Mixing
$k[x]/(x^n)$ module with finite free resolution is free
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How to show a $k[x]/(x^n)$ module with finite free resolution is free?
Suppose we have a exact sequence $k[x]/(x^n)^{oplus n_1}to k[x]/(x^n)^{oplus n_{0}}to Mto 0$, how do we get information to show $M$ is free?
commutative-algebra modules
edited Oct 19 '14 at 8:34
user26857
39.5k124283
asked Oct 19 '14 at 2:56
QixiaoQixiao
2,8721628
$endgroup$
add a comment |
$begingroup$
How to show a $k[x]/(x^n)$ module with finite free resolution is free?
Suppose we have a exact sequence $k[x]/(x^n)^{oplus n_1}to k[x]/(x^n)^{oplus n_{0}}to Mto 0$, how do we get information to show $M$ is free?
commutative-algebra modules
edited Oct 19 '14 at 8:34
user26857
39.5k124283
asked Oct 19 '14 at 2:56
QixiaoQixiao
2,8721628
$endgroup$
add a comment |
$begingroup$
How to show a $k[x]/(x^n)$ module with finite free resolution is free?
Suppose we have a exact sequence $k[x]/(x^n)^{oplus n_1}to k[x]/(x^n)^{oplus n_{0}}to Mto 0$, how do we get information to show $M$ is free?
commutative-algebra modules
edited Oct 19 '14 at 8:34
user26857
39.5k124283
asked Oct 19 '14 at 2:56
QixiaoQixiao
2,8721628
$endgroup$
How to show a $k[x]/(x^n)$ module with finite free resolution is free?
Suppose we have a exact sequence $k[x]/(x^n)^{oplus n_1}to k[x]/(x^n)^{oplus n_{0}}to Mto 0$, how do we get information to show $M$ is free?
commutative-algebra modules
commutative-algebra modules
edited Oct 19 '14 at 8:34
user26857
39.5k124283
asked Oct 19 '14 at 2:56
QixiaoQixiao
2,8721628
edited Oct 19 '14 at 8:34
user26857
39.5k124283
asked Oct 19 '14 at 2:56
QixiaoQixiao
2,8721628
edited Oct 19 '14 at 8:34
user26857
39.5k124283
edited Oct 19 '14 at 8:34
user26857
39.5k124283
edited Oct 19 '14 at 8:34
user26857
39.5k124283
39.5k124283
asked Oct 19 '14 at 2:56
QixiaoQixiao
2,8721628
asked Oct 19 '14 at 2:56
QixiaoQixiao
2,8721628
asked Oct 19 '14 at 2:56
QixiaoQixiao
2,8721628
2,8721628
add a comment |
add a comment |
4 Answers
4
active
oldest
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$R=k[x]/(x^n)$ is an artinian local ring. Let $M$ be an $R$-module having a finite free resolution, or equivalently $operatorname{pd}_RM<infty$. Now we can apply the Auslander-Buchsbaum formula and get $operatorname{pd}_RM=0$, that is, $M$ is projective, hence free.
answered Oct 19 '14 at 8:22
user26857user26857
39.5k124283
$endgroup$
$begingroup$
Great explanation! Actually I found it in Eisenbud's Commutative algebra Ex 1.20... before this claim we calculated there is a infinite free resolution of $R/(x^m)$ for $m<n$, is there any thoughts from this?
$endgroup$
– Qixiao
Oct 19 '14 at 17:25
$begingroup$
@mqx Do you want to find out an infinite free resolution for $R/(x^m)$ when $m<n$?
$endgroup$
– user26857
Oct 19 '14 at 18:11
$begingroup$
I found one, which is periodic $2$ and infinite, I was wondering if that helps to the claim... the exercise appears in the first chapter and may not assume the reader knows Auslander-Buchsbaum formula, though it is a clear explanation.
$endgroup$
– Qixiao
Oct 19 '14 at 19:12
$begingroup$
But the above explanation is clear and general, maybe it is not important to figure out what the arthor hints by mentioning a free resolution of $R/(x^n)$ :)
$endgroup$
– Qixiao
Oct 19 '14 at 19:15
$begingroup$
Any easier explanation is also welcome. I would like to see about how it works in the baby case.
$endgroup$
– Qixiao
Oct 19 '14 at 19:18
|
show 3 more comments
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Here's a more elementary explanation (from Exercise 19.7 in Eisenbud):
Proposition: Let $(R,P)$ be an Artinian local ring, not a field. Then no submodule of $PR^n$ is free.
Proof: $P$ is nilpotent in $R$, say $P^k = 0$, $P^{k-1} ne 0$ for some $k > 1$, so any submodule of $PR^n$ is annihilated by $P^{k-1}$, hence has nontrivial annihilator, and thus cannot be free (being non-faithful).
To apply this, suppose $M$ had a minimal free resolution which was finite of length $ge 1$:
$$0 to R^{n_m} xrightarrow{phi_m} R^{n_{m-1}} to ldots to R^{n_0} to M to 0$$
By minimality, $phi_m(R^{n_m}) subseteq PR^{n_{m-1}}$, but $phi_m$ is injective, so $phi_m(R^{n_m}) cong R^{n_m}$ is free, contradicting the proposition. Thus the finite free resolution must have length $0$, i.e. $0 to R^{n_0} to M to 0$, so $M cong R^{n_0}$ is free.
answered Oct 20 '14 at 22:05
zcnzcn
13.3k1540
$endgroup$
$begingroup$
This is, in fact, what I said in the first (deleted now) comment under my answer: a proof for a particular case of the Auslander-Buchsbaum formula, which is, at its turn, part of the proof of the theorem. (Of course, the proof deals with the $operatorname{depth}R=0$ case instead of $R$ artinian, but the reasoning is almost the same.)
$endgroup$
– user26857
Oct 20 '14 at 22:11
$begingroup$
@user26857: Which theorem are you referring to? Also, I see no reason why you would have deleted your comment - the OP seems to be looking for elementary explanations
$endgroup$
– zcn
Oct 20 '14 at 22:21
$begingroup$
Auslander-Buchsbaum formula = theorem. (My comment was like this: one can avoid using A-B formula by proving it in a particular case likewise $operatorname{depth}R=0$. Then the OP looked satisfied with the use of A-B formula, so I've deleted that comment.)
$endgroup$
– user26857
Oct 20 '14 at 22:49
$begingroup$
Actually the OP wants to see if the first part of exercise 1.23 can help proving the second part.
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– user26857
Oct 20 '14 at 22:50
add a comment |
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You needn't assume the resolution is of finitely generated free modules. You only need to assume it is free and finite, in the sense it is eventually $0$.
$R$ is an injective $R$-module. Indeed, any ideal of $R$ is of the form $(bar x^k)$ for $k=0,ldots,n$ since every element of $R$ can be written as $mu bar x^k$ for $mu$ a unit. Suppose we're given $eta:(bar x^k)to R$, and suppose $eta(bar x^k)=mu bar x^j$. Multiplication by $bar x^{n-k}$ yields that $n-k+jgeqslant n$ so $jgeqslant k$, say $j=k+l$. Define $tildeeta:Rto R$ by $bar 1to mu bar x^l$. Then $tildeeta$ extends $eta$. By Baer's criterion, $R$ is $R$-injective. Since $R$ is local, projective and free modules coincide, and since $R$ is noetherian the theorem of Bass Papp says any direct sum of copies of $R$ is injective, that is any projective $R$-module is injective. But then I claim any module that admits a finite projective resolution must itself be projective. Indeed, say we have a module that admits a resolution $$0to P_nto P_{n-1}to cdots to P_0to Mto 0$$
but no shorter resolution. We can assume $n=1$, since $M'=P_{n-1}/P_{n}$ admits $0to P_{n-1}to P_nto M'to 0$ but cannot itself be projective, else we would be able to shorten $M$'s resolution to one of the form $$0to M'to P_{n-2}tocdots P_0to Mto 0$$ contrary to our assumption. Then for $M'$ we have a SEC $$0to P_1'to P_0'to M'to 0$$ which cannot split. But it does, since $P_1'$ being projective is injective. The conclusion is that any $R$-module with a finite free resolution must in fact be free.
edited Jan 30 at 8:43
Aolong Li
887615
answered Feb 1 '15 at 6:39
Pedro Tamaroff♦Pedro Tamaroff
97.5k10153298
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You actually proved that $k[x]/(x^n)$ is a (quasi)Frobenius ring, and then used this argument.
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– user26857
Feb 1 '15 at 17:16
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@user26857 Thanks for the reference.
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– Pedro Tamaroff♦
Feb 1 '15 at 18:55
add a comment |
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You can use elementary tools to show this. In here elementary methods are used assuming the modules involved are finitely generated.
Set $R=k[x]/(x^n)$. To solve your problem let us prove the following:
Let $M$ be free a $R$-module. If $Bsubseteq M$ is a linearly independent set over $R$, then there is a basis of $M$ over $R$ that contains $B$.
As $M$ is free, there is a set $I$ such that
$$ Msimeqbigoplus_IR,$$
so we consider $M$ as this direct sum. Define the function
$$psi:Mrightarrowbigoplus_I k,$$
by letting for each $iin I$ and each $ain M$ $text{ }psi(a)_i$ to be the constant term of $a_i$. As for each $ain M$ we have $x^{n-1}a=x^{n-1}psi(a)(1)$, from the linear independence of $B$ over $R$ we get
that the set ${psi(x):xin B}$ is a linear independent subset of the $k$-vector space $bigoplus_I k$.
Thus there is a basis $C$ of the $k$-vector space $bigoplus_I k$ such that ${psi(x):xin B}subseteq C$. We claim the set
$$B'=(Csetminus {psi(x):xin B})cup B$$
is a basis of $M$ over $R$; notice $psi$ is $1-1$ on $B'$ and $psi(B')=C$.
Let us see by reverse induction on $0leq jleq n-1$ that for each $ain M$ with $deg(a_i)geq j$ or $a_i=0$ for each $iin I$ we have $ain$Span$_R(B')$
From $(1)$ and because of the construction of $C$ the case $j=n-1$ follows.
If for some $0<jleq n-1$ the property holds, pick any $ain M$ with $deg(a_i)geq j-1$ or $a_i=0$ for all $iin I$, then by the inductive hypothesis we may assume that for each $iin I$, $a_i=b_ix^{j-1}$ for some $b_iin k$. Now consider the set ${x^{j-1}b:bin B'}$, by the inductive hypothesis we may assume that each $x^{j-1}b$ equals $x^{j-1}psi(b)$, hence as $psi(B')=C$ and $psi$ is $1-1$ on $B'$ we get that $ain$Span$_R(B')$.
Finally, let's check $B'$ is linearly independent over $R$. Suppose $c_1,ldots, c_min B'$ are distinct and $p_1,ldots p_min R$ are such that
$$p_1c_1+cdots+p_mc_m=0.$$
Each $p_i$ can be written as $p_i=sum_{j=0}^{n-1}b_{i,j}x^j$. Let us see by induction on $0leq jleq n-1$ that $b_{i,j}=0$ for all $1leq ileq m.$
We have
$$0=x^{n-1}p_1c_1+cdots+x^{n-1}p_mc_m=x^{n-1}a_{1,0}psi(c_1)+cdots+x^{n-1}a_{1,m}psi(c_m),$$
thus as $psi$ is $1-1$ on $B'$ and $psi(B')=C$ we obtain $a_{1,i}=0$ for all $1leq ileq m.$
If for some $0leq j<n-1$ we have that $a_{j',i}=0$ for all $0leq j'leq j$ for all $i$, then
$$0=x^{n-2-j}p_1c_1+cdots+x^{n-2-j}p_mc_m=x^{n-2-j}a_{j+1,0}psi(c_1)+cdots+x^{n-2-j}a_{j+1,m}psi(c_m);$$
since $j+1+(n-2-j)=n-1$, and as we argued in the previous case we get $a_{j+1,i}=0$. Therefore $p_1=cdots=p_m=0$.
Therefore $B'$ is a $R$-basis of $M$ with $B'supseteq B$.
Now suppose $M$ is an $R$-module that has a free finite resolution
$$0rightarrow M_mrightarrow M_{m-1}cdotsrightarrow M_1rightarrow Mrightarrow 0,$$
where the map $M_jrightarrow M_{j-1}$ is denoted by $f_j$, and $M_0=M$.
Let us see by reverse induction on $1leq jleq m$ that $Im(f_j)$ is a free submodule of $M_{j-1}$. The case $j=m$ is trivial as $f_m$ is injective and $M_m$ is free.
So suppose $Im(f_j)$ is free for some $1<jleq m$. Then by what we prove above we know
$$Im(f_{j-1})simeq M_{j-1}/ker(f_{j-1}),$$
is free since $M_{j-1}$ is free and $ker(f_{j-1})=Im(f_j)$ is free.
Therefore $M$ is free.
edited Apr 13 '17 at 12:21
Community♦
1
answered Jul 5 '16 at 18:47
Camilo Arosemena-SerratoCamilo Arosemena-Serrato
5,72611849
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4 Answers
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4 Answers
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$begingroup$
$R=k[x]/(x^n)$ is an artinian local ring. Let $M$ be an $R$-module having a finite free resolution, or equivalently $operatorname{pd}_RM<infty$. Now we can apply the Auslander-Buchsbaum formula and get $operatorname{pd}_RM=0$, that is, $M$ is projective, hence free.
answered Oct 19 '14 at 8:22
user26857user26857
39.5k124283
$endgroup$
$begingroup$
Great explanation! Actually I found it in Eisenbud's Commutative algebra Ex 1.20... before this claim we calculated there is a infinite free resolution of $R/(x^m)$ for $m<n$, is there any thoughts from this?
$endgroup$
– Qixiao
Oct 19 '14 at 17:25
$begingroup$
@mqx Do you want to find out an infinite free resolution for $R/(x^m)$ when $m<n$?
$endgroup$
– user26857
Oct 19 '14 at 18:11
$begingroup$
I found one, which is periodic $2$ and infinite, I was wondering if that helps to the claim... the exercise appears in the first chapter and may not assume the reader knows Auslander-Buchsbaum formula, though it is a clear explanation.
$endgroup$
– Qixiao
Oct 19 '14 at 19:12
$begingroup$
But the above explanation is clear and general, maybe it is not important to figure out what the arthor hints by mentioning a free resolution of $R/(x^n)$ :)
$endgroup$
– Qixiao
Oct 19 '14 at 19:15
$begingroup$
Any easier explanation is also welcome. I would like to see about how it works in the baby case.
$endgroup$
– Qixiao
Oct 19 '14 at 19:18
|
show 3 more comments
$begingroup$
$R=k[x]/(x^n)$ is an artinian local ring. Let $M$ be an $R$-module having a finite free resolution, or equivalently $operatorname{pd}_RM<infty$. Now we can apply the Auslander-Buchsbaum formula and get $operatorname{pd}_RM=0$, that is, $M$ is projective, hence free.
answered Oct 19 '14 at 8:22
user26857user26857
39.5k124283
$endgroup$
$begingroup$
Great explanation! Actually I found it in Eisenbud's Commutative algebra Ex 1.20... before this claim we calculated there is a infinite free resolution of $R/(x^m)$ for $m<n$, is there any thoughts from this?
$endgroup$
– Qixiao
Oct 19 '14 at 17:25
$begingroup$
@mqx Do you want to find out an infinite free resolution for $R/(x^m)$ when $m<n$?
$endgroup$
– user26857
Oct 19 '14 at 18:11
$begingroup$
I found one, which is periodic $2$ and infinite, I was wondering if that helps to the claim... the exercise appears in the first chapter and may not assume the reader knows Auslander-Buchsbaum formula, though it is a clear explanation.
$endgroup$
– Qixiao
Oct 19 '14 at 19:12
$begingroup$
But the above explanation is clear and general, maybe it is not important to figure out what the arthor hints by mentioning a free resolution of $R/(x^n)$ :)
$endgroup$
– Qixiao
Oct 19 '14 at 19:15
$begingroup$
Any easier explanation is also welcome. I would like to see about how it works in the baby case.
$endgroup$
– Qixiao
Oct 19 '14 at 19:18
|
show 3 more comments
$begingroup$
$R=k[x]/(x^n)$ is an artinian local ring. Let $M$ be an $R$-module having a finite free resolution, or equivalently $operatorname{pd}_RM<infty$. Now we can apply the Auslander-Buchsbaum formula and get $operatorname{pd}_RM=0$, that is, $M$ is projective, hence free.
answered Oct 19 '14 at 8:22
user26857user26857
39.5k124283
$endgroup$
$R=k[x]/(x^n)$ is an artinian local ring. Let $M$ be an $R$-module having a finite free resolution, or equivalently $operatorname{pd}_RM<infty$. Now we can apply the Auslander-Buchsbaum formula and get $operatorname{pd}_RM=0$, that is, $M$ is projective, hence free.
answered Oct 19 '14 at 8:22
user26857user26857
39.5k124283
edited Oct 20 '14 at 16:04
answered Oct 19 '14 at 8:22
user26857user26857
39.5k124283
answered Oct 19 '14 at 8:22
user26857user26857
39.5k124283
answered Oct 19 '14 at 8:22
user26857user26857
39.5k124283
39.5k124283
$begingroup$
Great explanation! Actually I found it in Eisenbud's Commutative algebra Ex 1.20... before this claim we calculated there is a infinite free resolution of $R/(x^m)$ for $m<n$, is there any thoughts from this?
$endgroup$
– Qixiao
Oct 19 '14 at 17:25
$begingroup$
@mqx Do you want to find out an infinite free resolution for $R/(x^m)$ when $m<n$?
$endgroup$
– user26857
Oct 19 '14 at 18:11
$begingroup$
I found one, which is periodic $2$ and infinite, I was wondering if that helps to the claim... the exercise appears in the first chapter and may not assume the reader knows Auslander-Buchsbaum formula, though it is a clear explanation.
$endgroup$
– Qixiao
Oct 19 '14 at 19:12
$begingroup$
But the above explanation is clear and general, maybe it is not important to figure out what the arthor hints by mentioning a free resolution of $R/(x^n)$ :)
$endgroup$
– Qixiao
Oct 19 '14 at 19:15
$begingroup$
Any easier explanation is also welcome. I would like to see about how it works in the baby case.
$endgroup$
– Qixiao
Oct 19 '14 at 19:18
|
show 3 more comments
$begingroup$
Great explanation! Actually I found it in Eisenbud's Commutative algebra Ex 1.20... before this claim we calculated there is a infinite free resolution of $R/(x^m)$ for $m<n$, is there any thoughts from this?
$endgroup$
– Qixiao
Oct 19 '14 at 17:25
$begingroup$
@mqx Do you want to find out an infinite free resolution for $R/(x^m)$ when $m<n$?
$endgroup$
– user26857
Oct 19 '14 at 18:11
$begingroup$
I found one, which is periodic $2$ and infinite, I was wondering if that helps to the claim... the exercise appears in the first chapter and may not assume the reader knows Auslander-Buchsbaum formula, though it is a clear explanation.
$endgroup$
– Qixiao
Oct 19 '14 at 19:12
$begingroup$
But the above explanation is clear and general, maybe it is not important to figure out what the arthor hints by mentioning a free resolution of $R/(x^n)$ :)
$endgroup$
– Qixiao
Oct 19 '14 at 19:15
$begingroup$
Any easier explanation is also welcome. I would like to see about how it works in the baby case.
$endgroup$
– Qixiao
Oct 19 '14 at 19:18
$begingroup$
Great explanation! Actually I found it in Eisenbud's Commutative algebra Ex 1.20... before this claim we calculated there is a infinite free resolution of $R/(x^m)$ for $m<n$, is there any thoughts from this?
$endgroup$
– Qixiao
Oct 19 '14 at 17:25
$begingroup$
Great explanation! Actually I found it in Eisenbud's Commutative algebra Ex 1.20... before this claim we calculated there is a infinite free resolution of $R/(x^m)$ for $m<n$, is there any thoughts from this?
$endgroup$
– Qixiao
Oct 19 '14 at 17:25
$begingroup$
@mqx Do you want to find out an infinite free resolution for $R/(x^m)$ when $m<n$?
$endgroup$
– user26857
Oct 19 '14 at 18:11
$begingroup$
@mqx Do you want to find out an infinite free resolution for $R/(x^m)$ when $m<n$?
$endgroup$
– user26857
Oct 19 '14 at 18:11
$begingroup$
I found one, which is periodic $2$ and infinite, I was wondering if that helps to the claim... the exercise appears in the first chapter and may not assume the reader knows Auslander-Buchsbaum formula, though it is a clear explanation.
$endgroup$
– Qixiao
Oct 19 '14 at 19:12
$begingroup$
I found one, which is periodic $2$ and infinite, I was wondering if that helps to the claim... the exercise appears in the first chapter and may not assume the reader knows Auslander-Buchsbaum formula, though it is a clear explanation.
$endgroup$
– Qixiao
Oct 19 '14 at 19:12
$begingroup$
But the above explanation is clear and general, maybe it is not important to figure out what the arthor hints by mentioning a free resolution of $R/(x^n)$ :)
$endgroup$
– Qixiao
Oct 19 '14 at 19:15
$begingroup$
But the above explanation is clear and general, maybe it is not important to figure out what the arthor hints by mentioning a free resolution of $R/(x^n)$ :)
$endgroup$
– Qixiao
Oct 19 '14 at 19:15
$begingroup$
Any easier explanation is also welcome. I would like to see about how it works in the baby case.
$endgroup$
– Qixiao
Oct 19 '14 at 19:18
$begingroup$
Any easier explanation is also welcome. I would like to see about how it works in the baby case.
$endgroup$
– Qixiao
Oct 19 '14 at 19:18
|
show 3 more comments
$begingroup$
Here's a more elementary explanation (from Exercise 19.7 in Eisenbud):
Proposition: Let $(R,P)$ be an Artinian local ring, not a field. Then no submodule of $PR^n$ is free.
Proof: $P$ is nilpotent in $R$, say $P^k = 0$, $P^{k-1} ne 0$ for some $k > 1$, so any submodule of $PR^n$ is annihilated by $P^{k-1}$, hence has nontrivial annihilator, and thus cannot be free (being non-faithful).
To apply this, suppose $M$ had a minimal free resolution which was finite of length $ge 1$:
$$0 to R^{n_m} xrightarrow{phi_m} R^{n_{m-1}} to ldots to R^{n_0} to M to 0$$
By minimality, $phi_m(R^{n_m}) subseteq PR^{n_{m-1}}$, but $phi_m$ is injective, so $phi_m(R^{n_m}) cong R^{n_m}$ is free, contradicting the proposition. Thus the finite free resolution must have length $0$, i.e. $0 to R^{n_0} to M to 0$, so $M cong R^{n_0}$ is free.
answered Oct 20 '14 at 22:05
zcnzcn
13.3k1540
$endgroup$
$begingroup$
This is, in fact, what I said in the first (deleted now) comment under my answer: a proof for a particular case of the Auslander-Buchsbaum formula, which is, at its turn, part of the proof of the theorem. (Of course, the proof deals with the $operatorname{depth}R=0$ case instead of $R$ artinian, but the reasoning is almost the same.)
$endgroup$
– user26857
Oct 20 '14 at 22:11
$begingroup$
@user26857: Which theorem are you referring to? Also, I see no reason why you would have deleted your comment - the OP seems to be looking for elementary explanations
$endgroup$
– zcn
Oct 20 '14 at 22:21
$begingroup$
Auslander-Buchsbaum formula = theorem. (My comment was like this: one can avoid using A-B formula by proving it in a particular case likewise $operatorname{depth}R=0$. Then the OP looked satisfied with the use of A-B formula, so I've deleted that comment.)
$endgroup$
– user26857
Oct 20 '14 at 22:49
$begingroup$
Actually the OP wants to see if the first part of exercise 1.23 can help proving the second part.
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– user26857
Oct 20 '14 at 22:50
add a comment |
$begingroup$
Here's a more elementary explanation (from Exercise 19.7 in Eisenbud):
Proposition: Let $(R,P)$ be an Artinian local ring, not a field. Then no submodule of $PR^n$ is free.
Proof: $P$ is nilpotent in $R$, say $P^k = 0$, $P^{k-1} ne 0$ for some $k > 1$, so any submodule of $PR^n$ is annihilated by $P^{k-1}$, hence has nontrivial annihilator, and thus cannot be free (being non-faithful).
To apply this, suppose $M$ had a minimal free resolution which was finite of length $ge 1$:
$$0 to R^{n_m} xrightarrow{phi_m} R^{n_{m-1}} to ldots to R^{n_0} to M to 0$$
By minimality, $phi_m(R^{n_m}) subseteq PR^{n_{m-1}}$, but $phi_m$ is injective, so $phi_m(R^{n_m}) cong R^{n_m}$ is free, contradicting the proposition. Thus the finite free resolution must have length $0$, i.e. $0 to R^{n_0} to M to 0$, so $M cong R^{n_0}$ is free.
answered Oct 20 '14 at 22:05
zcnzcn
13.3k1540
$endgroup$
$begingroup$
This is, in fact, what I said in the first (deleted now) comment under my answer: a proof for a particular case of the Auslander-Buchsbaum formula, which is, at its turn, part of the proof of the theorem. (Of course, the proof deals with the $operatorname{depth}R=0$ case instead of $R$ artinian, but the reasoning is almost the same.)
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– user26857
Oct 20 '14 at 22:11
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@user26857: Which theorem are you referring to? Also, I see no reason why you would have deleted your comment - the OP seems to be looking for elementary explanations
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– zcn
Oct 20 '14 at 22:21
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Auslander-Buchsbaum formula = theorem. (My comment was like this: one can avoid using A-B formula by proving it in a particular case likewise $operatorname{depth}R=0$. Then the OP looked satisfied with the use of A-B formula, so I've deleted that comment.)
$endgroup$
– user26857
Oct 20 '14 at 22:49
$begingroup$
Actually the OP wants to see if the first part of exercise 1.23 can help proving the second part.
$endgroup$
– user26857
Oct 20 '14 at 22:50
add a comment |
$begingroup$
Here's a more elementary explanation (from Exercise 19.7 in Eisenbud):
Proposition: Let $(R,P)$ be an Artinian local ring, not a field. Then no submodule of $PR^n$ is free.
Proof: $P$ is nilpotent in $R$, say $P^k = 0$, $P^{k-1} ne 0$ for some $k > 1$, so any submodule of $PR^n$ is annihilated by $P^{k-1}$, hence has nontrivial annihilator, and thus cannot be free (being non-faithful).
To apply this, suppose $M$ had a minimal free resolution which was finite of length $ge 1$:
$$0 to R^{n_m} xrightarrow{phi_m} R^{n_{m-1}} to ldots to R^{n_0} to M to 0$$
By minimality, $phi_m(R^{n_m}) subseteq PR^{n_{m-1}}$, but $phi_m$ is injective, so $phi_m(R^{n_m}) cong R^{n_m}$ is free, contradicting the proposition. Thus the finite free resolution must have length $0$, i.e. $0 to R^{n_0} to M to 0$, so $M cong R^{n_0}$ is free.
answered Oct 20 '14 at 22:05
zcnzcn
13.3k1540
$endgroup$
Here's a more elementary explanation (from Exercise 19.7 in Eisenbud):
Proposition: Let $(R,P)$ be an Artinian local ring, not a field. Then no submodule of $PR^n$ is free.
Proof: $P$ is nilpotent in $R$, say $P^k = 0$, $P^{k-1} ne 0$ for some $k > 1$, so any submodule of $PR^n$ is annihilated by $P^{k-1}$, hence has nontrivial annihilator, and thus cannot be free (being non-faithful).
To apply this, suppose $M$ had a minimal free resolution which was finite of length $ge 1$:
$$0 to R^{n_m} xrightarrow{phi_m} R^{n_{m-1}} to ldots to R^{n_0} to M to 0$$
By minimality, $phi_m(R^{n_m}) subseteq PR^{n_{m-1}}$, but $phi_m$ is injective, so $phi_m(R^{n_m}) cong R^{n_m}$ is free, contradicting the proposition. Thus the finite free resolution must have length $0$, i.e. $0 to R^{n_0} to M to 0$, so $M cong R^{n_0}$ is free.
answered Oct 20 '14 at 22:05
zcnzcn
13.3k1540
edited Oct 20 '14 at 22:27
answered Oct 20 '14 at 22:05
zcnzcn
13.3k1540
answered Oct 20 '14 at 22:05
zcnzcn
13.3k1540
answered Oct 20 '14 at 22:05
zcnzcn
13.3k1540
13.3k1540
$begingroup$
This is, in fact, what I said in the first (deleted now) comment under my answer: a proof for a particular case of the Auslander-Buchsbaum formula, which is, at its turn, part of the proof of the theorem. (Of course, the proof deals with the $operatorname{depth}R=0$ case instead of $R$ artinian, but the reasoning is almost the same.)
$endgroup$
– user26857
Oct 20 '14 at 22:11
$begingroup$
@user26857: Which theorem are you referring to? Also, I see no reason why you would have deleted your comment - the OP seems to be looking for elementary explanations
$endgroup$
– zcn
Oct 20 '14 at 22:21
$begingroup$
Auslander-Buchsbaum formula = theorem. (My comment was like this: one can avoid using A-B formula by proving it in a particular case likewise $operatorname{depth}R=0$. Then the OP looked satisfied with the use of A-B formula, so I've deleted that comment.)
$endgroup$
– user26857
Oct 20 '14 at 22:49
$begingroup$
Actually the OP wants to see if the first part of exercise 1.23 can help proving the second part.
$endgroup$
– user26857
Oct 20 '14 at 22:50
add a comment |
$begingroup$
This is, in fact, what I said in the first (deleted now) comment under my answer: a proof for a particular case of the Auslander-Buchsbaum formula, which is, at its turn, part of the proof of the theorem. (Of course, the proof deals with the $operatorname{depth}R=0$ case instead of $R$ artinian, but the reasoning is almost the same.)
$endgroup$
– user26857
Oct 20 '14 at 22:11
$begingroup$
@user26857: Which theorem are you referring to? Also, I see no reason why you would have deleted your comment - the OP seems to be looking for elementary explanations
$endgroup$
– zcn
Oct 20 '14 at 22:21
$begingroup$
Auslander-Buchsbaum formula = theorem. (My comment was like this: one can avoid using A-B formula by proving it in a particular case likewise $operatorname{depth}R=0$. Then the OP looked satisfied with the use of A-B formula, so I've deleted that comment.)
$endgroup$
– user26857
Oct 20 '14 at 22:49
$begingroup$
Actually the OP wants to see if the first part of exercise 1.23 can help proving the second part.
$endgroup$
– user26857
Oct 20 '14 at 22:50
$begingroup$
This is, in fact, what I said in the first (deleted now) comment under my answer: a proof for a particular case of the Auslander-Buchsbaum formula, which is, at its turn, part of the proof of the theorem. (Of course, the proof deals with the $operatorname{depth}R=0$ case instead of $R$ artinian, but the reasoning is almost the same.)
$endgroup$
– user26857
Oct 20 '14 at 22:11
$begingroup$
This is, in fact, what I said in the first (deleted now) comment under my answer: a proof for a particular case of the Auslander-Buchsbaum formula, which is, at its turn, part of the proof of the theorem. (Of course, the proof deals with the $operatorname{depth}R=0$ case instead of $R$ artinian, but the reasoning is almost the same.)
$endgroup$
– user26857
Oct 20 '14 at 22:11
$begingroup$
@user26857: Which theorem are you referring to? Also, I see no reason why you would have deleted your comment - the OP seems to be looking for elementary explanations
$endgroup$
– zcn
Oct 20 '14 at 22:21
$begingroup$
@user26857: Which theorem are you referring to? Also, I see no reason why you would have deleted your comment - the OP seems to be looking for elementary explanations
$endgroup$
– zcn
Oct 20 '14 at 22:21
$begingroup$
Auslander-Buchsbaum formula = theorem. (My comment was like this: one can avoid using A-B formula by proving it in a particular case likewise $operatorname{depth}R=0$. Then the OP looked satisfied with the use of A-B formula, so I've deleted that comment.)
$endgroup$
– user26857
Oct 20 '14 at 22:49
$begingroup$
Auslander-Buchsbaum formula = theorem. (My comment was like this: one can avoid using A-B formula by proving it in a particular case likewise $operatorname{depth}R=0$. Then the OP looked satisfied with the use of A-B formula, so I've deleted that comment.)
$endgroup$
– user26857
Oct 20 '14 at 22:49
$begingroup$
Actually the OP wants to see if the first part of exercise 1.23 can help proving the second part.
$endgroup$
– user26857
Oct 20 '14 at 22:50
$begingroup$
Actually the OP wants to see if the first part of exercise 1.23 can help proving the second part.
$endgroup$
– user26857
Oct 20 '14 at 22:50
add a comment |
$begingroup$
You needn't assume the resolution is of finitely generated free modules. You only need to assume it is free and finite, in the sense it is eventually $0$.
$R$ is an injective $R$-module. Indeed, any ideal of $R$ is of the form $(bar x^k)$ for $k=0,ldots,n$ since every element of $R$ can be written as $mu bar x^k$ for $mu$ a unit. Suppose we're given $eta:(bar x^k)to R$, and suppose $eta(bar x^k)=mu bar x^j$. Multiplication by $bar x^{n-k}$ yields that $n-k+jgeqslant n$ so $jgeqslant k$, say $j=k+l$. Define $tildeeta:Rto R$ by $bar 1to mu bar x^l$. Then $tildeeta$ extends $eta$. By Baer's criterion, $R$ is $R$-injective. Since $R$ is local, projective and free modules coincide, and since $R$ is noetherian the theorem of Bass Papp says any direct sum of copies of $R$ is injective, that is any projective $R$-module is injective. But then I claim any module that admits a finite projective resolution must itself be projective. Indeed, say we have a module that admits a resolution $$0to P_nto P_{n-1}to cdots to P_0to Mto 0$$
but no shorter resolution. We can assume $n=1$, since $M'=P_{n-1}/P_{n}$ admits $0to P_{n-1}to P_nto M'to 0$ but cannot itself be projective, else we would be able to shorten $M$'s resolution to one of the form $$0to M'to P_{n-2}tocdots P_0to Mto 0$$ contrary to our assumption. Then for $M'$ we have a SEC $$0to P_1'to P_0'to M'to 0$$ which cannot split. But it does, since $P_1'$ being projective is injective. The conclusion is that any $R$-module with a finite free resolution must in fact be free.
edited Jan 30 at 8:43
Aolong Li
887615
answered Feb 1 '15 at 6:39
Pedro Tamaroff♦Pedro Tamaroff
97.5k10153298
$endgroup$
$begingroup$
You actually proved that $k[x]/(x^n)$ is a (quasi)Frobenius ring, and then used this argument.
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– user26857
Feb 1 '15 at 17:16
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@user26857 Thanks for the reference.
$endgroup$
– Pedro Tamaroff♦
Feb 1 '15 at 18:55
add a comment |
$begingroup$
You needn't assume the resolution is of finitely generated free modules. You only need to assume it is free and finite, in the sense it is eventually $0$.
$R$ is an injective $R$-module. Indeed, any ideal of $R$ is of the form $(bar x^k)$ for $k=0,ldots,n$ since every element of $R$ can be written as $mu bar x^k$ for $mu$ a unit. Suppose we're given $eta:(bar x^k)to R$, and suppose $eta(bar x^k)=mu bar x^j$. Multiplication by $bar x^{n-k}$ yields that $n-k+jgeqslant n$ so $jgeqslant k$, say $j=k+l$. Define $tildeeta:Rto R$ by $bar 1to mu bar x^l$. Then $tildeeta$ extends $eta$. By Baer's criterion, $R$ is $R$-injective. Since $R$ is local, projective and free modules coincide, and since $R$ is noetherian the theorem of Bass Papp says any direct sum of copies of $R$ is injective, that is any projective $R$-module is injective. But then I claim any module that admits a finite projective resolution must itself be projective. Indeed, say we have a module that admits a resolution $$0to P_nto P_{n-1}to cdots to P_0to Mto 0$$
but no shorter resolution. We can assume $n=1$, since $M'=P_{n-1}/P_{n}$ admits $0to P_{n-1}to P_nto M'to 0$ but cannot itself be projective, else we would be able to shorten $M$'s resolution to one of the form $$0to M'to P_{n-2}tocdots P_0to Mto 0$$ contrary to our assumption. Then for $M'$ we have a SEC $$0to P_1'to P_0'to M'to 0$$ which cannot split. But it does, since $P_1'$ being projective is injective. The conclusion is that any $R$-module with a finite free resolution must in fact be free.
edited Jan 30 at 8:43
Aolong Li
887615
answered Feb 1 '15 at 6:39
Pedro Tamaroff♦Pedro Tamaroff
97.5k10153298
$endgroup$
$begingroup$
You actually proved that $k[x]/(x^n)$ is a (quasi)Frobenius ring, and then used this argument.
$endgroup$
– user26857
Feb 1 '15 at 17:16
$begingroup$
@user26857 Thanks for the reference.
$endgroup$
– Pedro Tamaroff♦
Feb 1 '15 at 18:55
add a comment |
$begingroup$
You needn't assume the resolution is of finitely generated free modules. You only need to assume it is free and finite, in the sense it is eventually $0$.
$R$ is an injective $R$-module. Indeed, any ideal of $R$ is of the form $(bar x^k)$ for $k=0,ldots,n$ since every element of $R$ can be written as $mu bar x^k$ for $mu$ a unit. Suppose we're given $eta:(bar x^k)to R$, and suppose $eta(bar x^k)=mu bar x^j$. Multiplication by $bar x^{n-k}$ yields that $n-k+jgeqslant n$ so $jgeqslant k$, say $j=k+l$. Define $tildeeta:Rto R$ by $bar 1to mu bar x^l$. Then $tildeeta$ extends $eta$. By Baer's criterion, $R$ is $R$-injective. Since $R$ is local, projective and free modules coincide, and since $R$ is noetherian the theorem of Bass Papp says any direct sum of copies of $R$ is injective, that is any projective $R$-module is injective. But then I claim any module that admits a finite projective resolution must itself be projective. Indeed, say we have a module that admits a resolution $$0to P_nto P_{n-1}to cdots to P_0to Mto 0$$
but no shorter resolution. We can assume $n=1$, since $M'=P_{n-1}/P_{n}$ admits $0to P_{n-1}to P_nto M'to 0$ but cannot itself be projective, else we would be able to shorten $M$'s resolution to one of the form $$0to M'to P_{n-2}tocdots P_0to Mto 0$$ contrary to our assumption. Then for $M'$ we have a SEC $$0to P_1'to P_0'to M'to 0$$ which cannot split. But it does, since $P_1'$ being projective is injective. The conclusion is that any $R$-module with a finite free resolution must in fact be free.
edited Jan 30 at 8:43
Aolong Li
887615
answered Feb 1 '15 at 6:39
Pedro Tamaroff♦Pedro Tamaroff
97.5k10153298
$endgroup$
You needn't assume the resolution is of finitely generated free modules. You only need to assume it is free and finite, in the sense it is eventually $0$.
$R$ is an injective $R$-module. Indeed, any ideal of $R$ is of the form $(bar x^k)$ for $k=0,ldots,n$ since every element of $R$ can be written as $mu bar x^k$ for $mu$ a unit. Suppose we're given $eta:(bar x^k)to R$, and suppose $eta(bar x^k)=mu bar x^j$. Multiplication by $bar x^{n-k}$ yields that $n-k+jgeqslant n$ so $jgeqslant k$, say $j=k+l$. Define $tildeeta:Rto R$ by $bar 1to mu bar x^l$. Then $tildeeta$ extends $eta$. By Baer's criterion, $R$ is $R$-injective. Since $R$ is local, projective and free modules coincide, and since $R$ is noetherian the theorem of Bass Papp says any direct sum of copies of $R$ is injective, that is any projective $R$-module is injective. But then I claim any module that admits a finite projective resolution must itself be projective. Indeed, say we have a module that admits a resolution $$0to P_nto P_{n-1}to cdots to P_0to Mto 0$$
but no shorter resolution. We can assume $n=1$, since $M'=P_{n-1}/P_{n}$ admits $0to P_{n-1}to P_nto M'to 0$ but cannot itself be projective, else we would be able to shorten $M$'s resolution to one of the form $$0to M'to P_{n-2}tocdots P_0to Mto 0$$ contrary to our assumption. Then for $M'$ we have a SEC $$0to P_1'to P_0'to M'to 0$$ which cannot split. But it does, since $P_1'$ being projective is injective. The conclusion is that any $R$-module with a finite free resolution must in fact be free.
edited Jan 30 at 8:43
Aolong Li
887615
answered Feb 1 '15 at 6:39
Pedro Tamaroff♦Pedro Tamaroff
97.5k10153298
edited Jan 30 at 8:43
Aolong Li
887615
edited Jan 30 at 8:43
Aolong Li
887615
edited Jan 30 at 8:43
Aolong Li
887615
887615
answered Feb 1 '15 at 6:39
Pedro Tamaroff♦Pedro Tamaroff
97.5k10153298
answered Feb 1 '15 at 6:39
Pedro Tamaroff♦Pedro Tamaroff
97.5k10153298
answered Feb 1 '15 at 6:39
Pedro Tamaroff♦Pedro Tamaroff
97.5k10153298
97.5k10153298
$begingroup$
You actually proved that $k[x]/(x^n)$ is a (quasi)Frobenius ring, and then used this argument.
$endgroup$
– user26857
Feb 1 '15 at 17:16
$begingroup$
@user26857 Thanks for the reference.
$endgroup$
– Pedro Tamaroff♦
Feb 1 '15 at 18:55
add a comment |
$begingroup$
You actually proved that $k[x]/(x^n)$ is a (quasi)Frobenius ring, and then used this argument.
$endgroup$
– user26857
Feb 1 '15 at 17:16
$begingroup$
@user26857 Thanks for the reference.
$endgroup$
– Pedro Tamaroff♦
Feb 1 '15 at 18:55
$begingroup$
You actually proved that $k[x]/(x^n)$ is a (quasi)Frobenius ring, and then used this argument.
$endgroup$
– user26857
Feb 1 '15 at 17:16
$begingroup$
You actually proved that $k[x]/(x^n)$ is a (quasi)Frobenius ring, and then used this argument.
$endgroup$
– user26857
Feb 1 '15 at 17:16
$begingroup$
@user26857 Thanks for the reference.
$endgroup$
– Pedro Tamaroff♦
Feb 1 '15 at 18:55
$begingroup$
@user26857 Thanks for the reference.
$endgroup$
– Pedro Tamaroff♦
Feb 1 '15 at 18:55
add a comment |
$begingroup$
You can use elementary tools to show this. In here elementary methods are used assuming the modules involved are finitely generated.
Set $R=k[x]/(x^n)$. To solve your problem let us prove the following:
Let $M$ be free a $R$-module. If $Bsubseteq M$ is a linearly independent set over $R$, then there is a basis of $M$ over $R$ that contains $B$.
As $M$ is free, there is a set $I$ such that
$$ Msimeqbigoplus_IR,$$
so we consider $M$ as this direct sum. Define the function
$$psi:Mrightarrowbigoplus_I k,$$
by letting for each $iin I$ and each $ain M$ $text{ }psi(a)_i$ to be the constant term of $a_i$. As for each $ain M$ we have $x^{n-1}a=x^{n-1}psi(a)(1)$, from the linear independence of $B$ over $R$ we get
that the set ${psi(x):xin B}$ is a linear independent subset of the $k$-vector space $bigoplus_I k$.
Thus there is a basis $C$ of the $k$-vector space $bigoplus_I k$ such that ${psi(x):xin B}subseteq C$. We claim the set
$$B'=(Csetminus {psi(x):xin B})cup B$$
is a basis of $M$ over $R$; notice $psi$ is $1-1$ on $B'$ and $psi(B')=C$.
Let us see by reverse induction on $0leq jleq n-1$ that for each $ain M$ with $deg(a_i)geq j$ or $a_i=0$ for each $iin I$ we have $ain$Span$_R(B')$
From $(1)$ and because of the construction of $C$ the case $j=n-1$ follows.
If for some $0<jleq n-1$ the property holds, pick any $ain M$ with $deg(a_i)geq j-1$ or $a_i=0$ for all $iin I$, then by the inductive hypothesis we may assume that for each $iin I$, $a_i=b_ix^{j-1}$ for some $b_iin k$. Now consider the set ${x^{j-1}b:bin B'}$, by the inductive hypothesis we may assume that each $x^{j-1}b$ equals $x^{j-1}psi(b)$, hence as $psi(B')=C$ and $psi$ is $1-1$ on $B'$ we get that $ain$Span$_R(B')$.
Finally, let's check $B'$ is linearly independent over $R$. Suppose $c_1,ldots, c_min B'$ are distinct and $p_1,ldots p_min R$ are such that
$$p_1c_1+cdots+p_mc_m=0.$$
Each $p_i$ can be written as $p_i=sum_{j=0}^{n-1}b_{i,j}x^j$. Let us see by induction on $0leq jleq n-1$ that $b_{i,j}=0$ for all $1leq ileq m.$
We have
$$0=x^{n-1}p_1c_1+cdots+x^{n-1}p_mc_m=x^{n-1}a_{1,0}psi(c_1)+cdots+x^{n-1}a_{1,m}psi(c_m),$$
thus as $psi$ is $1-1$ on $B'$ and $psi(B')=C$ we obtain $a_{1,i}=0$ for all $1leq ileq m.$
If for some $0leq j<n-1$ we have that $a_{j',i}=0$ for all $0leq j'leq j$ for all $i$, then
$$0=x^{n-2-j}p_1c_1+cdots+x^{n-2-j}p_mc_m=x^{n-2-j}a_{j+1,0}psi(c_1)+cdots+x^{n-2-j}a_{j+1,m}psi(c_m);$$
since $j+1+(n-2-j)=n-1$, and as we argued in the previous case we get $a_{j+1,i}=0$. Therefore $p_1=cdots=p_m=0$.
Therefore $B'$ is a $R$-basis of $M$ with $B'supseteq B$.
Now suppose $M$ is an $R$-module that has a free finite resolution
$$0rightarrow M_mrightarrow M_{m-1}cdotsrightarrow M_1rightarrow Mrightarrow 0,$$
where the map $M_jrightarrow M_{j-1}$ is denoted by $f_j$, and $M_0=M$.
Let us see by reverse induction on $1leq jleq m$ that $Im(f_j)$ is a free submodule of $M_{j-1}$. The case $j=m$ is trivial as $f_m$ is injective and $M_m$ is free.
So suppose $Im(f_j)$ is free for some $1<jleq m$. Then by what we prove above we know
$$Im(f_{j-1})simeq M_{j-1}/ker(f_{j-1}),$$
is free since $M_{j-1}$ is free and $ker(f_{j-1})=Im(f_j)$ is free.
Therefore $M$ is free.
edited Apr 13 '17 at 12:21
Community♦
1
answered Jul 5 '16 at 18:47
Camilo Arosemena-SerratoCamilo Arosemena-Serrato
5,72611849
$endgroup$
add a comment |
$begingroup$
You can use elementary tools to show this. In here elementary methods are used assuming the modules involved are finitely generated.
Set $R=k[x]/(x^n)$. To solve your problem let us prove the following:
Let $M$ be free a $R$-module. If $Bsubseteq M$ is a linearly independent set over $R$, then there is a basis of $M$ over $R$ that contains $B$.
As $M$ is free, there is a set $I$ such that
$$ Msimeqbigoplus_IR,$$
so we consider $M$ as this direct sum. Define the function
$$psi:Mrightarrowbigoplus_I k,$$
by letting for each $iin I$ and each $ain M$ $text{ }psi(a)_i$ to be the constant term of $a_i$. As for each $ain M$ we have $x^{n-1}a=x^{n-1}psi(a)(1)$, from the linear independence of $B$ over $R$ we get
that the set ${psi(x):xin B}$ is a linear independent subset of the $k$-vector space $bigoplus_I k$.
Thus there is a basis $C$ of the $k$-vector space $bigoplus_I k$ such that ${psi(x):xin B}subseteq C$. We claim the set
$$B'=(Csetminus {psi(x):xin B})cup B$$
is a basis of $M$ over $R$; notice $psi$ is $1-1$ on $B'$ and $psi(B')=C$.
Let us see by reverse induction on $0leq jleq n-1$ that for each $ain M$ with $deg(a_i)geq j$ or $a_i=0$ for each $iin I$ we have $ain$Span$_R(B')$
From $(1)$ and because of the construction of $C$ the case $j=n-1$ follows.
If for some $0<jleq n-1$ the property holds, pick any $ain M$ with $deg(a_i)geq j-1$ or $a_i=0$ for all $iin I$, then by the inductive hypothesis we may assume that for each $iin I$, $a_i=b_ix^{j-1}$ for some $b_iin k$. Now consider the set ${x^{j-1}b:bin B'}$, by the inductive hypothesis we may assume that each $x^{j-1}b$ equals $x^{j-1}psi(b)$, hence as $psi(B')=C$ and $psi$ is $1-1$ on $B'$ we get that $ain$Span$_R(B')$.
Finally, let's check $B'$ is linearly independent over $R$. Suppose $c_1,ldots, c_min B'$ are distinct and $p_1,ldots p_min R$ are such that
$$p_1c_1+cdots+p_mc_m=0.$$
Each $p_i$ can be written as $p_i=sum_{j=0}^{n-1}b_{i,j}x^j$. Let us see by induction on $0leq jleq n-1$ that $b_{i,j}=0$ for all $1leq ileq m.$
We have
$$0=x^{n-1}p_1c_1+cdots+x^{n-1}p_mc_m=x^{n-1}a_{1,0}psi(c_1)+cdots+x^{n-1}a_{1,m}psi(c_m),$$
thus as $psi$ is $1-1$ on $B'$ and $psi(B')=C$ we obtain $a_{1,i}=0$ for all $1leq ileq m.$
If for some $0leq j<n-1$ we have that $a_{j',i}=0$ for all $0leq j'leq j$ for all $i$, then
$$0=x^{n-2-j}p_1c_1+cdots+x^{n-2-j}p_mc_m=x^{n-2-j}a_{j+1,0}psi(c_1)+cdots+x^{n-2-j}a_{j+1,m}psi(c_m);$$
since $j+1+(n-2-j)=n-1$, and as we argued in the previous case we get $a_{j+1,i}=0$. Therefore $p_1=cdots=p_m=0$.
Therefore $B'$ is a $R$-basis of $M$ with $B'supseteq B$.
Now suppose $M$ is an $R$-module that has a free finite resolution
$$0rightarrow M_mrightarrow M_{m-1}cdotsrightarrow M_1rightarrow Mrightarrow 0,$$
where the map $M_jrightarrow M_{j-1}$ is denoted by $f_j$, and $M_0=M$.
Let us see by reverse induction on $1leq jleq m$ that $Im(f_j)$ is a free submodule of $M_{j-1}$. The case $j=m$ is trivial as $f_m$ is injective and $M_m$ is free.
So suppose $Im(f_j)$ is free for some $1<jleq m$. Then by what we prove above we know
$$Im(f_{j-1})simeq M_{j-1}/ker(f_{j-1}),$$
is free since $M_{j-1}$ is free and $ker(f_{j-1})=Im(f_j)$ is free.
Therefore $M$ is free.
edited Apr 13 '17 at 12:21
Community♦
1
answered Jul 5 '16 at 18:47
Camilo Arosemena-SerratoCamilo Arosemena-Serrato
5,72611849
$endgroup$
add a comment |
$begingroup$
You can use elementary tools to show this. In here elementary methods are used assuming the modules involved are finitely generated.
Set $R=k[x]/(x^n)$. To solve your problem let us prove the following:
Let $M$ be free a $R$-module. If $Bsubseteq M$ is a linearly independent set over $R$, then there is a basis of $M$ over $R$ that contains $B$.
As $M$ is free, there is a set $I$ such that
$$ Msimeqbigoplus_IR,$$
so we consider $M$ as this direct sum. Define the function
$$psi:Mrightarrowbigoplus_I k,$$
by letting for each $iin I$ and each $ain M$ $text{ }psi(a)_i$ to be the constant term of $a_i$. As for each $ain M$ we have $x^{n-1}a=x^{n-1}psi(a)(1)$, from the linear independence of $B$ over $R$ we get
that the set ${psi(x):xin B}$ is a linear independent subset of the $k$-vector space $bigoplus_I k$.
Thus there is a basis $C$ of the $k$-vector space $bigoplus_I k$ such that ${psi(x):xin B}subseteq C$. We claim the set
$$B'=(Csetminus {psi(x):xin B})cup B$$
is a basis of $M$ over $R$; notice $psi$ is $1-1$ on $B'$ and $psi(B')=C$.
Let us see by reverse induction on $0leq jleq n-1$ that for each $ain M$ with $deg(a_i)geq j$ or $a_i=0$ for each $iin I$ we have $ain$Span$_R(B')$
From $(1)$ and because of the construction of $C$ the case $j=n-1$ follows.
If for some $0<jleq n-1$ the property holds, pick any $ain M$ with $deg(a_i)geq j-1$ or $a_i=0$ for all $iin I$, then by the inductive hypothesis we may assume that for each $iin I$, $a_i=b_ix^{j-1}$ for some $b_iin k$. Now consider the set ${x^{j-1}b:bin B'}$, by the inductive hypothesis we may assume that each $x^{j-1}b$ equals $x^{j-1}psi(b)$, hence as $psi(B')=C$ and $psi$ is $1-1$ on $B'$ we get that $ain$Span$_R(B')$.
Finally, let's check $B'$ is linearly independent over $R$. Suppose $c_1,ldots, c_min B'$ are distinct and $p_1,ldots p_min R$ are such that
$$p_1c_1+cdots+p_mc_m=0.$$
Each $p_i$ can be written as $p_i=sum_{j=0}^{n-1}b_{i,j}x^j$. Let us see by induction on $0leq jleq n-1$ that $b_{i,j}=0$ for all $1leq ileq m.$
We have
$$0=x^{n-1}p_1c_1+cdots+x^{n-1}p_mc_m=x^{n-1}a_{1,0}psi(c_1)+cdots+x^{n-1}a_{1,m}psi(c_m),$$
thus as $psi$ is $1-1$ on $B'$ and $psi(B')=C$ we obtain $a_{1,i}=0$ for all $1leq ileq m.$
If for some $0leq j<n-1$ we have that $a_{j',i}=0$ for all $0leq j'leq j$ for all $i$, then
$$0=x^{n-2-j}p_1c_1+cdots+x^{n-2-j}p_mc_m=x^{n-2-j}a_{j+1,0}psi(c_1)+cdots+x^{n-2-j}a_{j+1,m}psi(c_m);$$
since $j+1+(n-2-j)=n-1$, and as we argued in the previous case we get $a_{j+1,i}=0$. Therefore $p_1=cdots=p_m=0$.
Therefore $B'$ is a $R$-basis of $M$ with $B'supseteq B$.
Now suppose $M$ is an $R$-module that has a free finite resolution
$$0rightarrow M_mrightarrow M_{m-1}cdotsrightarrow M_1rightarrow Mrightarrow 0,$$
where the map $M_jrightarrow M_{j-1}$ is denoted by $f_j$, and $M_0=M$.
Let us see by reverse induction on $1leq jleq m$ that $Im(f_j)$ is a free submodule of $M_{j-1}$. The case $j=m$ is trivial as $f_m$ is injective and $M_m$ is free.
So suppose $Im(f_j)$ is free for some $1<jleq m$. Then by what we prove above we know
$$Im(f_{j-1})simeq M_{j-1}/ker(f_{j-1}),$$
is free since $M_{j-1}$ is free and $ker(f_{j-1})=Im(f_j)$ is free.
Therefore $M$ is free.
edited Apr 13 '17 at 12:21
Community♦
1
answered Jul 5 '16 at 18:47
Camilo Arosemena-SerratoCamilo Arosemena-Serrato
5,72611849
$endgroup$
You can use elementary tools to show this. In here elementary methods are used assuming the modules involved are finitely generated.
Set $R=k[x]/(x^n)$. To solve your problem let us prove the following:
Let $M$ be free a $R$-module. If $Bsubseteq M$ is a linearly independent set over $R$, then there is a basis of $M$ over $R$ that contains $B$.
As $M$ is free, there is a set $I$ such that
$$ Msimeqbigoplus_IR,$$
so we consider $M$ as this direct sum. Define the function
$$psi:Mrightarrowbigoplus_I k,$$
by letting for each $iin I$ and each $ain M$ $text{ }psi(a)_i$ to be the constant term of $a_i$. As for each $ain M$ we have $x^{n-1}a=x^{n-1}psi(a)(1)$, from the linear independence of $B$ over $R$ we get
that the set ${psi(x):xin B}$ is a linear independent subset of the $k$-vector space $bigoplus_I k$.
Thus there is a basis $C$ of the $k$-vector space $bigoplus_I k$ such that ${psi(x):xin B}subseteq C$. We claim the set
$$B'=(Csetminus {psi(x):xin B})cup B$$
is a basis of $M$ over $R$; notice $psi$ is $1-1$ on $B'$ and $psi(B')=C$.
Let us see by reverse induction on $0leq jleq n-1$ that for each $ain M$ with $deg(a_i)geq j$ or $a_i=0$ for each $iin I$ we have $ain$Span$_R(B')$
From $(1)$ and because of the construction of $C$ the case $j=n-1$ follows.
If for some $0<jleq n-1$ the property holds, pick any $ain M$ with $deg(a_i)geq j-1$ or $a_i=0$ for all $iin I$, then by the inductive hypothesis we may assume that for each $iin I$, $a_i=b_ix^{j-1}$ for some $b_iin k$. Now consider the set ${x^{j-1}b:bin B'}$, by the inductive hypothesis we may assume that each $x^{j-1}b$ equals $x^{j-1}psi(b)$, hence as $psi(B')=C$ and $psi$ is $1-1$ on $B'$ we get that $ain$Span$_R(B')$.
Finally, let's check $B'$ is linearly independent over $R$. Suppose $c_1,ldots, c_min B'$ are distinct and $p_1,ldots p_min R$ are such that
$$p_1c_1+cdots+p_mc_m=0.$$
Each $p_i$ can be written as $p_i=sum_{j=0}^{n-1}b_{i,j}x^j$. Let us see by induction on $0leq jleq n-1$ that $b_{i,j}=0$ for all $1leq ileq m.$
We have
$$0=x^{n-1}p_1c_1+cdots+x^{n-1}p_mc_m=x^{n-1}a_{1,0}psi(c_1)+cdots+x^{n-1}a_{1,m}psi(c_m),$$
thus as $psi$ is $1-1$ on $B'$ and $psi(B')=C$ we obtain $a_{1,i}=0$ for all $1leq ileq m.$
If for some $0leq j<n-1$ we have that $a_{j',i}=0$ for all $0leq j'leq j$ for all $i$, then
$$0=x^{n-2-j}p_1c_1+cdots+x^{n-2-j}p_mc_m=x^{n-2-j}a_{j+1,0}psi(c_1)+cdots+x^{n-2-j}a_{j+1,m}psi(c_m);$$
since $j+1+(n-2-j)=n-1$, and as we argued in the previous case we get $a_{j+1,i}=0$. Therefore $p_1=cdots=p_m=0$.
Therefore $B'$ is a $R$-basis of $M$ with $B'supseteq B$.
Now suppose $M$ is an $R$-module that has a free finite resolution
$$0rightarrow M_mrightarrow M_{m-1}cdotsrightarrow M_1rightarrow Mrightarrow 0,$$
where the map $M_jrightarrow M_{j-1}$ is denoted by $f_j$, and $M_0=M$.
Let us see by reverse induction on $1leq jleq m$ that $Im(f_j)$ is a free submodule of $M_{j-1}$. The case $j=m$ is trivial as $f_m$ is injective and $M_m$ is free.
So suppose $Im(f_j)$ is free for some $1<jleq m$. Then by what we prove above we know
$$Im(f_{j-1})simeq M_{j-1}/ker(f_{j-1}),$$
is free since $M_{j-1}$ is free and $ker(f_{j-1})=Im(f_j)$ is free.
Therefore $M$ is free.
edited Apr 13 '17 at 12:21
Community♦
1
answered Jul 5 '16 at 18:47
Camilo Arosemena-SerratoCamilo Arosemena-Serrato
5,72611849
edited Apr 13 '17 at 12:21
Community♦
1
edited Apr 13 '17 at 12:21
Community♦
1
edited Apr 13 '17 at 12:21
Community♦
1
1
answered Jul 5 '16 at 18:47
Camilo Arosemena-SerratoCamilo Arosemena-Serrato
5,72611849
answered Jul 5 '16 at 18:47
Camilo Arosemena-SerratoCamilo Arosemena-Serrato
5,72611849
answered Jul 5 '16 at 18:47
Camilo Arosemena-SerratoCamilo Arosemena-Serrato
5,72611849
5,72611849
add a comment |
add a comment |
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