Let $mathbb{C}[X]$ be a module over itself. Given $z in mathbb{C}$ define $M_z = { f in mathbb{C}[X] | f(z)...












0












$begingroup$



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







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$endgroup$












  • $begingroup$
    @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
















0












$begingroup$



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







share|cite|improve this question









$endgroup$












  • $begingroup$
    @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














0












0








0





$begingroup$



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







share|cite|improve this question









$endgroup$





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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 3 at 15:48









PerturbativePerturbative

4,39121553




4,39121553












  • $begingroup$
    @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




$begingroup$
@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










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$begingroup$

An element is in $M_z$ if $f=(X-z)P(x)$ so a basis is $(X-z)X^i$






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    active

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    2












    $begingroup$

    An element is in $M_z$ if $f=(X-z)P(x)$ so a basis is $(X-z)X^i$






    share|cite|improve this answer









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      $begingroup$

      An element is in $M_z$ if $f=(X-z)P(x)$ so a basis is $(X-z)X^i$






      share|cite|improve this answer









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        2








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        $begingroup$

        An element is in $M_z$ if $f=(X-z)P(x)$ so a basis is $(X-z)X^i$






        share|cite|improve this answer









        $endgroup$



        An element is in $M_z$ if $f=(X-z)P(x)$ so a basis is $(X-z)X^i$







        share|cite|improve this answer












        share|cite|improve this answer



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        answered Jan 3 at 15:50









        Tsemo AristideTsemo Aristide

        58.9k11445




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