Linear dependence as a binary relation












1












$begingroup$


Elements $a$ and $b$ from an $R$-module $M$ are linearly dependent if there are scalars $x$, $y$ in $R$, $x neq 0$ or $y neq 0$, such that $xa = yb$.



Let $M$ be a torsion-free module over an integral domain $R$.

In this case the binary relation has the following properties:

1. It is an equivalence relation on nonzero elements of $M$;

2. Each equivalence class combined with the zero element is a submodule of $M$.



Let's call the submodule an axis of module $M$.

Clearly, if $M$ is a free module over a principal ideal domain, all the axes are lines isomorphic to $R$ over itself.

But what if $M$ is not a free module or $R$ is not a PID?

There is an example of a torsion-free module over an integral domain where the only axis is the module itself, and it is not a line: $langle 2,x rangle$ over $mathbb{Z}[x]$ (Show that $langle 2,x rangle$ is not a principal ideal in $mathbb Z [x]$).



Questions:

1. Are the axes of a torsion-free module over an integral domain isomorphic to each other?

2. Are the axes of a free module over an integral domain lines?

3. If an axis $A$ of a free module over an integral domain is a line, is there a direct sum decomposition $M = A oplus M'$?






abstract-algebra modules direct-sum free-modules







share|cite|improve this question











$endgroup$












  • $begingroup$
    What do you call a "line" in the second question?
    $endgroup$
    – lisyarus
    Feb 10 at 23:06










  • $begingroup$
    @lisyarus: By line I mean a free module of rank 1.
    $endgroup$
    – Alex C
    Feb 10 at 23:10
















1












$begingroup$


Elements $a$ and $b$ from an $R$-module $M$ are linearly dependent if there are scalars $x$, $y$ in $R$, $x neq 0$ or $y neq 0$, such that $xa = yb$.



Let $M$ be a torsion-free module over an integral domain $R$.

In this case the binary relation has the following properties:

1. It is an equivalence relation on nonzero elements of $M$;

2. Each equivalence class combined with the zero element is a submodule of $M$.



Let's call the submodule an axis of module $M$.

Clearly, if $M$ is a free module over a principal ideal domain, all the axes are lines isomorphic to $R$ over itself.

But what if $M$ is not a free module or $R$ is not a PID?

There is an example of a torsion-free module over an integral domain where the only axis is the module itself, and it is not a line: $langle 2,x rangle$ over $mathbb{Z}[x]$ (Show that $langle 2,x rangle$ is not a principal ideal in $mathbb Z [x]$).



Questions:

1. Are the axes of a torsion-free module over an integral domain isomorphic to each other?

2. Are the axes of a free module over an integral domain lines?

3. If an axis $A$ of a free module over an integral domain is a line, is there a direct sum decomposition $M = A oplus M'$?






abstract-algebra modules direct-sum free-modules







share|cite|improve this question











$endgroup$












  • $begingroup$
    What do you call a "line" in the second question?
    $endgroup$
    – lisyarus
    Feb 10 at 23:06










  • $begingroup$
    @lisyarus: By line I mean a free module of rank 1.
    $endgroup$
    – Alex C
    Feb 10 at 23:10














1












1








1





$begingroup$


Elements $a$ and $b$ from an $R$-module $M$ are linearly dependent if there are scalars $x$, $y$ in $R$, $x neq 0$ or $y neq 0$, such that $xa = yb$.



Let $M$ be a torsion-free module over an integral domain $R$.

In this case the binary relation has the following properties:

1. It is an equivalence relation on nonzero elements of $M$;

2. Each equivalence class combined with the zero element is a submodule of $M$.



Let's call the submodule an axis of module $M$.

Clearly, if $M$ is a free module over a principal ideal domain, all the axes are lines isomorphic to $R$ over itself.

But what if $M$ is not a free module or $R$ is not a PID?

There is an example of a torsion-free module over an integral domain where the only axis is the module itself, and it is not a line: $langle 2,x rangle$ over $mathbb{Z}[x]$ (Show that $langle 2,x rangle$ is not a principal ideal in $mathbb Z [x]$).



Questions:

1. Are the axes of a torsion-free module over an integral domain isomorphic to each other?

2. Are the axes of a free module over an integral domain lines?

3. If an axis $A$ of a free module over an integral domain is a line, is there a direct sum decomposition $M = A oplus M'$?






abstract-algebra modules direct-sum free-modules







share|cite|improve this question











$endgroup$




Elements $a$ and $b$ from an $R$-module $M$ are linearly dependent if there are scalars $x$, $y$ in $R$, $x neq 0$ or $y neq 0$, such that $xa = yb$.



Let $M$ be a torsion-free module over an integral domain $R$.

In this case the binary relation has the following properties:

1. It is an equivalence relation on nonzero elements of $M$;

2. Each equivalence class combined with the zero element is a submodule of $M$.



Let's call the submodule an axis of module $M$.

Clearly, if $M$ is a free module over a principal ideal domain, all the axes are lines isomorphic to $R$ over itself.

But what if $M$ is not a free module or $R$ is not a PID?

There is an example of a torsion-free module over an integral domain where the only axis is the module itself, and it is not a line: $langle 2,x rangle$ over $mathbb{Z}[x]$ (Show that $langle 2,x rangle$ is not a principal ideal in $mathbb Z [x]$).



Questions:

1. Are the axes of a torsion-free module over an integral domain isomorphic to each other?

2. Are the axes of a free module over an integral domain lines?

3. If an axis $A$ of a free module over an integral domain is a line, is there a direct sum decomposition $M = A oplus M'$?






abstract-algebra modules direct-sum free-modules




abstract-algebra modules direct-sum free-modules






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Feb 22 at 10:53







Alex C

















asked Jan 9 at 6:56









Alex CAlex C

10718




10718












  • $begingroup$
    What do you call a "line" in the second question?
    $endgroup$
    – lisyarus
    Feb 10 at 23:06










  • $begingroup$
    @lisyarus: By line I mean a free module of rank 1.
    $endgroup$
    – Alex C
    Feb 10 at 23:10


















  • $begingroup$
    What do you call a "line" in the second question?
    $endgroup$
    – lisyarus
    Feb 10 at 23:06










  • $begingroup$
    @lisyarus: By line I mean a free module of rank 1.
    $endgroup$
    – Alex C
    Feb 10 at 23:10
















$begingroup$
What do you call a "line" in the second question?
$endgroup$
– lisyarus
Feb 10 at 23:06




$begingroup$
What do you call a "line" in the second question?
$endgroup$
– lisyarus
Feb 10 at 23:06












$begingroup$
@lisyarus: By line I mean a free module of rank 1.
$endgroup$
– Alex C
Feb 10 at 23:10




$begingroup$
@lisyarus: By line I mean a free module of rank 1.
$endgroup$
– Alex C
Feb 10 at 23:10










1 Answer
1






active

oldest

votes


















0












$begingroup$

It looks like the answer to the first question is negative:

in $mathbb{Z} times mathbb{R}$ over $mathbb{Z}$ axis $mathbb{Z}$ is not isomorphic to axis $mathbb{R}$.



Trying to find a counterexample for the second question.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    $mathbb R$ is not an axis: e.g. $sqrt 2$ and $sqrt 3$ are linearly independent over $mathbb Z$.
    $endgroup$
    – lisyarus
    Feb 10 at 23:01






  • 1




    $begingroup$
    On the other hand, in $mathbb Z times mathbb Q$ the $mathbb Q$ part is an axis, providing a counterexample.
    $endgroup$
    – lisyarus
    Feb 10 at 23:06










  • $begingroup$
    @lisyarus: Yes, you are right. Thank you!
    $endgroup$
    – Alex C
    Feb 10 at 23:13












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1 Answer
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1 Answer
1






active

oldest

votes









active

oldest

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active

oldest

votes









0












$begingroup$

It looks like the answer to the first question is negative:

in $mathbb{Z} times mathbb{R}$ over $mathbb{Z}$ axis $mathbb{Z}$ is not isomorphic to axis $mathbb{R}$.



Trying to find a counterexample for the second question.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    $mathbb R$ is not an axis: e.g. $sqrt 2$ and $sqrt 3$ are linearly independent over $mathbb Z$.
    $endgroup$
    – lisyarus
    Feb 10 at 23:01






  • 1




    $begingroup$
    On the other hand, in $mathbb Z times mathbb Q$ the $mathbb Q$ part is an axis, providing a counterexample.
    $endgroup$
    – lisyarus
    Feb 10 at 23:06










  • $begingroup$
    @lisyarus: Yes, you are right. Thank you!
    $endgroup$
    – Alex C
    Feb 10 at 23:13
















0












$begingroup$

It looks like the answer to the first question is negative:

in $mathbb{Z} times mathbb{R}$ over $mathbb{Z}$ axis $mathbb{Z}$ is not isomorphic to axis $mathbb{R}$.



Trying to find a counterexample for the second question.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    $mathbb R$ is not an axis: e.g. $sqrt 2$ and $sqrt 3$ are linearly independent over $mathbb Z$.
    $endgroup$
    – lisyarus
    Feb 10 at 23:01






  • 1




    $begingroup$
    On the other hand, in $mathbb Z times mathbb Q$ the $mathbb Q$ part is an axis, providing a counterexample.
    $endgroup$
    – lisyarus
    Feb 10 at 23:06










  • $begingroup$
    @lisyarus: Yes, you are right. Thank you!
    $endgroup$
    – Alex C
    Feb 10 at 23:13














0












0








0





$begingroup$

It looks like the answer to the first question is negative:

in $mathbb{Z} times mathbb{R}$ over $mathbb{Z}$ axis $mathbb{Z}$ is not isomorphic to axis $mathbb{R}$.



Trying to find a counterexample for the second question.






share|cite|improve this answer









$endgroup$



It looks like the answer to the first question is negative:

in $mathbb{Z} times mathbb{R}$ over $mathbb{Z}$ axis $mathbb{Z}$ is not isomorphic to axis $mathbb{R}$.



Trying to find a counterexample for the second question.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 11 at 23:06









Alex CAlex C

10718




10718












  • $begingroup$
    $mathbb R$ is not an axis: e.g. $sqrt 2$ and $sqrt 3$ are linearly independent over $mathbb Z$.
    $endgroup$
    – lisyarus
    Feb 10 at 23:01






  • 1




    $begingroup$
    On the other hand, in $mathbb Z times mathbb Q$ the $mathbb Q$ part is an axis, providing a counterexample.
    $endgroup$
    – lisyarus
    Feb 10 at 23:06










  • $begingroup$
    @lisyarus: Yes, you are right. Thank you!
    $endgroup$
    – Alex C
    Feb 10 at 23:13


















  • $begingroup$
    $mathbb R$ is not an axis: e.g. $sqrt 2$ and $sqrt 3$ are linearly independent over $mathbb Z$.
    $endgroup$
    – lisyarus
    Feb 10 at 23:01






  • 1




    $begingroup$
    On the other hand, in $mathbb Z times mathbb Q$ the $mathbb Q$ part is an axis, providing a counterexample.
    $endgroup$
    – lisyarus
    Feb 10 at 23:06










  • $begingroup$
    @lisyarus: Yes, you are right. Thank you!
    $endgroup$
    – Alex C
    Feb 10 at 23:13
















$begingroup$
$mathbb R$ is not an axis: e.g. $sqrt 2$ and $sqrt 3$ are linearly independent over $mathbb Z$.
$endgroup$
– lisyarus
Feb 10 at 23:01




$begingroup$
$mathbb R$ is not an axis: e.g. $sqrt 2$ and $sqrt 3$ are linearly independent over $mathbb Z$.
$endgroup$
– lisyarus
Feb 10 at 23:01




1




1




$begingroup$
On the other hand, in $mathbb Z times mathbb Q$ the $mathbb Q$ part is an axis, providing a counterexample.
$endgroup$
– lisyarus
Feb 10 at 23:06




$begingroup$
On the other hand, in $mathbb Z times mathbb Q$ the $mathbb Q$ part is an axis, providing a counterexample.
$endgroup$
– lisyarus
Feb 10 at 23:06












$begingroup$
@lisyarus: Yes, you are right. Thank you!
$endgroup$
– Alex C
Feb 10 at 23:13




$begingroup$
@lisyarus: Yes, you are right. Thank you!
$endgroup$
– Alex C
Feb 10 at 23:13


















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