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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

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

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asked Mar 14 at 13:37

bellcirclebellcircle

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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

asked Mar 14 at 13:37

bellcirclebellcircle

1,373411

$endgroup$

add a comment | 

$begingroup$

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

asked Mar 14 at 13:37

bellcirclebellcircle

1,373411

$endgroup$

share|cite|improve this question

asked Mar 14 at 13:37

bellcirclebellcircle

1,373411

share|cite|improve this question

asked Mar 14 at 13:37

bellcirclebellcircle

1,373411

asked Mar 14 at 13:37

bellcirclebellcircle

1,373411

asked Mar 14 at 13:37

bellcirclebellcircle

1,373411

1 Answer
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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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answered Mar 14 at 14:40

bellcirclebellcircle

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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

answered Mar 14 at 14:40

bellcirclebellcircle

1,373411

$endgroup$

add a comment | 

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

answered Mar 14 at 14:40

bellcirclebellcircle

1,373411

$endgroup$

add a comment | 

$begingroup$

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

answered Mar 14 at 14:40

bellcirclebellcircle

1,373411

$endgroup$

share|cite|improve this answer

answered Mar 14 at 14:40

bellcirclebellcircle

1,373411

answered Mar 14 at 14:40

bellcirclebellcircle

1,373411

answered Mar 14 at 14:40

bellcirclebellcircle

1,373411