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Why doesn't this work for showing that $A[x]$ is a flat $A$-algebra?
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$N$ is flat if and only if when $f: M' rightarrow M$ is injective then $f otimes textid_N : M' otimes N rightarrow M otimes N$ is injective ($N,M,M'$ are all $A$-modules). So consider $f: M' rightarrow M$ injective and tensor by the $A$-module $A[x]$. We get $f otimes textid_A[x] : M' otimes A[x] rightarrow M otimes A[x]$. To check if this is injective, we consider its kernel:
$$
beginalign ker(f otimes textid_A[x]) &= ker(f) otimes A[x] + M' otimes ker(textid_A[x]) \
&= 0 otimes A[x] + M' otimes 0 \
&=0+0 \
&=0.
endalign
$$
Here I have used the rule for kernels of tensor products of maps, that $f$ is injective and that the identity map is injective. So the tensored map is injective and so $A[x]$ is flat.
I'm not sure about this because it seems that you could prove any module is flat this way, and obviously not all modules are flat. Would anyone be able to shed some light on where my reasoning has gone wrong? I have indeed seen this way to do this question, but I'd still like to know why this method is wrong. Many thanks.
abstract-algebra ring-theory commutative-algebra modules algebras
asked Jul 16 at 19:02
mathphys
961415
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$N$ is flat if and only if when $f: M' rightarrow M$ is injective then $f otimes textid_N : M' otimes N rightarrow M otimes N$ is injective ($N,M,M'$ are all $A$-modules). So consider $f: M' rightarrow M$ injective and tensor by the $A$-module $A[x]$. We get $f otimes textid_A[x] : M' otimes A[x] rightarrow M otimes A[x]$. To check if this is injective, we consider its kernel:
$$
beginalign ker(f otimes textid_A[x]) &= ker(f) otimes A[x] + M' otimes ker(textid_A[x]) \
&= 0 otimes A[x] + M' otimes 0 \
&=0+0 \
&=0.
endalign
$$
Here I have used the rule for kernels of tensor products of maps, that $f$ is injective and that the identity map is injective. So the tensored map is injective and so $A[x]$ is flat.
I'm not sure about this because it seems that you could prove any module is flat this way, and obviously not all modules are flat. Would anyone be able to shed some light on where my reasoning has gone wrong? I have indeed seen this way to do this question, but I'd still like to know why this method is wrong. Many thanks.
abstract-algebra ring-theory commutative-algebra modules algebras
asked Jul 16 at 19:02
mathphys
961415
1
$A[x]$ is flat as an $A$-module; it's free, so it must be flat. So?
– Lord Shark the Unknown
Jul 16 at 19:04
I think this is the same as the answer I linked in my question. I agree, this is the best way to do the question. Is there an obvious reason as to why my attempt is wrong though? Thanks.
– mathphys
Jul 16 at 20:49
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
$N$ is flat if and only if when $f: M' rightarrow M$ is injective then $f otimes textid_N : M' otimes N rightarrow M otimes N$ is injective ($N,M,M'$ are all $A$-modules). So consider $f: M' rightarrow M$ injective and tensor by the $A$-module $A[x]$. We get $f otimes textid_A[x] : M' otimes A[x] rightarrow M otimes A[x]$. To check if this is injective, we consider its kernel:
$$
beginalign ker(f otimes textid_A[x]) &= ker(f) otimes A[x] + M' otimes ker(textid_A[x]) \
&= 0 otimes A[x] + M' otimes 0 \
&=0+0 \
&=0.
endalign
$$
Here I have used the rule for kernels of tensor products of maps, that $f$ is injective and that the identity map is injective. So the tensored map is injective and so $A[x]$ is flat.
I'm not sure about this because it seems that you could prove any module is flat this way, and obviously not all modules are flat. Would anyone be able to shed some light on where my reasoning has gone wrong? I have indeed seen this way to do this question, but I'd still like to know why this method is wrong. Many thanks.
abstract-algebra ring-theory commutative-algebra modules algebras
asked Jul 16 at 19:02
mathphys
961415
$N$ is flat if and only if when $f: M' rightarrow M$ is injective then $f otimes textid_N : M' otimes N rightarrow M otimes N$ is injective ($N,M,M'$ are all $A$-modules). So consider $f: M' rightarrow M$ injective and tensor by the $A$-module $A[x]$. We get $f otimes textid_A[x] : M' otimes A[x] rightarrow M otimes A[x]$. To check if this is injective, we consider its kernel:
$$
beginalign ker(f otimes textid_A[x]) &= ker(f) otimes A[x] + M' otimes ker(textid_A[x]) \
&= 0 otimes A[x] + M' otimes 0 \
&=0+0 \
&=0.
endalign
$$
Here I have used the rule for kernels of tensor products of maps, that $f$ is injective and that the identity map is injective. So the tensored map is injective and so $A[x]$ is flat.
I'm not sure about this because it seems that you could prove any module is flat this way, and obviously not all modules are flat. Would anyone be able to shed some light on where my reasoning has gone wrong? I have indeed seen this way to do this question, but I'd still like to know why this method is wrong. Many thanks.
abstract-algebra ring-theory commutative-algebra modules algebras
asked Jul 16 at 19:02
mathphys
961415
asked Jul 16 at 19:02
mathphys
961415
asked Jul 16 at 19:02
mathphys
961415
asked Jul 16 at 19:02
mathphys
961415
961415
1
$A[x]$ is flat as an $A$-module; it's free, so it must be flat. So?
– Lord Shark the Unknown
Jul 16 at 19:04
I think this is the same as the answer I linked in my question. I agree, this is the best way to do the question. Is there an obvious reason as to why my attempt is wrong though? Thanks.
– mathphys
Jul 16 at 20:49
add a comment |Â
1
$A[x]$ is flat as an $A$-module; it's free, so it must be flat. So?
– Lord Shark the Unknown
Jul 16 at 19:04
I think this is the same as the answer I linked in my question. I agree, this is the best way to do the question. Is there an obvious reason as to why my attempt is wrong though? Thanks.
– mathphys
Jul 16 at 20:49
1
1
$A[x]$ is flat as an $A$-module; it's free, so it must be flat. So?
– Lord Shark the Unknown
Jul 16 at 19:04
$A[x]$ is flat as an $A$-module; it's free, so it must be flat. So?
– Lord Shark the Unknown
Jul 16 at 19:04
I think this is the same as the answer I linked in my question. I agree, this is the best way to do the question. Is there an obvious reason as to why my attempt is wrong though? Thanks.
– mathphys
Jul 16 at 20:49
I think this is the same as the answer I linked in my question. I agree, this is the best way to do the question. Is there an obvious reason as to why my attempt is wrong though? Thanks.
– mathphys
Jul 16 at 20:49
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
The formula for the kernel of the tensor of maps that you refer to is only valid when both maps are surjective. Here's a reference (Theorem 2.19) to a proof in some free notes of Keith Conrad, and notably surjectivity plays a big role in the reasoning by allowing us to describe the form of elementary tensors in the image.
As you mentioned, if the kernel identity applied across the board, or even just to injective maps, or even just to injective maps tensored with isomorphisms, then every module would be flat, which is silly. Indeed the tensor of injective maps need not be injective, and modules need not be flat.
In fact, every integral domain which is not a field has lots of non-flat modules over it which are very easy to construct.
For a classic example with the integers, take $iota: mathbbZ rightarrow mathbbQ$ to be inclusion and $textid: mathbbZ/nmathbbZ rightarrow mathbbZ/nmathbbZ$, considering all as $mathbbZ$-modules. Sure enough, $(iota otimes textid)(1 otimes bar1)$. Both maps are injective and $1 otimes bar1$ is nonzero in $mathbbZ otimes mathbbZ/nmathbbZ$ but the tensor of maps is not injective since $(iota otimes textid)(1 otimes bar1) = frac1n otimes bar0$ in $mathbbQ otimes mathbbZ/nmathbbZ$.
answered Jul 16 at 22:22
Badam Baplan
3,331721
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The formula for the kernel of the tensor of maps that you refer to is only valid when both maps are surjective. Here's a reference (Theorem 2.19) to a proof in some free notes of Keith Conrad, and notably surjectivity plays a big role in the reasoning by allowing us to describe the form of elementary tensors in the image.
As you mentioned, if the kernel identity applied across the board, or even just to injective maps, or even just to injective maps tensored with isomorphisms, then every module would be flat, which is silly. Indeed the tensor of injective maps need not be injective, and modules need not be flat.
In fact, every integral domain which is not a field has lots of non-flat modules over it which are very easy to construct.
For a classic example with the integers, take $iota: mathbbZ rightarrow mathbbQ$ to be inclusion and $textid: mathbbZ/nmathbbZ rightarrow mathbbZ/nmathbbZ$, considering all as $mathbbZ$-modules. Sure enough, $(iota otimes textid)(1 otimes bar1)$. Both maps are injective and $1 otimes bar1$ is nonzero in $mathbbZ otimes mathbbZ/nmathbbZ$ but the tensor of maps is not injective since $(iota otimes textid)(1 otimes bar1) = frac1n otimes bar0$ in $mathbbQ otimes mathbbZ/nmathbbZ$.
answered Jul 16 at 22:22
Badam Baplan
3,331721
add a comment |Â
up vote
1
down vote
accepted
The formula for the kernel of the tensor of maps that you refer to is only valid when both maps are surjective. Here's a reference (Theorem 2.19) to a proof in some free notes of Keith Conrad, and notably surjectivity plays a big role in the reasoning by allowing us to describe the form of elementary tensors in the image.
As you mentioned, if the kernel identity applied across the board, or even just to injective maps, or even just to injective maps tensored with isomorphisms, then every module would be flat, which is silly. Indeed the tensor of injective maps need not be injective, and modules need not be flat.
In fact, every integral domain which is not a field has lots of non-flat modules over it which are very easy to construct.
For a classic example with the integers, take $iota: mathbbZ rightarrow mathbbQ$ to be inclusion and $textid: mathbbZ/nmathbbZ rightarrow mathbbZ/nmathbbZ$, considering all as $mathbbZ$-modules. Sure enough, $(iota otimes textid)(1 otimes bar1)$. Both maps are injective and $1 otimes bar1$ is nonzero in $mathbbZ otimes mathbbZ/nmathbbZ$ but the tensor of maps is not injective since $(iota otimes textid)(1 otimes bar1) = frac1n otimes bar0$ in $mathbbQ otimes mathbbZ/nmathbbZ$.
answered Jul 16 at 22:22
Badam Baplan
3,331721
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The formula for the kernel of the tensor of maps that you refer to is only valid when both maps are surjective. Here's a reference (Theorem 2.19) to a proof in some free notes of Keith Conrad, and notably surjectivity plays a big role in the reasoning by allowing us to describe the form of elementary tensors in the image.
As you mentioned, if the kernel identity applied across the board, or even just to injective maps, or even just to injective maps tensored with isomorphisms, then every module would be flat, which is silly. Indeed the tensor of injective maps need not be injective, and modules need not be flat.
In fact, every integral domain which is not a field has lots of non-flat modules over it which are very easy to construct.
For a classic example with the integers, take $iota: mathbbZ rightarrow mathbbQ$ to be inclusion and $textid: mathbbZ/nmathbbZ rightarrow mathbbZ/nmathbbZ$, considering all as $mathbbZ$-modules. Sure enough, $(iota otimes textid)(1 otimes bar1)$. Both maps are injective and $1 otimes bar1$ is nonzero in $mathbbZ otimes mathbbZ/nmathbbZ$ but the tensor of maps is not injective since $(iota otimes textid)(1 otimes bar1) = frac1n otimes bar0$ in $mathbbQ otimes mathbbZ/nmathbbZ$.
answered Jul 16 at 22:22
Badam Baplan
3,331721
The formula for the kernel of the tensor of maps that you refer to is only valid when both maps are surjective. Here's a reference (Theorem 2.19) to a proof in some free notes of Keith Conrad, and notably surjectivity plays a big role in the reasoning by allowing us to describe the form of elementary tensors in the image.
As you mentioned, if the kernel identity applied across the board, or even just to injective maps, or even just to injective maps tensored with isomorphisms, then every module would be flat, which is silly. Indeed the tensor of injective maps need not be injective, and modules need not be flat.
In fact, every integral domain which is not a field has lots of non-flat modules over it which are very easy to construct.
For a classic example with the integers, take $iota: mathbbZ rightarrow mathbbQ$ to be inclusion and $textid: mathbbZ/nmathbbZ rightarrow mathbbZ/nmathbbZ$, considering all as $mathbbZ$-modules. Sure enough, $(iota otimes textid)(1 otimes bar1)$. Both maps are injective and $1 otimes bar1$ is nonzero in $mathbbZ otimes mathbbZ/nmathbbZ$ but the tensor of maps is not injective since $(iota otimes textid)(1 otimes bar1) = frac1n otimes bar0$ in $mathbbQ otimes mathbbZ/nmathbbZ$.
answered Jul 16 at 22:22
Badam Baplan
3,331721
edited Jul 16 at 22:34
answered Jul 16 at 22:22
Badam Baplan
3,331721
answered Jul 16 at 22:22
Badam Baplan
3,331721
answered Jul 16 at 22:22
Badam Baplan
3,331721
3,331721
add a comment |Â
add a comment |Â
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1
$A[x]$ is flat as an $A$-module; it's free, so it must be flat. So?
– Lord Shark the Unknown
Jul 16 at 19:04
I think this is the same as the answer I linked in my question. I agree, this is the best way to do the question. Is there an obvious reason as to why my attempt is wrong though? Thanks.
– mathphys
Jul 16 at 20:49