Let $mathbb{C}[X]$ be a module over itself. Given $z in mathbb{C}$ define $M_z = { f in mathbb{C}[X] | f(z)...
$begingroup$
Let $mathbb{C}[X]$ be a module over itself. Given $z in mathbb{C}$ define $M_z = { f(X) in mathbb{C}[X] | f(z) = 0}$. $M_z$ is a submodule of $mathbb{C}[X]$. Give a $mathbb{C}$-basis for $M_z$.
Now the way I understand it, to provide a $mathbb{C}$-basis for $M_z$ means that we consider $M_z$ as a $mathbb{C}$-module and then provide a basis for it as a $mathbb{C}$ module. However I'm not sure if it's possible to do that.
I know that if we consider $M_z$ as a $mathbb{C}[X]$ module then $X-z$ is a basis for $M_z$, the reason for that is that any $f(X) in M_z$ can be expressed in the form $f(X) = (X-z)g(X)$ for some $g(X) in mathbb{C}[X]$. However when considering $M_z$ as a $mathbb{C}$-module I don't have any ideas as to what I basis should even look like.
Is it possible to obtain a $mathbb{C}$-basis for $M_z$?
abstract-algebra modules free-modules
asked Jan 3 at 15:48
PerturbativePerturbative
4,26811551
$endgroup$
add a comment |
$begingroup$
Let $mathbb{C}[X]$ be a module over itself. Given $z in mathbb{C}$ define $M_z = { f(X) in mathbb{C}[X] | f(z) = 0}$. $M_z$ is a submodule of $mathbb{C}[X]$. Give a $mathbb{C}$-basis for $M_z$.
Now the way I understand it, to provide a $mathbb{C}$-basis for $M_z$ means that we consider $M_z$ as a $mathbb{C}$-module and then provide a basis for it as a $mathbb{C}$ module. However I'm not sure if it's possible to do that.
I know that if we consider $M_z$ as a $mathbb{C}[X]$ module then $X-z$ is a basis for $M_z$, the reason for that is that any $f(X) in M_z$ can be expressed in the form $f(X) = (X-z)g(X)$ for some $g(X) in mathbb{C}[X]$. However when considering $M_z$ as a $mathbb{C}$-module I don't have any ideas as to what I basis should even look like.
Is it possible to obtain a $mathbb{C}$-basis for $M_z$?
abstract-algebra modules free-modules
asked Jan 3 at 15:48
PerturbativePerturbative
4,26811551
$endgroup$
$begingroup$
@Antonios-AlexandrosRobotis Ahh and every vector space has a basis so indeed $M_z$ must have a basis
$endgroup$
– Perturbative
Jan 3 at 15:52
add a comment |
$begingroup$
Let $mathbb{C}[X]$ be a module over itself. Given $z in mathbb{C}$ define $M_z = { f(X) in mathbb{C}[X] | f(z) = 0}$. $M_z$ is a submodule of $mathbb{C}[X]$. Give a $mathbb{C}$-basis for $M_z$.
Now the way I understand it, to provide a $mathbb{C}$-basis for $M_z$ means that we consider $M_z$ as a $mathbb{C}$-module and then provide a basis for it as a $mathbb{C}$ module. However I'm not sure if it's possible to do that.
I know that if we consider $M_z$ as a $mathbb{C}[X]$ module then $X-z$ is a basis for $M_z$, the reason for that is that any $f(X) in M_z$ can be expressed in the form $f(X) = (X-z)g(X)$ for some $g(X) in mathbb{C}[X]$. However when considering $M_z$ as a $mathbb{C}$-module I don't have any ideas as to what I basis should even look like.
Is it possible to obtain a $mathbb{C}$-basis for $M_z$?
abstract-algebra modules free-modules
asked Jan 3 at 15:48
PerturbativePerturbative
4,26811551
$endgroup$
Let $mathbb{C}[X]$ be a module over itself. Given $z in mathbb{C}$ define $M_z = { f(X) in mathbb{C}[X] | f(z) = 0}$. $M_z$ is a submodule of $mathbb{C}[X]$. Give a $mathbb{C}$-basis for $M_z$.
Now the way I understand it, to provide a $mathbb{C}$-basis for $M_z$ means that we consider $M_z$ as a $mathbb{C}$-module and then provide a basis for it as a $mathbb{C}$ module. However I'm not sure if it's possible to do that.
I know that if we consider $M_z$ as a $mathbb{C}[X]$ module then $X-z$ is a basis for $M_z$, the reason for that is that any $f(X) in M_z$ can be expressed in the form $f(X) = (X-z)g(X)$ for some $g(X) in mathbb{C}[X]$. However when considering $M_z$ as a $mathbb{C}$-module I don't have any ideas as to what I basis should even look like.
Is it possible to obtain a $mathbb{C}$-basis for $M_z$?
abstract-algebra modules free-modules
abstract-algebra modules free-modules
asked Jan 3 at 15:48
PerturbativePerturbative
4,26811551
asked Jan 3 at 15:48
PerturbativePerturbative
4,26811551
asked Jan 3 at 15:48
PerturbativePerturbative
4,26811551
asked Jan 3 at 15:48
PerturbativePerturbative
4,26811551
asked Jan 3 at 15:48
PerturbativePerturbative
4,26811551
4,26811551
$begingroup$
@Antonios-AlexandrosRobotis Ahh and every vector space has a basis so indeed $M_z$ must have a basis
$endgroup$
– Perturbative
Jan 3 at 15:52
add a comment |
$begingroup$
@Antonios-AlexandrosRobotis Ahh and every vector space has a basis so indeed $M_z$ must have a basis
$endgroup$
– Perturbative
Jan 3 at 15:52
$begingroup$
@Antonios-AlexandrosRobotis Ahh and every vector space has a basis so indeed $M_z$ must have a basis
$endgroup$
– Perturbative
Jan 3 at 15:52
$begingroup$
@Antonios-AlexandrosRobotis Ahh and every vector space has a basis so indeed $M_z$ must have a basis
$endgroup$
– Perturbative
Jan 3 at 15:52
add a comment |
1 Answer
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$begingroup$
An element is in $M_z$ if $f=(X-z)P(x)$ so a basis is $(X-z)X^i$
answered Jan 3 at 15:50
Tsemo AristideTsemo Aristide
57.5k11444
$endgroup$
add a comment |
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$begingroup$
An element is in $M_z$ if $f=(X-z)P(x)$ so a basis is $(X-z)X^i$
answered Jan 3 at 15:50
Tsemo AristideTsemo Aristide
57.5k11444
$endgroup$
add a comment |
$begingroup$
An element is in $M_z$ if $f=(X-z)P(x)$ so a basis is $(X-z)X^i$
answered Jan 3 at 15:50
Tsemo AristideTsemo Aristide
57.5k11444
$endgroup$
add a comment |
$begingroup$
An element is in $M_z$ if $f=(X-z)P(x)$ so a basis is $(X-z)X^i$
answered Jan 3 at 15:50
Tsemo AristideTsemo Aristide
57.5k11444
$endgroup$
An element is in $M_z$ if $f=(X-z)P(x)$ so a basis is $(X-z)X^i$
answered Jan 3 at 15:50
Tsemo AristideTsemo Aristide
57.5k11444
answered Jan 3 at 15:50
Tsemo AristideTsemo Aristide
57.5k11444
answered Jan 3 at 15:50
Tsemo AristideTsemo Aristide
57.5k11444
answered Jan 3 at 15:50
Tsemo AristideTsemo Aristide
57.5k11444
57.5k11444
add a comment |
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$begingroup$
@Antonios-AlexandrosRobotis Ahh and every vector space has a basis so indeed $M_z$ must have a basis
$endgroup$
– Perturbative
Jan 3 at 15:52