Decomposition a Module into a Pure Submodule and another Submodule
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I am having trouble doing the following exercise.
Let $F$ be a field and let $M$ be a finitely generated module over the polynomial ring $F[x]$. Let $N$ be a pure submodule of $M$. Then, there exists a submodule $L$ of $M$ such that $N+L = M$ and $N cap L =0$.
By pure, I mean that if an element $y$ in $N$ and and element $a$ in $F$ are such that there exists an element $x$ in $M$ where $ax=y$, then there exists an element $z$ in $N$ such that $az = y$.
linear-algebra abstract-algebra
asked Dec 4 '18 at 20:19
LinearGuyLinearGuy
13711
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add a comment |
$begingroup$
I am having trouble doing the following exercise.
Let $F$ be a field and let $M$ be a finitely generated module over the polynomial ring $F[x]$. Let $N$ be a pure submodule of $M$. Then, there exists a submodule $L$ of $M$ such that $N+L = M$ and $N cap L =0$.
By pure, I mean that if an element $y$ in $N$ and and element $a$ in $F$ are such that there exists an element $x$ in $M$ where $ax=y$, then there exists an element $z$ in $N$ such that $az = y$.
linear-algebra abstract-algebra
asked Dec 4 '18 at 20:19
LinearGuyLinearGuy
13711
$endgroup$
add a comment |
$begingroup$
I am having trouble doing the following exercise.
Let $F$ be a field and let $M$ be a finitely generated module over the polynomial ring $F[x]$. Let $N$ be a pure submodule of $M$. Then, there exists a submodule $L$ of $M$ such that $N+L = M$ and $N cap L =0$.
By pure, I mean that if an element $y$ in $N$ and and element $a$ in $F$ are such that there exists an element $x$ in $M$ where $ax=y$, then there exists an element $z$ in $N$ such that $az = y$.
linear-algebra abstract-algebra
asked Dec 4 '18 at 20:19
LinearGuyLinearGuy
13711
$endgroup$
I am having trouble doing the following exercise.
Let $F$ be a field and let $M$ be a finitely generated module over the polynomial ring $F[x]$. Let $N$ be a pure submodule of $M$. Then, there exists a submodule $L$ of $M$ such that $N+L = M$ and $N cap L =0$.
By pure, I mean that if an element $y$ in $N$ and and element $a$ in $F$ are such that there exists an element $x$ in $M$ where $ax=y$, then there exists an element $z$ in $N$ such that $az = y$.
linear-algebra abstract-algebra
linear-algebra abstract-algebra
asked Dec 4 '18 at 20:19
LinearGuyLinearGuy
13711
asked Dec 4 '18 at 20:19
LinearGuyLinearGuy
13711
edited Dec 12 '18 at 15:42
LinearGuy
asked Dec 4 '18 at 20:19
LinearGuyLinearGuy
13711
asked Dec 4 '18 at 20:19
LinearGuyLinearGuy
13711
asked Dec 4 '18 at 20:19
LinearGuyLinearGuy
13711
13711
add a comment |
add a comment |
1 Answer
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I will write down some hints. Firstly, since $F$ is a field we have that $F[x]$ is PID.
Consider the following short exact sequence: $0 longrightarrow N overset{i}longrightarrow M overset{pi}{longrightarrow} M / N longrightarrow 0$, where $i$ is the inclusion and $pi$ is the quotient map If this sequence splits then $M=N oplus M/N$.
In order to find such morphism $ucolon M/N to M$ such that it satisfies the $2$nd equivalent statement of Splitting Lemma consider $M/N$ as a finitely generated module over the PID $F[x]$ and aplpy the Structure theorem for finitely generated modules over a principal ideal domain and the use purity of $N$.
answered Dec 11 '18 at 16:28
CorneliusCornelius
1957
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If this answer does not help me let me know so i can enlighten you.
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– Cornelius
Dec 11 '18 at 16:28
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
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active
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votes
$begingroup$
I will write down some hints. Firstly, since $F$ is a field we have that $F[x]$ is PID.
Consider the following short exact sequence: $0 longrightarrow N overset{i}longrightarrow M overset{pi}{longrightarrow} M / N longrightarrow 0$, where $i$ is the inclusion and $pi$ is the quotient map If this sequence splits then $M=N oplus M/N$.
In order to find such morphism $ucolon M/N to M$ such that it satisfies the $2$nd equivalent statement of Splitting Lemma consider $M/N$ as a finitely generated module over the PID $F[x]$ and aplpy the Structure theorem for finitely generated modules over a principal ideal domain and the use purity of $N$.
answered Dec 11 '18 at 16:28
CorneliusCornelius
1957
$endgroup$
$begingroup$
If this answer does not help me let me know so i can enlighten you.
$endgroup$
– Cornelius
Dec 11 '18 at 16:28
add a comment |
$begingroup$
I will write down some hints. Firstly, since $F$ is a field we have that $F[x]$ is PID.
Consider the following short exact sequence: $0 longrightarrow N overset{i}longrightarrow M overset{pi}{longrightarrow} M / N longrightarrow 0$, where $i$ is the inclusion and $pi$ is the quotient map If this sequence splits then $M=N oplus M/N$.
In order to find such morphism $ucolon M/N to M$ such that it satisfies the $2$nd equivalent statement of Splitting Lemma consider $M/N$ as a finitely generated module over the PID $F[x]$ and aplpy the Structure theorem for finitely generated modules over a principal ideal domain and the use purity of $N$.
answered Dec 11 '18 at 16:28
CorneliusCornelius
1957
$endgroup$
$begingroup$
If this answer does not help me let me know so i can enlighten you.
$endgroup$
– Cornelius
Dec 11 '18 at 16:28
add a comment |
$begingroup$
I will write down some hints. Firstly, since $F$ is a field we have that $F[x]$ is PID.
Consider the following short exact sequence: $0 longrightarrow N overset{i}longrightarrow M overset{pi}{longrightarrow} M / N longrightarrow 0$, where $i$ is the inclusion and $pi$ is the quotient map If this sequence splits then $M=N oplus M/N$.
In order to find such morphism $ucolon M/N to M$ such that it satisfies the $2$nd equivalent statement of Splitting Lemma consider $M/N$ as a finitely generated module over the PID $F[x]$ and aplpy the Structure theorem for finitely generated modules over a principal ideal domain and the use purity of $N$.
answered Dec 11 '18 at 16:28
CorneliusCornelius
1957
$endgroup$
I will write down some hints. Firstly, since $F$ is a field we have that $F[x]$ is PID.
Consider the following short exact sequence: $0 longrightarrow N overset{i}longrightarrow M overset{pi}{longrightarrow} M / N longrightarrow 0$, where $i$ is the inclusion and $pi$ is the quotient map If this sequence splits then $M=N oplus M/N$.
In order to find such morphism $ucolon M/N to M$ such that it satisfies the $2$nd equivalent statement of Splitting Lemma consider $M/N$ as a finitely generated module over the PID $F[x]$ and aplpy the Structure theorem for finitely generated modules over a principal ideal domain and the use purity of $N$.
answered Dec 11 '18 at 16:28
CorneliusCornelius
1957
edited Dec 12 '18 at 8:02
answered Dec 11 '18 at 16:28
CorneliusCornelius
1957
answered Dec 11 '18 at 16:28
CorneliusCornelius
1957
answered Dec 11 '18 at 16:28
CorneliusCornelius
1957
1957
$begingroup$
If this answer does not help me let me know so i can enlighten you.
$endgroup$
– Cornelius
Dec 11 '18 at 16:28
add a comment |
$begingroup$
If this answer does not help me let me know so i can enlighten you.
$endgroup$
– Cornelius
Dec 11 '18 at 16:28
$begingroup$
If this answer does not help me let me know so i can enlighten you.
$endgroup$
– Cornelius
Dec 11 '18 at 16:28
$begingroup$
If this answer does not help me let me know so i can enlighten you.
$endgroup$
– Cornelius
Dec 11 '18 at 16:28
add a comment |
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