Extensions of indecomposable modules
$begingroup$
Let $R$ be a unital ring. Suppose that $A$, $B$, and $C$ are unitary left $R$-modules such that there exists a non-split exact sequence
$$0to A overset{alpha}{longrightarrow}Boverset{beta}{longrightarrow}Cto 0.$$
If $A$ and $C$ are indecomposable $R$-modules, does it follow that $B$ is also indecomposable? (What if $R$ is non-unital or the modules are not necessarily unitary $R$-modules?)
Edit: As Jeremy Rickard shows, my work below is faulty due to a bad assumption (italicized in the text below). Because of this, the statement is not true even when $A$ and $C$ are modules of finite length.
Wrong Attempt
I know the answer if $A$ and $C$ have finite length. In that case, $B$ has finite length. Since $operatorname{im}alphacong A$ is an indecomposable submodule of $B$, there exists by the Krull-Schmidt theorem a direct sum decomposition $$B=B_1oplus B_2oplusldots oplus B_n$$ of $B$, where each $B_i$ is indecomposable, such that $operatorname{im}alphasubseteq B_1$. Then, $$Ccong (B_1/operatorname{im}alpha)oplus B_2oplus ldotsoplus B_n.$$ By indecomposability of $C$, either $n=2$ and $B_1=operatorname{im}alpha$ which gives $Ccong B_2$, or $n=1$ and $B$ is indecomposable. However, in the case $n=2$ and $B_1=operatorname{im}alpha$, it follows that the exact sequence splits since we have a retraction from $B$ to $A$ (given by $B=B_1oplus B_2overset{operatorname{proj}_1}{ -!!!-!!!twoheadrightarrow } B_1overset{alpha^{-1}}{longrightarrow} A$). I am struggling to see whether this result extends to the infinite length cases.
abstract-algebra ring-theory modules exact-sequence direct-sum
$endgroup$
add a comment |
$begingroup$
Let $R$ be a unital ring. Suppose that $A$, $B$, and $C$ are unitary left $R$-modules such that there exists a non-split exact sequence
$$0to A overset{alpha}{longrightarrow}Boverset{beta}{longrightarrow}Cto 0.$$
If $A$ and $C$ are indecomposable $R$-modules, does it follow that $B$ is also indecomposable? (What if $R$ is non-unital or the modules are not necessarily unitary $R$-modules?)
Edit: As Jeremy Rickard shows, my work below is faulty due to a bad assumption (italicized in the text below). Because of this, the statement is not true even when $A$ and $C$ are modules of finite length.
Wrong Attempt
I know the answer if $A$ and $C$ have finite length. In that case, $B$ has finite length. Since $operatorname{im}alphacong A$ is an indecomposable submodule of $B$, there exists by the Krull-Schmidt theorem a direct sum decomposition $$B=B_1oplus B_2oplusldots oplus B_n$$ of $B$, where each $B_i$ is indecomposable, such that $operatorname{im}alphasubseteq B_1$. Then, $$Ccong (B_1/operatorname{im}alpha)oplus B_2oplus ldotsoplus B_n.$$ By indecomposability of $C$, either $n=2$ and $B_1=operatorname{im}alpha$ which gives $Ccong B_2$, or $n=1$ and $B$ is indecomposable. However, in the case $n=2$ and $B_1=operatorname{im}alpha$, it follows that the exact sequence splits since we have a retraction from $B$ to $A$ (given by $B=B_1oplus B_2overset{operatorname{proj}_1}{ -!!!-!!!twoheadrightarrow } B_1overset{alpha^{-1}}{longrightarrow} A$). I am struggling to see whether this result extends to the infinite length cases.
abstract-algebra ring-theory modules exact-sequence direct-sum
$endgroup$
add a comment |
$begingroup$
Let $R$ be a unital ring. Suppose that $A$, $B$, and $C$ are unitary left $R$-modules such that there exists a non-split exact sequence
$$0to A overset{alpha}{longrightarrow}Boverset{beta}{longrightarrow}Cto 0.$$
If $A$ and $C$ are indecomposable $R$-modules, does it follow that $B$ is also indecomposable? (What if $R$ is non-unital or the modules are not necessarily unitary $R$-modules?)
Edit: As Jeremy Rickard shows, my work below is faulty due to a bad assumption (italicized in the text below). Because of this, the statement is not true even when $A$ and $C$ are modules of finite length.
Wrong Attempt
I know the answer if $A$ and $C$ have finite length. In that case, $B$ has finite length. Since $operatorname{im}alphacong A$ is an indecomposable submodule of $B$, there exists by the Krull-Schmidt theorem a direct sum decomposition $$B=B_1oplus B_2oplusldots oplus B_n$$ of $B$, where each $B_i$ is indecomposable, such that $operatorname{im}alphasubseteq B_1$. Then, $$Ccong (B_1/operatorname{im}alpha)oplus B_2oplus ldotsoplus B_n.$$ By indecomposability of $C$, either $n=2$ and $B_1=operatorname{im}alpha$ which gives $Ccong B_2$, or $n=1$ and $B$ is indecomposable. However, in the case $n=2$ and $B_1=operatorname{im}alpha$, it follows that the exact sequence splits since we have a retraction from $B$ to $A$ (given by $B=B_1oplus B_2overset{operatorname{proj}_1}{ -!!!-!!!twoheadrightarrow } B_1overset{alpha^{-1}}{longrightarrow} A$). I am struggling to see whether this result extends to the infinite length cases.
abstract-algebra ring-theory modules exact-sequence direct-sum
$endgroup$
Let $R$ be a unital ring. Suppose that $A$, $B$, and $C$ are unitary left $R$-modules such that there exists a non-split exact sequence
$$0to A overset{alpha}{longrightarrow}Boverset{beta}{longrightarrow}Cto 0.$$
If $A$ and $C$ are indecomposable $R$-modules, does it follow that $B$ is also indecomposable? (What if $R$ is non-unital or the modules are not necessarily unitary $R$-modules?)
Edit: As Jeremy Rickard shows, my work below is faulty due to a bad assumption (italicized in the text below). Because of this, the statement is not true even when $A$ and $C$ are modules of finite length.
Wrong Attempt
I know the answer if $A$ and $C$ have finite length. In that case, $B$ has finite length. Since $operatorname{im}alphacong A$ is an indecomposable submodule of $B$, there exists by the Krull-Schmidt theorem a direct sum decomposition $$B=B_1oplus B_2oplusldots oplus B_n$$ of $B$, where each $B_i$ is indecomposable, such that $operatorname{im}alphasubseteq B_1$. Then, $$Ccong (B_1/operatorname{im}alpha)oplus B_2oplus ldotsoplus B_n.$$ By indecomposability of $C$, either $n=2$ and $B_1=operatorname{im}alpha$ which gives $Ccong B_2$, or $n=1$ and $B$ is indecomposable. However, in the case $n=2$ and $B_1=operatorname{im}alpha$, it follows that the exact sequence splits since we have a retraction from $B$ to $A$ (given by $B=B_1oplus B_2overset{operatorname{proj}_1}{ -!!!-!!!twoheadrightarrow } B_1overset{alpha^{-1}}{longrightarrow} A$). I am struggling to see whether this result extends to the infinite length cases.
abstract-algebra ring-theory modules exact-sequence direct-sum
abstract-algebra ring-theory modules exact-sequence direct-sum
edited Dec 3 '18 at 18:59
asked Dec 3 '18 at 18:02
user593746
add a comment |
add a comment |
1 Answer
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$begingroup$
Even for finite length modules, it's not true.
For example, ($R=mathbb{Z}$) there is a non-split exact sequence
$$0tomathbb{Z}/4mathbb{Z}stackrel{alpha}{to} mathbb{Z}/8mathbb{Z}oplusmathbb{Z}/2mathbb{Z}tomathbb{Z}/4mathbb{Z}to 0,$$
where $alpha(n)=(2n,n)$.
Your claim that $text{im}(alpha)$ must be contained in an indecomposable summand of $B$ isn't necessarily true.
answered Dec 3 '18 at 18:20
Jeremy RickardJeremy Rickard
16.1k11643
$endgroup$
$begingroup$
Thank you very much. Does my claim for the finite-length case at least hold if $R$ is an algebra over a field?
$endgroup$
– user593746
Dec 3 '18 at 18:40
1
$begingroup$
I just realized I could create a similar construction if $R$ is an algebra over $mathbb{R}$: $$0 to mathbb{R}[x]/big((x^2+1)^2big) overset{alpha}{longrightarrow} mathbb{R}[x]/big((x^2+1)^3big)oplus mathbb{R}[x]/big(x^2+1big) overset{beta}{longrightarrow} mathbb{R}[x]/big((x^2+1)^2big) to 0$$ via $alpha(f)=big((x^2+1)f,fbig)$ and $beta(f,g)=f-(x^2+1)g$.
$endgroup$
– user593746
Dec 3 '18 at 18:57
add a comment |
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$begingroup$
Even for finite length modules, it's not true.
For example, ($R=mathbb{Z}$) there is a non-split exact sequence
$$0tomathbb{Z}/4mathbb{Z}stackrel{alpha}{to} mathbb{Z}/8mathbb{Z}oplusmathbb{Z}/2mathbb{Z}tomathbb{Z}/4mathbb{Z}to 0,$$
where $alpha(n)=(2n,n)$.
Your claim that $text{im}(alpha)$ must be contained in an indecomposable summand of $B$ isn't necessarily true.
answered Dec 3 '18 at 18:20
Jeremy RickardJeremy Rickard
16.1k11643
$endgroup$
$begingroup$
Thank you very much. Does my claim for the finite-length case at least hold if $R$ is an algebra over a field?
$endgroup$
– user593746
Dec 3 '18 at 18:40
1
$begingroup$
I just realized I could create a similar construction if $R$ is an algebra over $mathbb{R}$: $$0 to mathbb{R}[x]/big((x^2+1)^2big) overset{alpha}{longrightarrow} mathbb{R}[x]/big((x^2+1)^3big)oplus mathbb{R}[x]/big(x^2+1big) overset{beta}{longrightarrow} mathbb{R}[x]/big((x^2+1)^2big) to 0$$ via $alpha(f)=big((x^2+1)f,fbig)$ and $beta(f,g)=f-(x^2+1)g$.
$endgroup$
– user593746
Dec 3 '18 at 18:57
add a comment |
$begingroup$
Even for finite length modules, it's not true.
For example, ($R=mathbb{Z}$) there is a non-split exact sequence
$$0tomathbb{Z}/4mathbb{Z}stackrel{alpha}{to} mathbb{Z}/8mathbb{Z}oplusmathbb{Z}/2mathbb{Z}tomathbb{Z}/4mathbb{Z}to 0,$$
where $alpha(n)=(2n,n)$.
Your claim that $text{im}(alpha)$ must be contained in an indecomposable summand of $B$ isn't necessarily true.
answered Dec 3 '18 at 18:20
Jeremy RickardJeremy Rickard
16.1k11643
$endgroup$
$begingroup$
Thank you very much. Does my claim for the finite-length case at least hold if $R$ is an algebra over a field?
$endgroup$
– user593746
Dec 3 '18 at 18:40
1
$begingroup$
I just realized I could create a similar construction if $R$ is an algebra over $mathbb{R}$: $$0 to mathbb{R}[x]/big((x^2+1)^2big) overset{alpha}{longrightarrow} mathbb{R}[x]/big((x^2+1)^3big)oplus mathbb{R}[x]/big(x^2+1big) overset{beta}{longrightarrow} mathbb{R}[x]/big((x^2+1)^2big) to 0$$ via $alpha(f)=big((x^2+1)f,fbig)$ and $beta(f,g)=f-(x^2+1)g$.
$endgroup$
– user593746
Dec 3 '18 at 18:57
add a comment |
$begingroup$
Even for finite length modules, it's not true.
For example, ($R=mathbb{Z}$) there is a non-split exact sequence
$$0tomathbb{Z}/4mathbb{Z}stackrel{alpha}{to} mathbb{Z}/8mathbb{Z}oplusmathbb{Z}/2mathbb{Z}tomathbb{Z}/4mathbb{Z}to 0,$$
where $alpha(n)=(2n,n)$.
Your claim that $text{im}(alpha)$ must be contained in an indecomposable summand of $B$ isn't necessarily true.
answered Dec 3 '18 at 18:20
Jeremy RickardJeremy Rickard
16.1k11643
$endgroup$
Even for finite length modules, it's not true.
For example, ($R=mathbb{Z}$) there is a non-split exact sequence
$$0tomathbb{Z}/4mathbb{Z}stackrel{alpha}{to} mathbb{Z}/8mathbb{Z}oplusmathbb{Z}/2mathbb{Z}tomathbb{Z}/4mathbb{Z}to 0,$$
where $alpha(n)=(2n,n)$.
Your claim that $text{im}(alpha)$ must be contained in an indecomposable summand of $B$ isn't necessarily true.
answered Dec 3 '18 at 18:20
Jeremy RickardJeremy Rickard
16.1k11643
answered Dec 3 '18 at 18:20
Jeremy RickardJeremy Rickard
16.1k11643
answered Dec 3 '18 at 18:20
Jeremy RickardJeremy Rickard
16.1k11643
answered Dec 3 '18 at 18:20
Jeremy RickardJeremy Rickard
16.1k11643
16.1k11643
$begingroup$
Thank you very much. Does my claim for the finite-length case at least hold if $R$ is an algebra over a field?
$endgroup$
– user593746
Dec 3 '18 at 18:40
1
$begingroup$
I just realized I could create a similar construction if $R$ is an algebra over $mathbb{R}$: $$0 to mathbb{R}[x]/big((x^2+1)^2big) overset{alpha}{longrightarrow} mathbb{R}[x]/big((x^2+1)^3big)oplus mathbb{R}[x]/big(x^2+1big) overset{beta}{longrightarrow} mathbb{R}[x]/big((x^2+1)^2big) to 0$$ via $alpha(f)=big((x^2+1)f,fbig)$ and $beta(f,g)=f-(x^2+1)g$.
$endgroup$
– user593746
Dec 3 '18 at 18:57
add a comment |
$begingroup$
Thank you very much. Does my claim for the finite-length case at least hold if $R$ is an algebra over a field?
$endgroup$
– user593746
Dec 3 '18 at 18:40
1
$begingroup$
I just realized I could create a similar construction if $R$ is an algebra over $mathbb{R}$: $$0 to mathbb{R}[x]/big((x^2+1)^2big) overset{alpha}{longrightarrow} mathbb{R}[x]/big((x^2+1)^3big)oplus mathbb{R}[x]/big(x^2+1big) overset{beta}{longrightarrow} mathbb{R}[x]/big((x^2+1)^2big) to 0$$ via $alpha(f)=big((x^2+1)f,fbig)$ and $beta(f,g)=f-(x^2+1)g$.
$endgroup$
– user593746
Dec 3 '18 at 18:57
$begingroup$
Thank you very much. Does my claim for the finite-length case at least hold if $R$ is an algebra over a field?
$endgroup$
– user593746
Dec 3 '18 at 18:40
$begingroup$
Thank you very much. Does my claim for the finite-length case at least hold if $R$ is an algebra over a field?
$endgroup$
– user593746
Dec 3 '18 at 18:40
1
1
$begingroup$
I just realized I could create a similar construction if $R$ is an algebra over $mathbb{R}$: $$0 to mathbb{R}[x]/big((x^2+1)^2big) overset{alpha}{longrightarrow} mathbb{R}[x]/big((x^2+1)^3big)oplus mathbb{R}[x]/big(x^2+1big) overset{beta}{longrightarrow} mathbb{R}[x]/big((x^2+1)^2big) to 0$$ via $alpha(f)=big((x^2+1)f,fbig)$ and $beta(f,g)=f-(x^2+1)g$.
$endgroup$
– user593746
Dec 3 '18 at 18:57
$begingroup$
I just realized I could create a similar construction if $R$ is an algebra over $mathbb{R}$: $$0 to mathbb{R}[x]/big((x^2+1)^2big) overset{alpha}{longrightarrow} mathbb{R}[x]/big((x^2+1)^3big)oplus mathbb{R}[x]/big(x^2+1big) overset{beta}{longrightarrow} mathbb{R}[x]/big((x^2+1)^2big) to 0$$ via $alpha(f)=big((x^2+1)f,fbig)$ and $beta(f,g)=f-(x^2+1)g$.
$endgroup$
– user593746
Dec 3 '18 at 18:57
add a comment |
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