Tensor product of modules.
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¡Hello! Please give me a hint for this problem.
" Let $R$ a ring, $L$ a left ideal of $R$, $I$ right ideal of $R$. Show if $M$ are a left $R$-module, then exists a bijective function of commutative groups.
$$ f colon (R/I) otimes_{R} M to M/IM$$ such that
$ f((r+I) otimes m) = rm + IM$. And show that as commutative groups
$$ (R/I) otimes _{R} (R/L) cong R/(I+L)$$
abstract-algebra modules tensor-products free-modules
edited Nov 20 at 18:39
Frank Murphy
1,1151823
asked Nov 20 at 18:31
Davis We
655
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0
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favorite
¡Hello! Please give me a hint for this problem.
" Let $R$ a ring, $L$ a left ideal of $R$, $I$ right ideal of $R$. Show if $M$ are a left $R$-module, then exists a bijective function of commutative groups.
$$ f colon (R/I) otimes_{R} M to M/IM$$ such that
$ f((r+I) otimes m) = rm + IM$. And show that as commutative groups
$$ (R/I) otimes _{R} (R/L) cong R/(I+L)$$
abstract-algebra modules tensor-products free-modules
edited Nov 20 at 18:39
Frank Murphy
1,1151823
asked Nov 20 at 18:31
Davis We
655
2
Show that $M/IM$ satisfies the universal property of $(R/I)otimes M$.
– Frank Murphy
Nov 20 at 18:39
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
¡Hello! Please give me a hint for this problem.
" Let $R$ a ring, $L$ a left ideal of $R$, $I$ right ideal of $R$. Show if $M$ are a left $R$-module, then exists a bijective function of commutative groups.
$$ f colon (R/I) otimes_{R} M to M/IM$$ such that
$ f((r+I) otimes m) = rm + IM$. And show that as commutative groups
$$ (R/I) otimes _{R} (R/L) cong R/(I+L)$$
abstract-algebra modules tensor-products free-modules
edited Nov 20 at 18:39
Frank Murphy
1,1151823
asked Nov 20 at 18:31
Davis We
655
¡Hello! Please give me a hint for this problem.
" Let $R$ a ring, $L$ a left ideal of $R$, $I$ right ideal of $R$. Show if $M$ are a left $R$-module, then exists a bijective function of commutative groups.
$$ f colon (R/I) otimes_{R} M to M/IM$$ such that
$ f((r+I) otimes m) = rm + IM$. And show that as commutative groups
$$ (R/I) otimes _{R} (R/L) cong R/(I+L)$$
abstract-algebra modules tensor-products free-modules
abstract-algebra modules tensor-products free-modules
edited Nov 20 at 18:39
Frank Murphy
1,1151823
asked Nov 20 at 18:31
Davis We
655
edited Nov 20 at 18:39
Frank Murphy
1,1151823
asked Nov 20 at 18:31
Davis We
655
edited Nov 20 at 18:39
Frank Murphy
1,1151823
edited Nov 20 at 18:39
Frank Murphy
1,1151823
edited Nov 20 at 18:39
Frank Murphy
1,1151823
1,1151823
asked Nov 20 at 18:31
Davis We
655
asked Nov 20 at 18:31
Davis We
655
asked Nov 20 at 18:31
Davis We
655
655
2
Show that $M/IM$ satisfies the universal property of $(R/I)otimes M$.
– Frank Murphy
Nov 20 at 18:39
add a comment |
2
Show that $M/IM$ satisfies the universal property of $(R/I)otimes M$.
– Frank Murphy
Nov 20 at 18:39
2
2
Show that $M/IM$ satisfies the universal property of $(R/I)otimes M$.
– Frank Murphy
Nov 20 at 18:39
Show that $M/IM$ satisfies the universal property of $(R/I)otimes M$.
– Frank Murphy
Nov 20 at 18:39
add a comment |
1 Answer
1
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0
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If we tensor with $M$ the short exact sequence $0to Ito Rto R/Ito 0$ and consider the exact sequence $0to IMto Mto M/IMto 0$, we obtain a commutative diagram of abelian groups with exact rows
$$require{AMScd}
begin{CD}
{} @. Iotimes_RM @>>> Rotimes_RM @>>> (R/I)otimes_R M @>>> 0 \
@. @VaVV @VbVV @VcVV @. \
0 @>>> IM @>>> M @>>> M/IM @>>> 0
end{CD}
$$
We have $a(rotimes x)=rx$, $b(rotimes x)=rx$ and $c((r+I)otimes x)=rx+IM$. Note that $a$ is surjective and $b$ is an isomorphism. A very simple diagram chasing shows that $c$ is an isomorphism as well.
For the second case, we need to compute
$$
I(R/L)=(IR+L)/L=(I+L)/L
$$
and so
$$
(R/L)big/I(R/L)=(R/L)big/((I+L)/L)cong R/(I+L)
$$
by the homomorphism theorem.
answered Nov 20 at 20:48
egreg
175k1383198
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
If we tensor with $M$ the short exact sequence $0to Ito Rto R/Ito 0$ and consider the exact sequence $0to IMto Mto M/IMto 0$, we obtain a commutative diagram of abelian groups with exact rows
$$require{AMScd}
begin{CD}
{} @. Iotimes_RM @>>> Rotimes_RM @>>> (R/I)otimes_R M @>>> 0 \
@. @VaVV @VbVV @VcVV @. \
0 @>>> IM @>>> M @>>> M/IM @>>> 0
end{CD}
$$
We have $a(rotimes x)=rx$, $b(rotimes x)=rx$ and $c((r+I)otimes x)=rx+IM$. Note that $a$ is surjective and $b$ is an isomorphism. A very simple diagram chasing shows that $c$ is an isomorphism as well.
For the second case, we need to compute
$$
I(R/L)=(IR+L)/L=(I+L)/L
$$
and so
$$
(R/L)big/I(R/L)=(R/L)big/((I+L)/L)cong R/(I+L)
$$
by the homomorphism theorem.
answered Nov 20 at 20:48
egreg
175k1383198
add a comment |
up vote
0
down vote
If we tensor with $M$ the short exact sequence $0to Ito Rto R/Ito 0$ and consider the exact sequence $0to IMto Mto M/IMto 0$, we obtain a commutative diagram of abelian groups with exact rows
$$require{AMScd}
begin{CD}
{} @. Iotimes_RM @>>> Rotimes_RM @>>> (R/I)otimes_R M @>>> 0 \
@. @VaVV @VbVV @VcVV @. \
0 @>>> IM @>>> M @>>> M/IM @>>> 0
end{CD}
$$
We have $a(rotimes x)=rx$, $b(rotimes x)=rx$ and $c((r+I)otimes x)=rx+IM$. Note that $a$ is surjective and $b$ is an isomorphism. A very simple diagram chasing shows that $c$ is an isomorphism as well.
For the second case, we need to compute
$$
I(R/L)=(IR+L)/L=(I+L)/L
$$
and so
$$
(R/L)big/I(R/L)=(R/L)big/((I+L)/L)cong R/(I+L)
$$
by the homomorphism theorem.
answered Nov 20 at 20:48
egreg
175k1383198
add a comment |
up vote
0
down vote
up vote
0
down vote
If we tensor with $M$ the short exact sequence $0to Ito Rto R/Ito 0$ and consider the exact sequence $0to IMto Mto M/IMto 0$, we obtain a commutative diagram of abelian groups with exact rows
$$require{AMScd}
begin{CD}
{} @. Iotimes_RM @>>> Rotimes_RM @>>> (R/I)otimes_R M @>>> 0 \
@. @VaVV @VbVV @VcVV @. \
0 @>>> IM @>>> M @>>> M/IM @>>> 0
end{CD}
$$
We have $a(rotimes x)=rx$, $b(rotimes x)=rx$ and $c((r+I)otimes x)=rx+IM$. Note that $a$ is surjective and $b$ is an isomorphism. A very simple diagram chasing shows that $c$ is an isomorphism as well.
For the second case, we need to compute
$$
I(R/L)=(IR+L)/L=(I+L)/L
$$
and so
$$
(R/L)big/I(R/L)=(R/L)big/((I+L)/L)cong R/(I+L)
$$
by the homomorphism theorem.
answered Nov 20 at 20:48
egreg
175k1383198
If we tensor with $M$ the short exact sequence $0to Ito Rto R/Ito 0$ and consider the exact sequence $0to IMto Mto M/IMto 0$, we obtain a commutative diagram of abelian groups with exact rows
$$require{AMScd}
begin{CD}
{} @. Iotimes_RM @>>> Rotimes_RM @>>> (R/I)otimes_R M @>>> 0 \
@. @VaVV @VbVV @VcVV @. \
0 @>>> IM @>>> M @>>> M/IM @>>> 0
end{CD}
$$
We have $a(rotimes x)=rx$, $b(rotimes x)=rx$ and $c((r+I)otimes x)=rx+IM$. Note that $a$ is surjective and $b$ is an isomorphism. A very simple diagram chasing shows that $c$ is an isomorphism as well.
For the second case, we need to compute
$$
I(R/L)=(IR+L)/L=(I+L)/L
$$
and so
$$
(R/L)big/I(R/L)=(R/L)big/((I+L)/L)cong R/(I+L)
$$
by the homomorphism theorem.
answered Nov 20 at 20:48
egreg
175k1383198
answered Nov 20 at 20:48
egreg
175k1383198
answered Nov 20 at 20:48
egreg
175k1383198
answered Nov 20 at 20:48
egreg
175k1383198
175k1383198
add a comment |
add a comment |
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2
Show that $M/IM$ satisfies the universal property of $(R/I)otimes M$.
– Frank Murphy
Nov 20 at 18:39