Generalizing an implicit function theorem for formal power series$p$ prime, Group of order $p^n$ is cyclic...

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Generalizing an implicit function theorem for formal power series


$p$ prime, Group of order $p^n$ is cyclic iff it is an abelian group having a unique subgroup of order $p$Finitely generated modules over a PIR (structure theorem application)An Exercise from Lang's AlgebraAlgebraic Formal Power SeriesGroup Presentation of the Direct Product.Power Series Arithmetic through Formal Power SeriesIf $A$ is a subalgebra of $M_n(mathbb{C})$, the $C^*$ modules of $A$ are semisimpleidentity of a formal power seriesTaylor expansion for formal power seriesProving that the given element is irreducible but not prime













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$begingroup$


This exercise is from Greuel & Lossen & Shustin's Introduction to Singularities and Deformations, Exercise 1.2.5.




Let $finmathbf{C}langle mathbf{x},yrangle$, where $mathbf{C}langle mathbf{x},yrangle$ is the ring of convergent power series with $n+1$ variables, namely $mathbf{x}=(x_1,ldots,x_n)$ and $y$. If $fin langle yrangle +langle mathbf{x}rangle^ksubsetmathbf{C}langle mathbf{x},yrangle$ and $frac{partial f}{partial y}(mathbf{0})neq 0,$ i.e. the coefficient of $y$ is nonzero, then prove that there exists $Yin langle mathbf{x}rangle^ksubsetmathbf{C}langle mathbf{x}rangle$ such that $f(mathbf{x},Y)=0$.




My attempt: If $k=1$, then it is merely the direct consequence of implicit function theorem. However, the problem requires to prove much stronger result. First I attempted as follows:



Let $f=yg+h$ where $gin mathbf{C}langle mathbf{x},yrangle$ and $hin langle mathbf{x}rangle^k.$ Then I simply let $Y=-g^{-1}h$, but I found that $Y$ is not necessarily in $mathbf{C}langle mathbf{x}rangle$. How should I find $Y$ such that $Yin mathbf{C}langle mathbf{x}rangle$ satisfying $f(mathbf{x},Y)=0$?



Thanks in advance!






abstract-algebra analytic-geometry analytic-functions formal-power-series







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    This exercise is from Greuel & Lossen & Shustin's Introduction to Singularities and Deformations, Exercise 1.2.5.




    Let $finmathbf{C}langle mathbf{x},yrangle$, where $mathbf{C}langle mathbf{x},yrangle$ is the ring of convergent power series with $n+1$ variables, namely $mathbf{x}=(x_1,ldots,x_n)$ and $y$. If $fin langle yrangle +langle mathbf{x}rangle^ksubsetmathbf{C}langle mathbf{x},yrangle$ and $frac{partial f}{partial y}(mathbf{0})neq 0,$ i.e. the coefficient of $y$ is nonzero, then prove that there exists $Yin langle mathbf{x}rangle^ksubsetmathbf{C}langle mathbf{x}rangle$ such that $f(mathbf{x},Y)=0$.




    My attempt: If $k=1$, then it is merely the direct consequence of implicit function theorem. However, the problem requires to prove much stronger result. First I attempted as follows:



    Let $f=yg+h$ where $gin mathbf{C}langle mathbf{x},yrangle$ and $hin langle mathbf{x}rangle^k.$ Then I simply let $Y=-g^{-1}h$, but I found that $Y$ is not necessarily in $mathbf{C}langle mathbf{x}rangle$. How should I find $Y$ such that $Yin mathbf{C}langle mathbf{x}rangle$ satisfying $f(mathbf{x},Y)=0$?



    Thanks in advance!






    abstract-algebra analytic-geometry analytic-functions formal-power-series







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      This exercise is from Greuel & Lossen & Shustin's Introduction to Singularities and Deformations, Exercise 1.2.5.




      Let $finmathbf{C}langle mathbf{x},yrangle$, where $mathbf{C}langle mathbf{x},yrangle$ is the ring of convergent power series with $n+1$ variables, namely $mathbf{x}=(x_1,ldots,x_n)$ and $y$. If $fin langle yrangle +langle mathbf{x}rangle^ksubsetmathbf{C}langle mathbf{x},yrangle$ and $frac{partial f}{partial y}(mathbf{0})neq 0,$ i.e. the coefficient of $y$ is nonzero, then prove that there exists $Yin langle mathbf{x}rangle^ksubsetmathbf{C}langle mathbf{x}rangle$ such that $f(mathbf{x},Y)=0$.




      My attempt: If $k=1$, then it is merely the direct consequence of implicit function theorem. However, the problem requires to prove much stronger result. First I attempted as follows:



      Let $f=yg+h$ where $gin mathbf{C}langle mathbf{x},yrangle$ and $hin langle mathbf{x}rangle^k.$ Then I simply let $Y=-g^{-1}h$, but I found that $Y$ is not necessarily in $mathbf{C}langle mathbf{x}rangle$. How should I find $Y$ such that $Yin mathbf{C}langle mathbf{x}rangle$ satisfying $f(mathbf{x},Y)=0$?



      Thanks in advance!






      abstract-algebra analytic-geometry analytic-functions formal-power-series







      share|cite|improve this question









      $endgroup$




      This exercise is from Greuel & Lossen & Shustin's Introduction to Singularities and Deformations, Exercise 1.2.5.




      Let $finmathbf{C}langle mathbf{x},yrangle$, where $mathbf{C}langle mathbf{x},yrangle$ is the ring of convergent power series with $n+1$ variables, namely $mathbf{x}=(x_1,ldots,x_n)$ and $y$. If $fin langle yrangle +langle mathbf{x}rangle^ksubsetmathbf{C}langle mathbf{x},yrangle$ and $frac{partial f}{partial y}(mathbf{0})neq 0,$ i.e. the coefficient of $y$ is nonzero, then prove that there exists $Yin langle mathbf{x}rangle^ksubsetmathbf{C}langle mathbf{x}rangle$ such that $f(mathbf{x},Y)=0$.




      My attempt: If $k=1$, then it is merely the direct consequence of implicit function theorem. However, the problem requires to prove much stronger result. First I attempted as follows:



      Let $f=yg+h$ where $gin mathbf{C}langle mathbf{x},yrangle$ and $hin langle mathbf{x}rangle^k.$ Then I simply let $Y=-g^{-1}h$, but I found that $Y$ is not necessarily in $mathbf{C}langle mathbf{x}rangle$. How should I find $Y$ such that $Yin mathbf{C}langle mathbf{x}rangle$ satisfying $f(mathbf{x},Y)=0$?



      Thanks in advance!






      abstract-algebra analytic-geometry analytic-functions formal-power-series




      abstract-algebra analytic-geometry analytic-functions formal-power-series






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




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      asked Mar 14 at 13:37









      bellcirclebellcircle

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      1,373411






















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          $begingroup$

          Let $f(mathbf{x},y)=yg(mathbf{x},y)+h(mathbf{x},y),$ where $gin mathbf{C}langlemathbf{x},yrangle$ and $hin langlemathbf{x}rangle^k$. Then $f(mathbf{0},0)=0$ and by assumption $frac{partial f}{partial y}(mathbf{0},0)neq 0.$ Therefore, by implicit function theorem, there exists a $Y(mathbf{x})in langlemathbf{x}ranglesubsetmathbf{C}langle mathbf{x}rangle$ such that $f(mathbf{x},Y(mathbf{x}))=0.$ This means $Y(mathbf{x})g(mathbf{x},Y(mathbf{x}))+h(mathbf{x},Y(mathbf{x}))=0.$ Note that $g(mathbf{x},Y(mathbf{x}))$ is a unit and $h(mathbf{x},Y(mathbf{x}))in langle xrangle^k.$ Then it must be the case that $Y(mathbf{x})in langle mathbf{x}rangle^k$.






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            $begingroup$

            Let $f(mathbf{x},y)=yg(mathbf{x},y)+h(mathbf{x},y),$ where $gin mathbf{C}langlemathbf{x},yrangle$ and $hin langlemathbf{x}rangle^k$. Then $f(mathbf{0},0)=0$ and by assumption $frac{partial f}{partial y}(mathbf{0},0)neq 0.$ Therefore, by implicit function theorem, there exists a $Y(mathbf{x})in langlemathbf{x}ranglesubsetmathbf{C}langle mathbf{x}rangle$ such that $f(mathbf{x},Y(mathbf{x}))=0.$ This means $Y(mathbf{x})g(mathbf{x},Y(mathbf{x}))+h(mathbf{x},Y(mathbf{x}))=0.$ Note that $g(mathbf{x},Y(mathbf{x}))$ is a unit and $h(mathbf{x},Y(mathbf{x}))in langle xrangle^k.$ Then it must be the case that $Y(mathbf{x})in langle mathbf{x}rangle^k$.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              Let $f(mathbf{x},y)=yg(mathbf{x},y)+h(mathbf{x},y),$ where $gin mathbf{C}langlemathbf{x},yrangle$ and $hin langlemathbf{x}rangle^k$. Then $f(mathbf{0},0)=0$ and by assumption $frac{partial f}{partial y}(mathbf{0},0)neq 0.$ Therefore, by implicit function theorem, there exists a $Y(mathbf{x})in langlemathbf{x}ranglesubsetmathbf{C}langle mathbf{x}rangle$ such that $f(mathbf{x},Y(mathbf{x}))=0.$ This means $Y(mathbf{x})g(mathbf{x},Y(mathbf{x}))+h(mathbf{x},Y(mathbf{x}))=0.$ Note that $g(mathbf{x},Y(mathbf{x}))$ is a unit and $h(mathbf{x},Y(mathbf{x}))in langle xrangle^k.$ Then it must be the case that $Y(mathbf{x})in langle mathbf{x}rangle^k$.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                Let $f(mathbf{x},y)=yg(mathbf{x},y)+h(mathbf{x},y),$ where $gin mathbf{C}langlemathbf{x},yrangle$ and $hin langlemathbf{x}rangle^k$. Then $f(mathbf{0},0)=0$ and by assumption $frac{partial f}{partial y}(mathbf{0},0)neq 0.$ Therefore, by implicit function theorem, there exists a $Y(mathbf{x})in langlemathbf{x}ranglesubsetmathbf{C}langle mathbf{x}rangle$ such that $f(mathbf{x},Y(mathbf{x}))=0.$ This means $Y(mathbf{x})g(mathbf{x},Y(mathbf{x}))+h(mathbf{x},Y(mathbf{x}))=0.$ Note that $g(mathbf{x},Y(mathbf{x}))$ is a unit and $h(mathbf{x},Y(mathbf{x}))in langle xrangle^k.$ Then it must be the case that $Y(mathbf{x})in langle mathbf{x}rangle^k$.






                share|cite|improve this answer









                $endgroup$



                Let $f(mathbf{x},y)=yg(mathbf{x},y)+h(mathbf{x},y),$ where $gin mathbf{C}langlemathbf{x},yrangle$ and $hin langlemathbf{x}rangle^k$. Then $f(mathbf{0},0)=0$ and by assumption $frac{partial f}{partial y}(mathbf{0},0)neq 0.$ Therefore, by implicit function theorem, there exists a $Y(mathbf{x})in langlemathbf{x}ranglesubsetmathbf{C}langle mathbf{x}rangle$ such that $f(mathbf{x},Y(mathbf{x}))=0.$ This means $Y(mathbf{x})g(mathbf{x},Y(mathbf{x}))+h(mathbf{x},Y(mathbf{x}))=0.$ Note that $g(mathbf{x},Y(mathbf{x}))$ is a unit and $h(mathbf{x},Y(mathbf{x}))in langle xrangle^k.$ Then it must be the case that $Y(mathbf{x})in langle mathbf{x}rangle^k$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Mar 14 at 14:40









                bellcirclebellcircle

                1,373411




                1,373411






























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