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If $R^n$ is generated by $e_1, cdots e_m$ as an $R$-module, then $nleq m$
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Given a commutative ring $R$, I want to show
If $R^n$ is generated by $e_1, cdots e_m$ as an $R$-module, then $m geq n$.
This follows from the standard result that if there exists a surjection from $R^m rightarrow R^n$, then $mgeq n$. Then standard proof is tensoring with $R/mathfrakm$ which gives a surjection of vector spaces $(R/mathfrakm)^m rightarrow R/mathfrakm)^n$.
Other argument will need to use Cayley-Hamilton.
Like a vector space $R^n$ has a free basis $k_1, cdots, k_n$.
What is the main obstacle that there cannot be a simple argument like in vector spaces?
Edit:
For vector space, the argument I had in mind is using basis. Given $mathbb F^n$, we know any basis has $n$ elements, but if $n>m$, and some set $k_1, cdots k_m$ spans $mathbb F^n$, then this is a contradiction. Now any (free) basis of $R^n$ have $n$ elements, if there exists a small set which spans $R^n$, isn't this a contradiction?
abstract-algebra commutative-algebra modules free-modules
asked Jul 28 at 16:40
Xiao
4,32411333
add a comment |Â
up vote
1
down vote
favorite
Given a commutative ring $R$, I want to show
If $R^n$ is generated by $e_1, cdots e_m$ as an $R$-module, then $m geq n$.
This follows from the standard result that if there exists a surjection from $R^m rightarrow R^n$, then $mgeq n$. Then standard proof is tensoring with $R/mathfrakm$ which gives a surjection of vector spaces $(R/mathfrakm)^m rightarrow R/mathfrakm)^n$.
Other argument will need to use Cayley-Hamilton.
Like a vector space $R^n$ has a free basis $k_1, cdots, k_n$.
What is the main obstacle that there cannot be a simple argument like in vector spaces?
Edit:
For vector space, the argument I had in mind is using basis. Given $mathbb F^n$, we know any basis has $n$ elements, but if $n>m$, and some set $k_1, cdots k_m$ spans $mathbb F^n$, then this is a contradiction. Now any (free) basis of $R^n$ have $n$ elements, if there exists a small set which spans $R^n$, isn't this a contradiction?
abstract-algebra commutative-algebra modules free-modules
asked Jul 28 at 16:40
Xiao
4,32411333
Have you written out an argument for the analogous result for vector spaces? I'm guessing several obstructions would leap out at you, depending on what path you chose. From my point of view, the essential thing is: linear relations in a vector space are much more powerful/flexible, because you can divide by any coefficient. This is, of course, not the case for modules over a commutative ring. But I guess I can be more explicit:
– Alex Wertheim
Jul 28 at 17:51
To prove this result for vector spaces, I would probably use rank nullity; how does one prove this? One "higher level" view would be: given a map of finite-dimensional vector spaces $f: V to W$, there is a short exact sequence of vector spaces $0 to ker(f) to V to W to 0$. One can then prove an alternating sum formula for dimensions in a short exact sequences of vector spaces, which gives the result. Alternatively, one can be more hands on: take a basis for $ker(f)$, and extend it to a basis for $V$. Then show that $im(f)$ has basis spanned by the images of non-kernel vectors.
– Alex Wertheim
Jul 28 at 17:58
You can see where these arguments go wrong. In the exact sequence argument, all vector spaces are free (including $ker(f)$), so there is no problem with dimension. In the commutative ring argument, there is a notion of dimension for free things (which one has to prove is well-defined!), but even then, there's no reason why the kernel of the surjection should be free, and one would still have to prove this alternating sum formula for dimensions. Similarly, the argument of extending a subset of linearly independent vectors to a basis does not carry over well to the commutative ring setting.
– Alex Wertheim
Jul 28 at 18:01
@AlexWertheim Thank you for the reply. For vector space, the argument I had in mind is using basis. Given $mathbb F^n$, we know any basis has $n$ elements, but if $n>m$, and some set $k_1, cdots k_m$ spans $mathbb F^n$, then this is a contradiction. Now any (free) basis of $R^n$ have $n$ elements, if there exists a small set which spans $R^n$, isn't this a contradiction?
– Xiao
Jul 28 at 18:55
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Given a commutative ring $R$, I want to show
If $R^n$ is generated by $e_1, cdots e_m$ as an $R$-module, then $m geq n$.
This follows from the standard result that if there exists a surjection from $R^m rightarrow R^n$, then $mgeq n$. Then standard proof is tensoring with $R/mathfrakm$ which gives a surjection of vector spaces $(R/mathfrakm)^m rightarrow R/mathfrakm)^n$.
Other argument will need to use Cayley-Hamilton.
Like a vector space $R^n$ has a free basis $k_1, cdots, k_n$.
What is the main obstacle that there cannot be a simple argument like in vector spaces?
Edit:
For vector space, the argument I had in mind is using basis. Given $mathbb F^n$, we know any basis has $n$ elements, but if $n>m$, and some set $k_1, cdots k_m$ spans $mathbb F^n$, then this is a contradiction. Now any (free) basis of $R^n$ have $n$ elements, if there exists a small set which spans $R^n$, isn't this a contradiction?
abstract-algebra commutative-algebra modules free-modules
asked Jul 28 at 16:40
Xiao
4,32411333
Given a commutative ring $R$, I want to show
If $R^n$ is generated by $e_1, cdots e_m$ as an $R$-module, then $m geq n$.
This follows from the standard result that if there exists a surjection from $R^m rightarrow R^n$, then $mgeq n$. Then standard proof is tensoring with $R/mathfrakm$ which gives a surjection of vector spaces $(R/mathfrakm)^m rightarrow R/mathfrakm)^n$.
Other argument will need to use Cayley-Hamilton.
Like a vector space $R^n$ has a free basis $k_1, cdots, k_n$.
What is the main obstacle that there cannot be a simple argument like in vector spaces?
Edit:
For vector space, the argument I had in mind is using basis. Given $mathbb F^n$, we know any basis has $n$ elements, but if $n>m$, and some set $k_1, cdots k_m$ spans $mathbb F^n$, then this is a contradiction. Now any (free) basis of $R^n$ have $n$ elements, if there exists a small set which spans $R^n$, isn't this a contradiction?
abstract-algebra commutative-algebra modules free-modules
asked Jul 28 at 16:40
Xiao
4,32411333
edited Jul 28 at 18:58
asked Jul 28 at 16:40
Xiao
4,32411333
asked Jul 28 at 16:40
Xiao
4,32411333
asked Jul 28 at 16:40
Xiao
4,32411333
4,32411333
Have you written out an argument for the analogous result for vector spaces? I'm guessing several obstructions would leap out at you, depending on what path you chose. From my point of view, the essential thing is: linear relations in a vector space are much more powerful/flexible, because you can divide by any coefficient. This is, of course, not the case for modules over a commutative ring. But I guess I can be more explicit:
– Alex Wertheim
Jul 28 at 17:51
To prove this result for vector spaces, I would probably use rank nullity; how does one prove this? One "higher level" view would be: given a map of finite-dimensional vector spaces $f: V to W$, there is a short exact sequence of vector spaces $0 to ker(f) to V to W to 0$. One can then prove an alternating sum formula for dimensions in a short exact sequences of vector spaces, which gives the result. Alternatively, one can be more hands on: take a basis for $ker(f)$, and extend it to a basis for $V$. Then show that $im(f)$ has basis spanned by the images of non-kernel vectors.
– Alex Wertheim
Jul 28 at 17:58
You can see where these arguments go wrong. In the exact sequence argument, all vector spaces are free (including $ker(f)$), so there is no problem with dimension. In the commutative ring argument, there is a notion of dimension for free things (which one has to prove is well-defined!), but even then, there's no reason why the kernel of the surjection should be free, and one would still have to prove this alternating sum formula for dimensions. Similarly, the argument of extending a subset of linearly independent vectors to a basis does not carry over well to the commutative ring setting.
– Alex Wertheim
Jul 28 at 18:01
@AlexWertheim Thank you for the reply. For vector space, the argument I had in mind is using basis. Given $mathbb F^n$, we know any basis has $n$ elements, but if $n>m$, and some set $k_1, cdots k_m$ spans $mathbb F^n$, then this is a contradiction. Now any (free) basis of $R^n$ have $n$ elements, if there exists a small set which spans $R^n$, isn't this a contradiction?
– Xiao
Jul 28 at 18:55
add a comment |Â
Have you written out an argument for the analogous result for vector spaces? I'm guessing several obstructions would leap out at you, depending on what path you chose. From my point of view, the essential thing is: linear relations in a vector space are much more powerful/flexible, because you can divide by any coefficient. This is, of course, not the case for modules over a commutative ring. But I guess I can be more explicit:
– Alex Wertheim
Jul 28 at 17:51
To prove this result for vector spaces, I would probably use rank nullity; how does one prove this? One "higher level" view would be: given a map of finite-dimensional vector spaces $f: V to W$, there is a short exact sequence of vector spaces $0 to ker(f) to V to W to 0$. One can then prove an alternating sum formula for dimensions in a short exact sequences of vector spaces, which gives the result. Alternatively, one can be more hands on: take a basis for $ker(f)$, and extend it to a basis for $V$. Then show that $im(f)$ has basis spanned by the images of non-kernel vectors.
– Alex Wertheim
Jul 28 at 17:58
You can see where these arguments go wrong. In the exact sequence argument, all vector spaces are free (including $ker(f)$), so there is no problem with dimension. In the commutative ring argument, there is a notion of dimension for free things (which one has to prove is well-defined!), but even then, there's no reason why the kernel of the surjection should be free, and one would still have to prove this alternating sum formula for dimensions. Similarly, the argument of extending a subset of linearly independent vectors to a basis does not carry over well to the commutative ring setting.
– Alex Wertheim
Jul 28 at 18:01
@AlexWertheim Thank you for the reply. For vector space, the argument I had in mind is using basis. Given $mathbb F^n$, we know any basis has $n$ elements, but if $n>m$, and some set $k_1, cdots k_m$ spans $mathbb F^n$, then this is a contradiction. Now any (free) basis of $R^n$ have $n$ elements, if there exists a small set which spans $R^n$, isn't this a contradiction?
– Xiao
Jul 28 at 18:55
Have you written out an argument for the analogous result for vector spaces? I'm guessing several obstructions would leap out at you, depending on what path you chose. From my point of view, the essential thing is: linear relations in a vector space are much more powerful/flexible, because you can divide by any coefficient. This is, of course, not the case for modules over a commutative ring. But I guess I can be more explicit:
– Alex Wertheim
Jul 28 at 17:51
Have you written out an argument for the analogous result for vector spaces? I'm guessing several obstructions would leap out at you, depending on what path you chose. From my point of view, the essential thing is: linear relations in a vector space are much more powerful/flexible, because you can divide by any coefficient. This is, of course, not the case for modules over a commutative ring. But I guess I can be more explicit:
– Alex Wertheim
Jul 28 at 17:51
To prove this result for vector spaces, I would probably use rank nullity; how does one prove this? One "higher level" view would be: given a map of finite-dimensional vector spaces $f: V to W$, there is a short exact sequence of vector spaces $0 to ker(f) to V to W to 0$. One can then prove an alternating sum formula for dimensions in a short exact sequences of vector spaces, which gives the result. Alternatively, one can be more hands on: take a basis for $ker(f)$, and extend it to a basis for $V$. Then show that $im(f)$ has basis spanned by the images of non-kernel vectors.
– Alex Wertheim
Jul 28 at 17:58
To prove this result for vector spaces, I would probably use rank nullity; how does one prove this? One "higher level" view would be: given a map of finite-dimensional vector spaces $f: V to W$, there is a short exact sequence of vector spaces $0 to ker(f) to V to W to 0$. One can then prove an alternating sum formula for dimensions in a short exact sequences of vector spaces, which gives the result. Alternatively, one can be more hands on: take a basis for $ker(f)$, and extend it to a basis for $V$. Then show that $im(f)$ has basis spanned by the images of non-kernel vectors.
– Alex Wertheim
Jul 28 at 17:58
You can see where these arguments go wrong. In the exact sequence argument, all vector spaces are free (including $ker(f)$), so there is no problem with dimension. In the commutative ring argument, there is a notion of dimension for free things (which one has to prove is well-defined!), but even then, there's no reason why the kernel of the surjection should be free, and one would still have to prove this alternating sum formula for dimensions. Similarly, the argument of extending a subset of linearly independent vectors to a basis does not carry over well to the commutative ring setting.
– Alex Wertheim
Jul 28 at 18:01
You can see where these arguments go wrong. In the exact sequence argument, all vector spaces are free (including $ker(f)$), so there is no problem with dimension. In the commutative ring argument, there is a notion of dimension for free things (which one has to prove is well-defined!), but even then, there's no reason why the kernel of the surjection should be free, and one would still have to prove this alternating sum formula for dimensions. Similarly, the argument of extending a subset of linearly independent vectors to a basis does not carry over well to the commutative ring setting.
– Alex Wertheim
Jul 28 at 18:01
@AlexWertheim Thank you for the reply. For vector space, the argument I had in mind is using basis. Given $mathbb F^n$, we know any basis has $n$ elements, but if $n>m$, and some set $k_1, cdots k_m$ spans $mathbb F^n$, then this is a contradiction. Now any (free) basis of $R^n$ have $n$ elements, if there exists a small set which spans $R^n$, isn't this a contradiction?
– Xiao
Jul 28 at 18:55
@AlexWertheim Thank you for the reply. For vector space, the argument I had in mind is using basis. Given $mathbb F^n$, we know any basis has $n$ elements, but if $n>m$, and some set $k_1, cdots k_m$ spans $mathbb F^n$, then this is a contradiction. Now any (free) basis of $R^n$ have $n$ elements, if there exists a small set which spans $R^n$, isn't this a contradiction?
– Xiao
Jul 28 at 18:55
add a comment |Â
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Have you written out an argument for the analogous result for vector spaces? I'm guessing several obstructions would leap out at you, depending on what path you chose. From my point of view, the essential thing is: linear relations in a vector space are much more powerful/flexible, because you can divide by any coefficient. This is, of course, not the case for modules over a commutative ring. But I guess I can be more explicit:
– Alex Wertheim
Jul 28 at 17:51
To prove this result for vector spaces, I would probably use rank nullity; how does one prove this? One "higher level" view would be: given a map of finite-dimensional vector spaces $f: V to W$, there is a short exact sequence of vector spaces $0 to ker(f) to V to W to 0$. One can then prove an alternating sum formula for dimensions in a short exact sequences of vector spaces, which gives the result. Alternatively, one can be more hands on: take a basis for $ker(f)$, and extend it to a basis for $V$. Then show that $im(f)$ has basis spanned by the images of non-kernel vectors.
– Alex Wertheim
Jul 28 at 17:58
You can see where these arguments go wrong. In the exact sequence argument, all vector spaces are free (including $ker(f)$), so there is no problem with dimension. In the commutative ring argument, there is a notion of dimension for free things (which one has to prove is well-defined!), but even then, there's no reason why the kernel of the surjection should be free, and one would still have to prove this alternating sum formula for dimensions. Similarly, the argument of extending a subset of linearly independent vectors to a basis does not carry over well to the commutative ring setting.
– Alex Wertheim
Jul 28 at 18:01
@AlexWertheim Thank you for the reply. For vector space, the argument I had in mind is using basis. Given $mathbb F^n$, we know any basis has $n$ elements, but if $n>m$, and some set $k_1, cdots k_m$ spans $mathbb F^n$, then this is a contradiction. Now any (free) basis of $R^n$ have $n$ elements, if there exists a small set which spans $R^n$, isn't this a contradiction?
– Xiao
Jul 28 at 18:55