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















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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  • @Antonios-AlexandrosRobotis Ahh and every vector space has a basis so indeed $M_z$ must have a basis
    – Perturbative
    2 days ago
















0















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






















  • @Antonios-AlexandrosRobotis Ahh and every vector space has a basis so indeed $M_z$ must have a basis
    – Perturbative
    2 days ago














0












0








0








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














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 2 days ago









Perturbative

4,13511450




4,13511450












  • @Antonios-AlexandrosRobotis Ahh and every vector space has a basis so indeed $M_z$ must have a basis
    – Perturbative
    2 days ago


















  • @Antonios-AlexandrosRobotis Ahh and every vector space has a basis so indeed $M_z$ must have a basis
    – Perturbative
    2 days ago
















@Antonios-AlexandrosRobotis Ahh and every vector space has a basis so indeed $M_z$ must have a basis
– Perturbative
2 days ago




@Antonios-AlexandrosRobotis Ahh and every vector space has a basis so indeed $M_z$ must have a basis
– Perturbative
2 days ago










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






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    1 Answer
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    active

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    1 Answer
    1






    active

    oldest

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    active

    oldest

    votes






    active

    oldest

    votes









    2














    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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      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








        2






        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












        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 2 days ago









        Tsemo Aristide

        56.1k11444




        56.1k11444






























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