Why does $underline{cd}:Gleq silp:Glequnderline{cd}+1$ hold?
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I read the about text in a book but don't understand how or why this inequality is "straightforward" can anybody explain this to me?
homology-cohomology group-cohomology
asked Nov 22 at 16:40
Rhoswyn
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I read the about text in a book but don't understand how or why this inequality is "straightforward" can anybody explain this to me?
homology-cohomology group-cohomology
asked Nov 22 at 16:40
Rhoswyn
381210
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I read the about text in a book but don't understand how or why this inequality is "straightforward" can anybody explain this to me?
homology-cohomology group-cohomology
asked Nov 22 at 16:40
Rhoswyn
381210
I read the about text in a book but don't understand how or why this inequality is "straightforward" can anybody explain this to me?
homology-cohomology group-cohomology
homology-cohomology group-cohomology
asked Nov 22 at 16:40
Rhoswyn
381210
asked Nov 22 at 16:40
Rhoswyn
381210
edited Nov 22 at 16:51
asked Nov 22 at 16:40
Rhoswyn
381210
asked Nov 22 at 16:40
Rhoswyn
381210
asked Nov 22 at 16:40
Rhoswyn
381210
381210
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add a comment |
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First, you can show that in the definition of $underline{cd}$, you can use ${mathbb Z}G$-projective modules in the second slot instead of ${mathbb Z}G$-free modules.
The first inequality then follows from the fact that the second sup is taken over all ${mathbb Z}G$ modules rather than just ${mathbb Z}$-free modules.
For the second inequality, take an arbitrary ${mathbb Z}G$ module $A$ and a free ${mathbb Z}G$ module $F$ mapping onto it, with kernel $K$:
$$0 to K to F to A to 0$$
Since $F$ is ${mathbb Z}$-free and $K$ is a subgroup of $F$ as an abelian group, it's also ${mathbb Z}$-free. The inequality then follows by looking at the long exact sequence for Ext.
answered Nov 22 at 17:52
Bruce Ikenaga
1962
Ahh okay, I was on the right lines for part of it. Can you expand on your last statement though? I'm not sure I follow it.
– Rhoswyn
Nov 22 at 18:01
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1 Answer
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First, you can show that in the definition of $underline{cd}$, you can use ${mathbb Z}G$-projective modules in the second slot instead of ${mathbb Z}G$-free modules.
The first inequality then follows from the fact that the second sup is taken over all ${mathbb Z}G$ modules rather than just ${mathbb Z}$-free modules.
For the second inequality, take an arbitrary ${mathbb Z}G$ module $A$ and a free ${mathbb Z}G$ module $F$ mapping onto it, with kernel $K$:
$$0 to K to F to A to 0$$
Since $F$ is ${mathbb Z}$-free and $K$ is a subgroup of $F$ as an abelian group, it's also ${mathbb Z}$-free. The inequality then follows by looking at the long exact sequence for Ext.
answered Nov 22 at 17:52
Bruce Ikenaga
1962
Ahh okay, I was on the right lines for part of it. Can you expand on your last statement though? I'm not sure I follow it.
– Rhoswyn
Nov 22 at 18:01
add a comment |
up vote
0
down vote
First, you can show that in the definition of $underline{cd}$, you can use ${mathbb Z}G$-projective modules in the second slot instead of ${mathbb Z}G$-free modules.
The first inequality then follows from the fact that the second sup is taken over all ${mathbb Z}G$ modules rather than just ${mathbb Z}$-free modules.
For the second inequality, take an arbitrary ${mathbb Z}G$ module $A$ and a free ${mathbb Z}G$ module $F$ mapping onto it, with kernel $K$:
$$0 to K to F to A to 0$$
Since $F$ is ${mathbb Z}$-free and $K$ is a subgroup of $F$ as an abelian group, it's also ${mathbb Z}$-free. The inequality then follows by looking at the long exact sequence for Ext.
answered Nov 22 at 17:52
Bruce Ikenaga
1962
Ahh okay, I was on the right lines for part of it. Can you expand on your last statement though? I'm not sure I follow it.
– Rhoswyn
Nov 22 at 18:01
add a comment |
up vote
0
down vote
up vote
0
down vote
First, you can show that in the definition of $underline{cd}$, you can use ${mathbb Z}G$-projective modules in the second slot instead of ${mathbb Z}G$-free modules.
The first inequality then follows from the fact that the second sup is taken over all ${mathbb Z}G$ modules rather than just ${mathbb Z}$-free modules.
For the second inequality, take an arbitrary ${mathbb Z}G$ module $A$ and a free ${mathbb Z}G$ module $F$ mapping onto it, with kernel $K$:
$$0 to K to F to A to 0$$
Since $F$ is ${mathbb Z}$-free and $K$ is a subgroup of $F$ as an abelian group, it's also ${mathbb Z}$-free. The inequality then follows by looking at the long exact sequence for Ext.
answered Nov 22 at 17:52
Bruce Ikenaga
1962
First, you can show that in the definition of $underline{cd}$, you can use ${mathbb Z}G$-projective modules in the second slot instead of ${mathbb Z}G$-free modules.
The first inequality then follows from the fact that the second sup is taken over all ${mathbb Z}G$ modules rather than just ${mathbb Z}$-free modules.
For the second inequality, take an arbitrary ${mathbb Z}G$ module $A$ and a free ${mathbb Z}G$ module $F$ mapping onto it, with kernel $K$:
$$0 to K to F to A to 0$$
Since $F$ is ${mathbb Z}$-free and $K$ is a subgroup of $F$ as an abelian group, it's also ${mathbb Z}$-free. The inequality then follows by looking at the long exact sequence for Ext.
answered Nov 22 at 17:52
Bruce Ikenaga
1962
answered Nov 22 at 17:52
Bruce Ikenaga
1962
answered Nov 22 at 17:52
Bruce Ikenaga
1962
answered Nov 22 at 17:52
Bruce Ikenaga
1962
1962
Ahh okay, I was on the right lines for part of it. Can you expand on your last statement though? I'm not sure I follow it.
– Rhoswyn
Nov 22 at 18:01
add a comment |
Ahh okay, I was on the right lines for part of it. Can you expand on your last statement though? I'm not sure I follow it.
– Rhoswyn
Nov 22 at 18:01
Ahh okay, I was on the right lines for part of it. Can you expand on your last statement though? I'm not sure I follow it.
– Rhoswyn
Nov 22 at 18:01
Ahh okay, I was on the right lines for part of it. Can you expand on your last statement though? I'm not sure I follow it.
– Rhoswyn
Nov 22 at 18:01
add a comment |
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