is a basis of $N$ where $a_1, a_2, ..., a_m$ are nonzero elements of $R$ with the divisibility relations
$$a_1|a_2| cdots | a_m$$
Somewhere in the middle of the proof, it says:
We now verify that this element $y_1$ can be taken as one element in a basis for $M$ and that $a_1y_1$ can be taken as one element in a basis for $N$, namely that we have
$(a)$ $M=Ry_1 oplus ker v$, and
$(b)$ $N = Ra_1y_1 oplus (N cap ker v)$
This is the part I have a question about. It seems like there is a hidden theorem here which says
($pm$ some assumptions of $R$ being an integral domain or a P.I.D.)
If $M$ is a free $R$-module, $N$ is a submodule, and $M = Ry oplus N$, then there exists a basis of $M$ containing $y$.
However, I do not know how to prove this. If we already assumed the result, it would not be difficult; just take any basis for $N$ and adjoin it to $y$. However, without the result I am stuck.
Question: Did I identify the hidden theorem correctly? If so, how do we prove it?
abstract-algebra modules free-modules
abstract-algebra modules free-modules
asked Nov 17 at 15:26
Ovi
12.1k938108
asked Nov 17 at 15:26
Ovi
12.1k938108
asked Nov 17 at 15:26
Ovi
12.1k938108
asked Nov 17 at 15:26
Ovi
12.1k938108
asked Nov 17 at 15:26
Ovi
12.1k938108
12.1k938108
We can take any element of M and construct a basis. As an example, think of a vector space V. To construct a basis for V. One can start with any non-zero vector v$_1$ and then take another vector v$_2$ $notin$ Span(v$_1$). Then take v$_3$ $notin$ Span(v$_1$,v$_2$). Continue this process. A similar thing happens with free modules in general.
– Joel Pereira
Nov 17 at 15:59
@JoelPereira I understand why for later they might want to show that $M=Ry_1 oplus ker v$, but why mention explicitly that "we will show that $y_1$ can be taken as a basis for M"? Do you think they are just being redundant?
– Ovi
Nov 17 at 16:03
It's not true in general that $any$ y can be be used as part of the basis for BOTH M and N. It's true for M always, but not necessarily the case for a general submodule.
– Joel Pereira
Nov 17 at 16:07
@JoelPereira But as long as that $y$ is an element of both the module and the submodule, then it can be used for a basis, right?
– Ovi
Nov 18 at 14:04
Yes. But the proof is also saying that the a$_i$ have that divisibility condition.
– Joel Pereira
Nov 18 at 22:11
add a comment |
We can take any element of M and construct a basis. As an example, think of a vector space V. To construct a basis for V. One can start with any non-zero vector v$_1$ and then take another vector v$_2$ $notin$ Span(v$_1$). Then take v$_3$ $notin$ Span(v$_1$,v$_2$). Continue this process. A similar thing happens with free modules in general.
– Joel Pereira
Nov 17 at 15:59
@JoelPereira I understand why for later they might want to show that $M=Ry_1 oplus ker v$, but why mention explicitly that "we will show that $y_1$ can be taken as a basis for M"? Do you think they are just being redundant?
– Ovi
Nov 17 at 16:03
It's not true in general that $any$ y can be be used as part of the basis for BOTH M and N. It's true for M always, but not necessarily the case for a general submodule.
– Joel Pereira
Nov 17 at 16:07
@JoelPereira But as long as that $y$ is an element of both the module and the submodule, then it can be used for a basis, right?
– Ovi
Nov 18 at 14:04
Yes. But the proof is also saying that the a$_i$ have that divisibility condition.
– Joel Pereira
Nov 18 at 22:11
We can take any element of M and construct a basis. As an example, think of a vector space V. To construct a basis for V. One can start with any non-zero vector v$_1$ and then take another vector v$_2$ $notin$ Span(v$_1$). Then take v$_3$ $notin$ Span(v$_1$,v$_2$). Continue this process. A similar thing happens with free modules in general.
– Joel Pereira
Nov 17 at 15:59
We can take any element of M and construct a basis. As an example, think of a vector space V. To construct a basis for V. One can start with any non-zero vector v$_1$ and then take another vector v$_2$ $notin$ Span(v$_1$). Then take v$_3$ $notin$ Span(v$_1$,v$_2$). Continue this process. A similar thing happens with free modules in general.
– Joel Pereira
Nov 17 at 15:59
@JoelPereira I understand why for later they might want to show that $M=Ry_1 oplus ker v$, but why mention explicitly that "we will show that $y_1$ can be taken as a basis for M"? Do you think they are just being redundant?
– Ovi
Nov 17 at 16:03
@JoelPereira I understand why for later they might want to show that $M=Ry_1 oplus ker v$, but why mention explicitly that "we will show that $y_1$ can be taken as a basis for M"? Do you think they are just being redundant?
– Ovi
Nov 17 at 16:03
It's not true in general that $any$ y can be be used as part of the basis for BOTH M and N. It's true for M always, but not necessarily the case for a general submodule.
– Joel Pereira
Nov 17 at 16:07
It's not true in general that $any$ y can be be used as part of the basis for BOTH M and N. It's true for M always, but not necessarily the case for a general submodule.
– Joel Pereira
Nov 17 at 16:07
@JoelPereira But as long as that $y$ is an element of both the module and the submodule, then it can be used for a basis, right?
– Ovi
Nov 18 at 14:04
@JoelPereira But as long as that $y$ is an element of both the module and the submodule, then it can be used for a basis, right?
– Ovi
Nov 18 at 14:04
Yes. But the proof is also saying that the a$_i$ have that divisibility condition.
– Joel Pereira
Nov 18 at 22:11
Yes. But the proof is also saying that the a$_i$ have that divisibility condition.
– Joel Pereira
Nov 18 at 22:11
add a comment |
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We can take any element of M and construct a basis. As an example, think of a vector space V. To construct a basis for V. One can start with any non-zero vector v$_1$ and then take another vector v$_2$ $notin$ Span(v$_1$). Then take v$_3$ $notin$ Span(v$_1$,v$_2$). Continue this process. A similar thing happens with free modules in general.
– Joel Pereira
Nov 17 at 15:59
@JoelPereira I understand why for later they might want to show that $M=Ry_1 oplus ker v$, but why mention explicitly that "we will show that $y_1$ can be taken as a basis for M"? Do you think they are just being redundant?
– Ovi
Nov 17 at 16:03
It's not true in general that $any$ y can be be used as part of the basis for BOTH M and N. It's true for M always, but not necessarily the case for a general submodule.
– Joel Pereira
Nov 17 at 16:07
@JoelPereira But as long as that $y$ is an element of both the module and the submodule, then it can be used for a basis, right?
– Ovi
Nov 18 at 14:04
Yes. But the proof is also saying that the a$_i$ have that divisibility condition.
– Joel Pereira
Nov 18 at 22:11