free resolution of $mathbb{Z}$ as an $mathbb{Z}[x]/(x^n-1)$-module
$begingroup$
As the tittle says, I am having a bad time thinking on the construction of a free resolution of $mathbb{Z}$ as an $(mathbb{Z}[x]/(x^n-1))$-module.
I know that I should give an exact sequence of the form (if $R= mathbb{Z}[x]/(x^n-1)$):
$... longrightarrow R^{I_n} longrightarrow ...longrightarrow R^{I_1} longrightarrow R^{I_0} longrightarrow mathbb{Z} longrightarrow 0$
being
$d_i : R^{I_i} longrightarrow R^{I_{i-1}}$ for $i ge 1 $
$d_0: R^{I_0} longrightarrow mathbb{Z}$
$e: mathbb{Z} longrightarrow 0$.
and $R^{I_i}$ being free $R$-modules.
Moreover they should satisfy the exact sequence condition: $Im(d_i)=Ker(d_{i-1})$.
I began thinking on the first application $d_0: R^{I_0} longrightarrow mathbb{Z}$. But I had some troubles finding it. I thought sending
$a_0 + a_1 bar{x} + a_2 bar{x}^2+...+a_{n-1} bar{x}^{n-1} mapsto a_0$
because I need $Im(d_0)=Ker(e) = mathbb{Z}$. But it gets harder to find $d_1$.
Then I thought:
$a_0 + a_1 bar{x} + a_2 bar{x}^2+...+a_{n-1} bar{x}^{n-1} mapsto sum a_i$
but again it gets hard.
Any hint/help?
EDIT: I chose $R^{I_0}$ to be $R$.
abstract-algebra commutative-algebra free-modules
asked Dec 23 '18 at 12:36
idriskameniidriskameni
738321
$endgroup$
add a comment |
$begingroup$
As the tittle says, I am having a bad time thinking on the construction of a free resolution of $mathbb{Z}$ as an $(mathbb{Z}[x]/(x^n-1))$-module.
I know that I should give an exact sequence of the form (if $R= mathbb{Z}[x]/(x^n-1)$):
$... longrightarrow R^{I_n} longrightarrow ...longrightarrow R^{I_1} longrightarrow R^{I_0} longrightarrow mathbb{Z} longrightarrow 0$
being
$d_i : R^{I_i} longrightarrow R^{I_{i-1}}$ for $i ge 1 $
$d_0: R^{I_0} longrightarrow mathbb{Z}$
$e: mathbb{Z} longrightarrow 0$.
and $R^{I_i}$ being free $R$-modules.
Moreover they should satisfy the exact sequence condition: $Im(d_i)=Ker(d_{i-1})$.
I began thinking on the first application $d_0: R^{I_0} longrightarrow mathbb{Z}$. But I had some troubles finding it. I thought sending
$a_0 + a_1 bar{x} + a_2 bar{x}^2+...+a_{n-1} bar{x}^{n-1} mapsto a_0$
because I need $Im(d_0)=Ker(e) = mathbb{Z}$. But it gets harder to find $d_1$.
Then I thought:
$a_0 + a_1 bar{x} + a_2 bar{x}^2+...+a_{n-1} bar{x}^{n-1} mapsto sum a_i$
but again it gets hard.
Any hint/help?
EDIT: I chose $R^{I_0}$ to be $R$.
abstract-algebra commutative-algebra free-modules
asked Dec 23 '18 at 12:36
idriskameniidriskameni
738321
$endgroup$
1
$begingroup$
Your second guess is the right one. However, it might be better to say that $d_0$ sends $f(overline{x})$ to $f(1)$. What would then be the kernel of $d_1$?
$endgroup$
– Greg
Dec 23 '18 at 12:43
$begingroup$
Do you mean $d_0$ sends $f(bar{x})$ to $f(1)$?
$endgroup$
– idriskameni
Dec 23 '18 at 12:45
1
$begingroup$
Correct, sorry. I have editted the mistake
$endgroup$
– Greg
Dec 23 '18 at 12:45
1
$begingroup$
Agree with Greg. Basically this comes from how do you view $Bbb{Z}$ as an $R$-module. Because $x^n=1$ in $R$, the monomial $x^n$ must act as the identity. This places severe restrictions on how $x$ can act on $Bbb{Z}$. If $n$ is even, then $x$ can act via either $+1$ or $-1$, but if $n$ is odd, then you must have $xcdot m=m$ for all $minBbb{Z}$. Because $d_0$ must be a homomorphism of $R$-modules, it follows that either $x^imapsto d_0(1)$ or $x^imapsto (-1)^id_0(1)$. The latter case corresponding to the possibility that $n$ is even, and $xcdot m=-m$ for all $minBbb{Z}$.
$endgroup$
– Jyrki Lahtonen
Dec 23 '18 at 12:54
add a comment |
$begingroup$
As the tittle says, I am having a bad time thinking on the construction of a free resolution of $mathbb{Z}$ as an $(mathbb{Z}[x]/(x^n-1))$-module.
I know that I should give an exact sequence of the form (if $R= mathbb{Z}[x]/(x^n-1)$):
$... longrightarrow R^{I_n} longrightarrow ...longrightarrow R^{I_1} longrightarrow R^{I_0} longrightarrow mathbb{Z} longrightarrow 0$
being
$d_i : R^{I_i} longrightarrow R^{I_{i-1}}$ for $i ge 1 $
$d_0: R^{I_0} longrightarrow mathbb{Z}$
$e: mathbb{Z} longrightarrow 0$.
and $R^{I_i}$ being free $R$-modules.
Moreover they should satisfy the exact sequence condition: $Im(d_i)=Ker(d_{i-1})$.
I began thinking on the first application $d_0: R^{I_0} longrightarrow mathbb{Z}$. But I had some troubles finding it. I thought sending
$a_0 + a_1 bar{x} + a_2 bar{x}^2+...+a_{n-1} bar{x}^{n-1} mapsto a_0$
because I need $Im(d_0)=Ker(e) = mathbb{Z}$. But it gets harder to find $d_1$.
Then I thought:
$a_0 + a_1 bar{x} + a_2 bar{x}^2+...+a_{n-1} bar{x}^{n-1} mapsto sum a_i$
but again it gets hard.
Any hint/help?
EDIT: I chose $R^{I_0}$ to be $R$.
abstract-algebra commutative-algebra free-modules
asked Dec 23 '18 at 12:36
idriskameniidriskameni
738321
$endgroup$
As the tittle says, I am having a bad time thinking on the construction of a free resolution of $mathbb{Z}$ as an $(mathbb{Z}[x]/(x^n-1))$-module.
I know that I should give an exact sequence of the form (if $R= mathbb{Z}[x]/(x^n-1)$):
$... longrightarrow R^{I_n} longrightarrow ...longrightarrow R^{I_1} longrightarrow R^{I_0} longrightarrow mathbb{Z} longrightarrow 0$
being
$d_i : R^{I_i} longrightarrow R^{I_{i-1}}$ for $i ge 1 $
$d_0: R^{I_0} longrightarrow mathbb{Z}$
$e: mathbb{Z} longrightarrow 0$.
and $R^{I_i}$ being free $R$-modules.
Moreover they should satisfy the exact sequence condition: $Im(d_i)=Ker(d_{i-1})$.
I began thinking on the first application $d_0: R^{I_0} longrightarrow mathbb{Z}$. But I had some troubles finding it. I thought sending
$a_0 + a_1 bar{x} + a_2 bar{x}^2+...+a_{n-1} bar{x}^{n-1} mapsto a_0$
because I need $Im(d_0)=Ker(e) = mathbb{Z}$. But it gets harder to find $d_1$.
Then I thought:
$a_0 + a_1 bar{x} + a_2 bar{x}^2+...+a_{n-1} bar{x}^{n-1} mapsto sum a_i$
but again it gets hard.
Any hint/help?
EDIT: I chose $R^{I_0}$ to be $R$.
abstract-algebra commutative-algebra free-modules
abstract-algebra commutative-algebra free-modules
asked Dec 23 '18 at 12:36
idriskameniidriskameni
738321
asked Dec 23 '18 at 12:36
idriskameniidriskameni
738321
edited Dec 23 '18 at 12:43
idriskameni
asked Dec 23 '18 at 12:36
idriskameniidriskameni
738321
asked Dec 23 '18 at 12:36
idriskameniidriskameni
738321
asked Dec 23 '18 at 12:36
idriskameniidriskameni
738321
738321
1
$begingroup$
Your second guess is the right one. However, it might be better to say that $d_0$ sends $f(overline{x})$ to $f(1)$. What would then be the kernel of $d_1$?
$endgroup$
– Greg
Dec 23 '18 at 12:43
$begingroup$
Do you mean $d_0$ sends $f(bar{x})$ to $f(1)$?
$endgroup$
– idriskameni
Dec 23 '18 at 12:45
1
$begingroup$
Correct, sorry. I have editted the mistake
$endgroup$
– Greg
Dec 23 '18 at 12:45
1
$begingroup$
Agree with Greg. Basically this comes from how do you view $Bbb{Z}$ as an $R$-module. Because $x^n=1$ in $R$, the monomial $x^n$ must act as the identity. This places severe restrictions on how $x$ can act on $Bbb{Z}$. If $n$ is even, then $x$ can act via either $+1$ or $-1$, but if $n$ is odd, then you must have $xcdot m=m$ for all $minBbb{Z}$. Because $d_0$ must be a homomorphism of $R$-modules, it follows that either $x^imapsto d_0(1)$ or $x^imapsto (-1)^id_0(1)$. The latter case corresponding to the possibility that $n$ is even, and $xcdot m=-m$ for all $minBbb{Z}$.
$endgroup$
– Jyrki Lahtonen
Dec 23 '18 at 12:54
add a comment |
1
$begingroup$
Your second guess is the right one. However, it might be better to say that $d_0$ sends $f(overline{x})$ to $f(1)$. What would then be the kernel of $d_1$?
$endgroup$
– Greg
Dec 23 '18 at 12:43
$begingroup$
Do you mean $d_0$ sends $f(bar{x})$ to $f(1)$?
$endgroup$
– idriskameni
Dec 23 '18 at 12:45
1
$begingroup$
Correct, sorry. I have editted the mistake
$endgroup$
– Greg
Dec 23 '18 at 12:45
1
$begingroup$
Agree with Greg. Basically this comes from how do you view $Bbb{Z}$ as an $R$-module. Because $x^n=1$ in $R$, the monomial $x^n$ must act as the identity. This places severe restrictions on how $x$ can act on $Bbb{Z}$. If $n$ is even, then $x$ can act via either $+1$ or $-1$, but if $n$ is odd, then you must have $xcdot m=m$ for all $minBbb{Z}$. Because $d_0$ must be a homomorphism of $R$-modules, it follows that either $x^imapsto d_0(1)$ or $x^imapsto (-1)^id_0(1)$. The latter case corresponding to the possibility that $n$ is even, and $xcdot m=-m$ for all $minBbb{Z}$.
$endgroup$
– Jyrki Lahtonen
Dec 23 '18 at 12:54
1
1
$begingroup$
Your second guess is the right one. However, it might be better to say that $d_0$ sends $f(overline{x})$ to $f(1)$. What would then be the kernel of $d_1$?
$endgroup$
– Greg
Dec 23 '18 at 12:43
$begingroup$
Your second guess is the right one. However, it might be better to say that $d_0$ sends $f(overline{x})$ to $f(1)$. What would then be the kernel of $d_1$?
$endgroup$
– Greg
Dec 23 '18 at 12:43
$begingroup$
Do you mean $d_0$ sends $f(bar{x})$ to $f(1)$?
$endgroup$
– idriskameni
Dec 23 '18 at 12:45
$begingroup$
Do you mean $d_0$ sends $f(bar{x})$ to $f(1)$?
$endgroup$
– idriskameni
Dec 23 '18 at 12:45
1
1
$begingroup$
Correct, sorry. I have editted the mistake
$endgroup$
– Greg
Dec 23 '18 at 12:45
$begingroup$
Correct, sorry. I have editted the mistake
$endgroup$
– Greg
Dec 23 '18 at 12:45
1
1
$begingroup$
Agree with Greg. Basically this comes from how do you view $Bbb{Z}$ as an $R$-module. Because $x^n=1$ in $R$, the monomial $x^n$ must act as the identity. This places severe restrictions on how $x$ can act on $Bbb{Z}$. If $n$ is even, then $x$ can act via either $+1$ or $-1$, but if $n$ is odd, then you must have $xcdot m=m$ for all $minBbb{Z}$. Because $d_0$ must be a homomorphism of $R$-modules, it follows that either $x^imapsto d_0(1)$ or $x^imapsto (-1)^id_0(1)$. The latter case corresponding to the possibility that $n$ is even, and $xcdot m=-m$ for all $minBbb{Z}$.
$endgroup$
– Jyrki Lahtonen
Dec 23 '18 at 12:54
$begingroup$
Agree with Greg. Basically this comes from how do you view $Bbb{Z}$ as an $R$-module. Because $x^n=1$ in $R$, the monomial $x^n$ must act as the identity. This places severe restrictions on how $x$ can act on $Bbb{Z}$. If $n$ is even, then $x$ can act via either $+1$ or $-1$, but if $n$ is odd, then you must have $xcdot m=m$ for all $minBbb{Z}$. Because $d_0$ must be a homomorphism of $R$-modules, it follows that either $x^imapsto d_0(1)$ or $x^imapsto (-1)^id_0(1)$. The latter case corresponding to the possibility that $n$ is even, and $xcdot m=-m$ for all $minBbb{Z}$.
$endgroup$
– Jyrki Lahtonen
Dec 23 '18 at 12:54
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Consider the following sequence of $R$-modules:
$$ ldots longrightarrow R longrightarrow R longrightarrow R longrightarrow mathbb{Z} rightarrow 0 . $$
We denote by $d_i$ the usual map.
We define $d_0: R rightarrow mathbb{Z}: f(overline{x}) rightarrow f(1).$
Then, the kernel of $d_0$ corresponds to the polynomials in $R$ such that $f(1) = 0$.
Consider now $d_1: R rightarrow R: g(overline{x})rightarrow (overline{x-1})g(overline{x}). $
Notice that, per construction, the image of $d_1$ is the kernel of $d_0$. The kernel of $d_1$ is the set of functions such that $g(overline{x})(overline{x-1})=0$.
Try to complete the sequence, using the fact that $(1+x+ldots + x^{n-1})(x-1) = x^n-1.$
answered Dec 23 '18 at 12:53
GregGreg
183112
$endgroup$
1
$begingroup$
As a sidenote: $mathbb{Z}$ can be regarded as an $R$-module by letting $x$ act as $1$ on $mathbb{Z}$.
$endgroup$
– Greg
Dec 23 '18 at 12:57
2
$begingroup$
Oh yeah! The resolution continues periodically (assuming $x$ acts as the identity).
$endgroup$
– Jyrki Lahtonen
Dec 23 '18 at 12:57
1
$begingroup$
Exactly, @JyrkiLahtonen !
$endgroup$
– Greg
Dec 23 '18 at 13:14
1
$begingroup$
Yes, you can. As long as the sequence stays exact!
$endgroup$
– Greg
Dec 23 '18 at 13:26
1
$begingroup$
And by the usual map, I just meant that $d_i$ is the map from $R^{I_i} rightarrow R^{I_{i-1}}$.
$endgroup$
– Greg
Dec 23 '18 at 13:27
|
show 5 more comments
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
Consider the following sequence of $R$-modules:
$$ ldots longrightarrow R longrightarrow R longrightarrow R longrightarrow mathbb{Z} rightarrow 0 . $$
We denote by $d_i$ the usual map.
We define $d_0: R rightarrow mathbb{Z}: f(overline{x}) rightarrow f(1).$
Then, the kernel of $d_0$ corresponds to the polynomials in $R$ such that $f(1) = 0$.
Consider now $d_1: R rightarrow R: g(overline{x})rightarrow (overline{x-1})g(overline{x}). $
Notice that, per construction, the image of $d_1$ is the kernel of $d_0$. The kernel of $d_1$ is the set of functions such that $g(overline{x})(overline{x-1})=0$.
Try to complete the sequence, using the fact that $(1+x+ldots + x^{n-1})(x-1) = x^n-1.$
answered Dec 23 '18 at 12:53
GregGreg
183112
$endgroup$
1
$begingroup$
As a sidenote: $mathbb{Z}$ can be regarded as an $R$-module by letting $x$ act as $1$ on $mathbb{Z}$.
$endgroup$
– Greg
Dec 23 '18 at 12:57
2
$begingroup$
Oh yeah! The resolution continues periodically (assuming $x$ acts as the identity).
$endgroup$
– Jyrki Lahtonen
Dec 23 '18 at 12:57
1
$begingroup$
Exactly, @JyrkiLahtonen !
$endgroup$
– Greg
Dec 23 '18 at 13:14
1
$begingroup$
Yes, you can. As long as the sequence stays exact!
$endgroup$
– Greg
Dec 23 '18 at 13:26
1
$begingroup$
And by the usual map, I just meant that $d_i$ is the map from $R^{I_i} rightarrow R^{I_{i-1}}$.
$endgroup$
– Greg
Dec 23 '18 at 13:27
|
show 5 more comments
$begingroup$
Consider the following sequence of $R$-modules:
$$ ldots longrightarrow R longrightarrow R longrightarrow R longrightarrow mathbb{Z} rightarrow 0 . $$
We denote by $d_i$ the usual map.
We define $d_0: R rightarrow mathbb{Z}: f(overline{x}) rightarrow f(1).$
Then, the kernel of $d_0$ corresponds to the polynomials in $R$ such that $f(1) = 0$.
Consider now $d_1: R rightarrow R: g(overline{x})rightarrow (overline{x-1})g(overline{x}). $
Notice that, per construction, the image of $d_1$ is the kernel of $d_0$. The kernel of $d_1$ is the set of functions such that $g(overline{x})(overline{x-1})=0$.
Try to complete the sequence, using the fact that $(1+x+ldots + x^{n-1})(x-1) = x^n-1.$
answered Dec 23 '18 at 12:53
GregGreg
183112
$endgroup$
1
$begingroup$
As a sidenote: $mathbb{Z}$ can be regarded as an $R$-module by letting $x$ act as $1$ on $mathbb{Z}$.
$endgroup$
– Greg
Dec 23 '18 at 12:57
2
$begingroup$
Oh yeah! The resolution continues periodically (assuming $x$ acts as the identity).
$endgroup$
– Jyrki Lahtonen
Dec 23 '18 at 12:57
1
$begingroup$
Exactly, @JyrkiLahtonen !
$endgroup$
– Greg
Dec 23 '18 at 13:14
1
$begingroup$
Yes, you can. As long as the sequence stays exact!
$endgroup$
– Greg
Dec 23 '18 at 13:26
1
$begingroup$
And by the usual map, I just meant that $d_i$ is the map from $R^{I_i} rightarrow R^{I_{i-1}}$.
$endgroup$
– Greg
Dec 23 '18 at 13:27
|
show 5 more comments
$begingroup$
Consider the following sequence of $R$-modules:
$$ ldots longrightarrow R longrightarrow R longrightarrow R longrightarrow mathbb{Z} rightarrow 0 . $$
We denote by $d_i$ the usual map.
We define $d_0: R rightarrow mathbb{Z}: f(overline{x}) rightarrow f(1).$
Then, the kernel of $d_0$ corresponds to the polynomials in $R$ such that $f(1) = 0$.
Consider now $d_1: R rightarrow R: g(overline{x})rightarrow (overline{x-1})g(overline{x}). $
Notice that, per construction, the image of $d_1$ is the kernel of $d_0$. The kernel of $d_1$ is the set of functions such that $g(overline{x})(overline{x-1})=0$.
Try to complete the sequence, using the fact that $(1+x+ldots + x^{n-1})(x-1) = x^n-1.$
answered Dec 23 '18 at 12:53
GregGreg
183112
$endgroup$
Consider the following sequence of $R$-modules:
$$ ldots longrightarrow R longrightarrow R longrightarrow R longrightarrow mathbb{Z} rightarrow 0 . $$
We denote by $d_i$ the usual map.
We define $d_0: R rightarrow mathbb{Z}: f(overline{x}) rightarrow f(1).$
Then, the kernel of $d_0$ corresponds to the polynomials in $R$ such that $f(1) = 0$.
Consider now $d_1: R rightarrow R: g(overline{x})rightarrow (overline{x-1})g(overline{x}). $
Notice that, per construction, the image of $d_1$ is the kernel of $d_0$. The kernel of $d_1$ is the set of functions such that $g(overline{x})(overline{x-1})=0$.
Try to complete the sequence, using the fact that $(1+x+ldots + x^{n-1})(x-1) = x^n-1.$
answered Dec 23 '18 at 12:53
GregGreg
183112
answered Dec 23 '18 at 12:53
GregGreg
183112
answered Dec 23 '18 at 12:53
GregGreg
183112
answered Dec 23 '18 at 12:53
GregGreg
183112
183112
1
$begingroup$
As a sidenote: $mathbb{Z}$ can be regarded as an $R$-module by letting $x$ act as $1$ on $mathbb{Z}$.
$endgroup$
– Greg
Dec 23 '18 at 12:57
2
$begingroup$
Oh yeah! The resolution continues periodically (assuming $x$ acts as the identity).
$endgroup$
– Jyrki Lahtonen
Dec 23 '18 at 12:57
1
$begingroup$
Exactly, @JyrkiLahtonen !
$endgroup$
– Greg
Dec 23 '18 at 13:14
1
$begingroup$
Yes, you can. As long as the sequence stays exact!
$endgroup$
– Greg
Dec 23 '18 at 13:26
1
$begingroup$
And by the usual map, I just meant that $d_i$ is the map from $R^{I_i} rightarrow R^{I_{i-1}}$.
$endgroup$
– Greg
Dec 23 '18 at 13:27
|
show 5 more comments
1
$begingroup$
As a sidenote: $mathbb{Z}$ can be regarded as an $R$-module by letting $x$ act as $1$ on $mathbb{Z}$.
$endgroup$
– Greg
Dec 23 '18 at 12:57
2
$begingroup$
Oh yeah! The resolution continues periodically (assuming $x$ acts as the identity).
$endgroup$
– Jyrki Lahtonen
Dec 23 '18 at 12:57
1
$begingroup$
Exactly, @JyrkiLahtonen !
$endgroup$
– Greg
Dec 23 '18 at 13:14
1
$begingroup$
Yes, you can. As long as the sequence stays exact!
$endgroup$
– Greg
Dec 23 '18 at 13:26
1
$begingroup$
And by the usual map, I just meant that $d_i$ is the map from $R^{I_i} rightarrow R^{I_{i-1}}$.
$endgroup$
– Greg
Dec 23 '18 at 13:27
1
1
$begingroup$
As a sidenote: $mathbb{Z}$ can be regarded as an $R$-module by letting $x$ act as $1$ on $mathbb{Z}$.
$endgroup$
– Greg
Dec 23 '18 at 12:57
$begingroup$
As a sidenote: $mathbb{Z}$ can be regarded as an $R$-module by letting $x$ act as $1$ on $mathbb{Z}$.
$endgroup$
– Greg
Dec 23 '18 at 12:57
2
2
$begingroup$
Oh yeah! The resolution continues periodically (assuming $x$ acts as the identity).
$endgroup$
– Jyrki Lahtonen
Dec 23 '18 at 12:57
$begingroup$
Oh yeah! The resolution continues periodically (assuming $x$ acts as the identity).
$endgroup$
– Jyrki Lahtonen
Dec 23 '18 at 12:57
1
1
$begingroup$
Exactly, @JyrkiLahtonen !
$endgroup$
– Greg
Dec 23 '18 at 13:14
$begingroup$
Exactly, @JyrkiLahtonen !
$endgroup$
– Greg
Dec 23 '18 at 13:14
1
1
$begingroup$
Yes, you can. As long as the sequence stays exact!
$endgroup$
– Greg
Dec 23 '18 at 13:26
$begingroup$
Yes, you can. As long as the sequence stays exact!
$endgroup$
– Greg
Dec 23 '18 at 13:26
1
1
$begingroup$
And by the usual map, I just meant that $d_i$ is the map from $R^{I_i} rightarrow R^{I_{i-1}}$.
$endgroup$
– Greg
Dec 23 '18 at 13:27
$begingroup$
And by the usual map, I just meant that $d_i$ is the map from $R^{I_i} rightarrow R^{I_{i-1}}$.
$endgroup$
– Greg
Dec 23 '18 at 13:27
|
show 5 more comments
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1
$begingroup$
Your second guess is the right one. However, it might be better to say that $d_0$ sends $f(overline{x})$ to $f(1)$. What would then be the kernel of $d_1$?
$endgroup$
– Greg
Dec 23 '18 at 12:43
$begingroup$
Do you mean $d_0$ sends $f(bar{x})$ to $f(1)$?
$endgroup$
– idriskameni
Dec 23 '18 at 12:45
1
$begingroup$
Correct, sorry. I have editted the mistake
$endgroup$
– Greg
Dec 23 '18 at 12:45
1
$begingroup$
Agree with Greg. Basically this comes from how do you view $Bbb{Z}$ as an $R$-module. Because $x^n=1$ in $R$, the monomial $x^n$ must act as the identity. This places severe restrictions on how $x$ can act on $Bbb{Z}$. If $n$ is even, then $x$ can act via either $+1$ or $-1$, but if $n$ is odd, then you must have $xcdot m=m$ for all $minBbb{Z}$. Because $d_0$ must be a homomorphism of $R$-modules, it follows that either $x^imapsto d_0(1)$ or $x^imapsto (-1)^id_0(1)$. The latter case corresponding to the possibility that $n$ is even, and $xcdot m=-m$ for all $minBbb{Z}$.
$endgroup$
– Jyrki Lahtonen
Dec 23 '18 at 12:54