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Question about homomorphisms between free modules
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Let $(A,m,k)$ be a Noetherian local ring and let $g:Lrightarrow L'$ be a homomorphism between free finitely generated $A$-modules. I want to prove that $g$ is inversible to the left if, and only if, the induced homomorphism $h:L/mLrightarrow L'/mL'$ is injective. It is clear that if $g$ has an inverse to the left, then $h$ is injective, but the converse is hard to me.
Working in the converse I note that if I find basis to $L$ ad $L'$ then is easy define an inverse to the left to $g$. So other question: basis of $L/mL$ (like a $k$-vector space) induces a basis of $L$?
commutative-algebra free-modules
asked Jul 25 at 22:19
Rafael Holanda
2,425522
add a comment |Â
up vote
1
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Let $(A,m,k)$ be a Noetherian local ring and let $g:Lrightarrow L'$ be a homomorphism between free finitely generated $A$-modules. I want to prove that $g$ is inversible to the left if, and only if, the induced homomorphism $h:L/mLrightarrow L'/mL'$ is injective. It is clear that if $g$ has an inverse to the left, then $h$ is injective, but the converse is hard to me.
Working in the converse I note that if I find basis to $L$ ad $L'$ then is easy define an inverse to the left to $g$. So other question: basis of $L/mL$ (like a $k$-vector space) induces a basis of $L$?
commutative-algebra free-modules
asked Jul 25 at 22:19
Rafael Holanda
2,425522
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $(A,m,k)$ be a Noetherian local ring and let $g:Lrightarrow L'$ be a homomorphism between free finitely generated $A$-modules. I want to prove that $g$ is inversible to the left if, and only if, the induced homomorphism $h:L/mLrightarrow L'/mL'$ is injective. It is clear that if $g$ has an inverse to the left, then $h$ is injective, but the converse is hard to me.
Working in the converse I note that if I find basis to $L$ ad $L'$ then is easy define an inverse to the left to $g$. So other question: basis of $L/mL$ (like a $k$-vector space) induces a basis of $L$?
commutative-algebra free-modules
asked Jul 25 at 22:19
Rafael Holanda
2,425522
Let $(A,m,k)$ be a Noetherian local ring and let $g:Lrightarrow L'$ be a homomorphism between free finitely generated $A$-modules. I want to prove that $g$ is inversible to the left if, and only if, the induced homomorphism $h:L/mLrightarrow L'/mL'$ is injective. It is clear that if $g$ has an inverse to the left, then $h$ is injective, but the converse is hard to me.
Working in the converse I note that if I find basis to $L$ ad $L'$ then is easy define an inverse to the left to $g$. So other question: basis of $L/mL$ (like a $k$-vector space) induces a basis of $L$?
commutative-algebra free-modules
asked Jul 25 at 22:19
Rafael Holanda
2,425522
asked Jul 25 at 22:19
Rafael Holanda
2,425522
asked Jul 25 at 22:19
Rafael Holanda
2,425522
asked Jul 25 at 22:19
Rafael Holanda
2,425522
2,425522
add a comment |Â
add a comment |Â
2 Answers
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up vote
2
down vote
Yes a basis of $L/mathfrak mL$ can be lifted to a basis of $L$ by Nakayama's lemma:
Let $R$ be a commutative ring, $I$ an ideal of $R$, $M$ a finitely generated $R$-module, $Nsubset M$ a submodule.
If $Msubset N+IM$, there exists an element $ain I$ such that $(1+a)Msubset N$.
In the present; you consider vectors $u_1, dots u_nin L$ such that their images in $L/mathfrak mL$ are a basis of this $A/mathfrak m$-vector space, and denote $N=langle u_1, dots u_nrangle$. By hypothesis, we have
$$Lsubset N+mathfrak mL,$$
so $(1+a)Lsubset N$ for some $ainmathfrak m$? As $A$ is local with maximal ideal $mathfrak m$, $1+a$ is a unit in $A$, so actually $L=N$.
Checking $ u_1, dots u_n$ are linearly independent is easy (always with Nakayama).
answered Jul 25 at 22:50
Bernard
110k635103
If $a_1,...a_n$ are such that $a_1u_1+...a_nu_n=0$, then all $a_i$ belongs to $m$. How proceed?
– Rafael Holanda
Jul 25 at 23:03
Map $A^n$ onto $L$ (via the basis) and consider the submodule $K$ of $A^n$ made up of the linear relations between the generators. You can show $K=mathfrak m K$, and as it's finitely generated, it implies $K=0$.
– Bernard
Jul 25 at 23:50
add a comment |Â
up vote
0
down vote
First let's prove that if $L$ has rank $n$ then any set with $n$ generators of $L$ is a basis of $L$. For this, let $x_1,..., x_n$ be a set of generators of $L$ and consider the canonical exact sequence $0rightarrow Krightarrow Lrightarrow Lrightarrow0$ where $Lrightarrow L$ maps $e_i$ into $x_i$ and $e_i$ is the canonical basis of $L$. Since $L$ is free, we have exact sequence between $k$-vector spaces of finite dimension.
$$0rightarrow K/mKrightarrow L/mLrightarrow L/mLrightarrow0$$
Therefore $L/mLrightarrow L/mL$ is an isomorphism and $K=mK$.By Nakayama's lemma we have $K=0$ and we conclude that $Lrightarrow L$ is an isomorphism, i.e., $x_1,...,x_n$ is a basis of $L$.
Now, if $h:L/mLrightarrow L'/mL'$ is injective, let $overlinex_1,...,overlinex_n$ be a basis of $L/mL$ and let $overlineg(x_1),...,overlineg(x_n),overliney_1,...,overliney_m$ be a basis of $L'/mL'$. By Nakayama these sets span $L$ and $L'$ and we see above that $x_1,...,x_n$ and $g(x_1),...,g(x_n),y_1,...,y_m$ are basis of $L$ and $L'$, respectively. Thus the map $f:L'rightarrow L$ defined by $f(g(x_i))=x_i$ and $f(y_j)=0$ is an inverse to the left to $g$.
answered Aug 7 at 3:44
Rafael Holanda
2,425522
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Yes a basis of $L/mathfrak mL$ can be lifted to a basis of $L$ by Nakayama's lemma:
Let $R$ be a commutative ring, $I$ an ideal of $R$, $M$ a finitely generated $R$-module, $Nsubset M$ a submodule.
If $Msubset N+IM$, there exists an element $ain I$ such that $(1+a)Msubset N$.
In the present; you consider vectors $u_1, dots u_nin L$ such that their images in $L/mathfrak mL$ are a basis of this $A/mathfrak m$-vector space, and denote $N=langle u_1, dots u_nrangle$. By hypothesis, we have
$$Lsubset N+mathfrak mL,$$
so $(1+a)Lsubset N$ for some $ainmathfrak m$? As $A$ is local with maximal ideal $mathfrak m$, $1+a$ is a unit in $A$, so actually $L=N$.
Checking $ u_1, dots u_n$ are linearly independent is easy (always with Nakayama).
answered Jul 25 at 22:50
Bernard
110k635103
If $a_1,...a_n$ are such that $a_1u_1+...a_nu_n=0$, then all $a_i$ belongs to $m$. How proceed?
– Rafael Holanda
Jul 25 at 23:03
Map $A^n$ onto $L$ (via the basis) and consider the submodule $K$ of $A^n$ made up of the linear relations between the generators. You can show $K=mathfrak m K$, and as it's finitely generated, it implies $K=0$.
– Bernard
Jul 25 at 23:50
add a comment |Â
up vote
2
down vote
Yes a basis of $L/mathfrak mL$ can be lifted to a basis of $L$ by Nakayama's lemma:
Let $R$ be a commutative ring, $I$ an ideal of $R$, $M$ a finitely generated $R$-module, $Nsubset M$ a submodule.
If $Msubset N+IM$, there exists an element $ain I$ such that $(1+a)Msubset N$.
In the present; you consider vectors $u_1, dots u_nin L$ such that their images in $L/mathfrak mL$ are a basis of this $A/mathfrak m$-vector space, and denote $N=langle u_1, dots u_nrangle$. By hypothesis, we have
$$Lsubset N+mathfrak mL,$$
so $(1+a)Lsubset N$ for some $ainmathfrak m$? As $A$ is local with maximal ideal $mathfrak m$, $1+a$ is a unit in $A$, so actually $L=N$.
Checking $ u_1, dots u_n$ are linearly independent is easy (always with Nakayama).
answered Jul 25 at 22:50
Bernard
110k635103
If $a_1,...a_n$ are such that $a_1u_1+...a_nu_n=0$, then all $a_i$ belongs to $m$. How proceed?
– Rafael Holanda
Jul 25 at 23:03
Map $A^n$ onto $L$ (via the basis) and consider the submodule $K$ of $A^n$ made up of the linear relations between the generators. You can show $K=mathfrak m K$, and as it's finitely generated, it implies $K=0$.
– Bernard
Jul 25 at 23:50
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Yes a basis of $L/mathfrak mL$ can be lifted to a basis of $L$ by Nakayama's lemma:
Let $R$ be a commutative ring, $I$ an ideal of $R$, $M$ a finitely generated $R$-module, $Nsubset M$ a submodule.
If $Msubset N+IM$, there exists an element $ain I$ such that $(1+a)Msubset N$.
In the present; you consider vectors $u_1, dots u_nin L$ such that their images in $L/mathfrak mL$ are a basis of this $A/mathfrak m$-vector space, and denote $N=langle u_1, dots u_nrangle$. By hypothesis, we have
$$Lsubset N+mathfrak mL,$$
so $(1+a)Lsubset N$ for some $ainmathfrak m$? As $A$ is local with maximal ideal $mathfrak m$, $1+a$ is a unit in $A$, so actually $L=N$.
Checking $ u_1, dots u_n$ are linearly independent is easy (always with Nakayama).
answered Jul 25 at 22:50
Bernard
110k635103
Yes a basis of $L/mathfrak mL$ can be lifted to a basis of $L$ by Nakayama's lemma:
Let $R$ be a commutative ring, $I$ an ideal of $R$, $M$ a finitely generated $R$-module, $Nsubset M$ a submodule.
If $Msubset N+IM$, there exists an element $ain I$ such that $(1+a)Msubset N$.
In the present; you consider vectors $u_1, dots u_nin L$ such that their images in $L/mathfrak mL$ are a basis of this $A/mathfrak m$-vector space, and denote $N=langle u_1, dots u_nrangle$. By hypothesis, we have
$$Lsubset N+mathfrak mL,$$
so $(1+a)Lsubset N$ for some $ainmathfrak m$? As $A$ is local with maximal ideal $mathfrak m$, $1+a$ is a unit in $A$, so actually $L=N$.
Checking $ u_1, dots u_n$ are linearly independent is easy (always with Nakayama).
answered Jul 25 at 22:50
Bernard
110k635103
answered Jul 25 at 22:50
Bernard
110k635103
answered Jul 25 at 22:50
Bernard
110k635103
answered Jul 25 at 22:50
Bernard
110k635103
110k635103
If $a_1,...a_n$ are such that $a_1u_1+...a_nu_n=0$, then all $a_i$ belongs to $m$. How proceed?
– Rafael Holanda
Jul 25 at 23:03
Map $A^n$ onto $L$ (via the basis) and consider the submodule $K$ of $A^n$ made up of the linear relations between the generators. You can show $K=mathfrak m K$, and as it's finitely generated, it implies $K=0$.
– Bernard
Jul 25 at 23:50
add a comment |Â
If $a_1,...a_n$ are such that $a_1u_1+...a_nu_n=0$, then all $a_i$ belongs to $m$. How proceed?
– Rafael Holanda
Jul 25 at 23:03
Map $A^n$ onto $L$ (via the basis) and consider the submodule $K$ of $A^n$ made up of the linear relations between the generators. You can show $K=mathfrak m K$, and as it's finitely generated, it implies $K=0$.
– Bernard
Jul 25 at 23:50
If $a_1,...a_n$ are such that $a_1u_1+...a_nu_n=0$, then all $a_i$ belongs to $m$. How proceed?
– Rafael Holanda
Jul 25 at 23:03
If $a_1,...a_n$ are such that $a_1u_1+...a_nu_n=0$, then all $a_i$ belongs to $m$. How proceed?
– Rafael Holanda
Jul 25 at 23:03
Map $A^n$ onto $L$ (via the basis) and consider the submodule $K$ of $A^n$ made up of the linear relations between the generators. You can show $K=mathfrak m K$, and as it's finitely generated, it implies $K=0$.
– Bernard
Jul 25 at 23:50
Map $A^n$ onto $L$ (via the basis) and consider the submodule $K$ of $A^n$ made up of the linear relations between the generators. You can show $K=mathfrak m K$, and as it's finitely generated, it implies $K=0$.
– Bernard
Jul 25 at 23:50
add a comment |Â
up vote
0
down vote
First let's prove that if $L$ has rank $n$ then any set with $n$ generators of $L$ is a basis of $L$. For this, let $x_1,..., x_n$ be a set of generators of $L$ and consider the canonical exact sequence $0rightarrow Krightarrow Lrightarrow Lrightarrow0$ where $Lrightarrow L$ maps $e_i$ into $x_i$ and $e_i$ is the canonical basis of $L$. Since $L$ is free, we have exact sequence between $k$-vector spaces of finite dimension.
$$0rightarrow K/mKrightarrow L/mLrightarrow L/mLrightarrow0$$
Therefore $L/mLrightarrow L/mL$ is an isomorphism and $K=mK$.By Nakayama's lemma we have $K=0$ and we conclude that $Lrightarrow L$ is an isomorphism, i.e., $x_1,...,x_n$ is a basis of $L$.
Now, if $h:L/mLrightarrow L'/mL'$ is injective, let $overlinex_1,...,overlinex_n$ be a basis of $L/mL$ and let $overlineg(x_1),...,overlineg(x_n),overliney_1,...,overliney_m$ be a basis of $L'/mL'$. By Nakayama these sets span $L$ and $L'$ and we see above that $x_1,...,x_n$ and $g(x_1),...,g(x_n),y_1,...,y_m$ are basis of $L$ and $L'$, respectively. Thus the map $f:L'rightarrow L$ defined by $f(g(x_i))=x_i$ and $f(y_j)=0$ is an inverse to the left to $g$.
answered Aug 7 at 3:44
Rafael Holanda
2,425522
add a comment |Â
up vote
0
down vote
First let's prove that if $L$ has rank $n$ then any set with $n$ generators of $L$ is a basis of $L$. For this, let $x_1,..., x_n$ be a set of generators of $L$ and consider the canonical exact sequence $0rightarrow Krightarrow Lrightarrow Lrightarrow0$ where $Lrightarrow L$ maps $e_i$ into $x_i$ and $e_i$ is the canonical basis of $L$. Since $L$ is free, we have exact sequence between $k$-vector spaces of finite dimension.
$$0rightarrow K/mKrightarrow L/mLrightarrow L/mLrightarrow0$$
Therefore $L/mLrightarrow L/mL$ is an isomorphism and $K=mK$.By Nakayama's lemma we have $K=0$ and we conclude that $Lrightarrow L$ is an isomorphism, i.e., $x_1,...,x_n$ is a basis of $L$.
Now, if $h:L/mLrightarrow L'/mL'$ is injective, let $overlinex_1,...,overlinex_n$ be a basis of $L/mL$ and let $overlineg(x_1),...,overlineg(x_n),overliney_1,...,overliney_m$ be a basis of $L'/mL'$. By Nakayama these sets span $L$ and $L'$ and we see above that $x_1,...,x_n$ and $g(x_1),...,g(x_n),y_1,...,y_m$ are basis of $L$ and $L'$, respectively. Thus the map $f:L'rightarrow L$ defined by $f(g(x_i))=x_i$ and $f(y_j)=0$ is an inverse to the left to $g$.
answered Aug 7 at 3:44
Rafael Holanda
2,425522
add a comment |Â
up vote
0
down vote
up vote
0
down vote
First let's prove that if $L$ has rank $n$ then any set with $n$ generators of $L$ is a basis of $L$. For this, let $x_1,..., x_n$ be a set of generators of $L$ and consider the canonical exact sequence $0rightarrow Krightarrow Lrightarrow Lrightarrow0$ where $Lrightarrow L$ maps $e_i$ into $x_i$ and $e_i$ is the canonical basis of $L$. Since $L$ is free, we have exact sequence between $k$-vector spaces of finite dimension.
$$0rightarrow K/mKrightarrow L/mLrightarrow L/mLrightarrow0$$
Therefore $L/mLrightarrow L/mL$ is an isomorphism and $K=mK$.By Nakayama's lemma we have $K=0$ and we conclude that $Lrightarrow L$ is an isomorphism, i.e., $x_1,...,x_n$ is a basis of $L$.
Now, if $h:L/mLrightarrow L'/mL'$ is injective, let $overlinex_1,...,overlinex_n$ be a basis of $L/mL$ and let $overlineg(x_1),...,overlineg(x_n),overliney_1,...,overliney_m$ be a basis of $L'/mL'$. By Nakayama these sets span $L$ and $L'$ and we see above that $x_1,...,x_n$ and $g(x_1),...,g(x_n),y_1,...,y_m$ are basis of $L$ and $L'$, respectively. Thus the map $f:L'rightarrow L$ defined by $f(g(x_i))=x_i$ and $f(y_j)=0$ is an inverse to the left to $g$.
answered Aug 7 at 3:44
Rafael Holanda
2,425522
First let's prove that if $L$ has rank $n$ then any set with $n$ generators of $L$ is a basis of $L$. For this, let $x_1,..., x_n$ be a set of generators of $L$ and consider the canonical exact sequence $0rightarrow Krightarrow Lrightarrow Lrightarrow0$ where $Lrightarrow L$ maps $e_i$ into $x_i$ and $e_i$ is the canonical basis of $L$. Since $L$ is free, we have exact sequence between $k$-vector spaces of finite dimension.
$$0rightarrow K/mKrightarrow L/mLrightarrow L/mLrightarrow0$$
Therefore $L/mLrightarrow L/mL$ is an isomorphism and $K=mK$.By Nakayama's lemma we have $K=0$ and we conclude that $Lrightarrow L$ is an isomorphism, i.e., $x_1,...,x_n$ is a basis of $L$.
Now, if $h:L/mLrightarrow L'/mL'$ is injective, let $overlinex_1,...,overlinex_n$ be a basis of $L/mL$ and let $overlineg(x_1),...,overlineg(x_n),overliney_1,...,overliney_m$ be a basis of $L'/mL'$. By Nakayama these sets span $L$ and $L'$ and we see above that $x_1,...,x_n$ and $g(x_1),...,g(x_n),y_1,...,y_m$ are basis of $L$ and $L'$, respectively. Thus the map $f:L'rightarrow L$ defined by $f(g(x_i))=x_i$ and $f(y_j)=0$ is an inverse to the left to $g$.
answered Aug 7 at 3:44
Rafael Holanda
2,425522
answered Aug 7 at 3:44
Rafael Holanda
2,425522
answered Aug 7 at 3:44
Rafael Holanda
2,425522
answered Aug 7 at 3:44
Rafael Holanda
2,425522
2,425522
add a comment |Â
add a comment |Â
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