what would be the solution by using Hensel's Lemma?, p-adic numbers












1












$begingroup$


Point out the main difference or relation between Newton's polygon and Hensel's lemma when it comes to find solution of the two variable polynomial $ f(x,y)=y^6-5xy^5+x^3y^4-7x^2y^2+6x^3+x^4=0$.



$ text{Using Newton's polygon}:$



Plot the points (exponents $(a,b)$) as given below:



$(6,0), (5,1), (4,3), (2,2), (0,3), (0,4)$.



The plot is given below:



enter image description here



Thus the vertices of the Newton polygon are $ (0,3), ( 2,2), (6,0)$.



The line segment joining these points is $ x+2y=6$.



Now the monomial s corresponding to the vertices $ (0,3), ( 2,2), (6,0)$ are



$y^6-7x^2y^2+6x^3=0, .......(1)$.



This can be solved easily.



Divide the equation $(1)$ by $x^3$ both sides,



$ frac{y^6}{x^3}-7 frac{y^2}{x}+6=0$



or, $ Y^6-7Y^2+6=0$, where $ Y=frac{y}{sqrt x}$,



This can be written as



$ (Y^2-1)(Y^2-2)(Y^2+3)=0 \ Rightarrow Y=pm 1, pm sqrt 2, pm sqrt{-3} . $



i.e., $ y=pm sqrt x, y=pm sqrt {2x}, y=pm sqrt{-3x}$.



These are the solutions by using $ text{Newton Polygon}$.



But what would be the solution by using Hensel's Lemma?



What would be the difference?



Please help me using Hensel's lemma and pointing out the difference or relation in both method.






p-adic-number-theory hensels-lemma







share|cite|improve this question









$endgroup$












  • $begingroup$
    $f(x,y) in mathbb{Z}_p[x,y], g(x,y ) = partial_x f(x,y) in mathbb{Z}_p[x,y]$. Given $b in mathbb{Z}_p$ and $A bmod p$, if $f(A bmod p,b bmod p) = 0 bmod p$ and $g(A bmod p,b bmod p) ne 0 bmod p$ then you can find $a in A+pmathbb{Z}_p$ such that $f(a,b)=0$ and that $a$ is unique.
    $endgroup$
    – reuns
    Dec 10 '18 at 18:44












  • $begingroup$
    @reuns, Please give a short answer so that I will get you. As i am new to this topic, I will not get by hints.
    $endgroup$
    – M. A. SARKAR
    Dec 11 '18 at 6:15










  • $begingroup$
    Well look at en.wikipedia.org/wiki/Hensel%27s_lemma
    $endgroup$
    – reuns
    Dec 11 '18 at 14:35
















1












$begingroup$


Point out the main difference or relation between Newton's polygon and Hensel's lemma when it comes to find solution of the two variable polynomial $ f(x,y)=y^6-5xy^5+x^3y^4-7x^2y^2+6x^3+x^4=0$.



$ text{Using Newton's polygon}:$



Plot the points (exponents $(a,b)$) as given below:



$(6,0), (5,1), (4,3), (2,2), (0,3), (0,4)$.



The plot is given below:



enter image description here



Thus the vertices of the Newton polygon are $ (0,3), ( 2,2), (6,0)$.



The line segment joining these points is $ x+2y=6$.



Now the monomial s corresponding to the vertices $ (0,3), ( 2,2), (6,0)$ are



$y^6-7x^2y^2+6x^3=0, .......(1)$.



This can be solved easily.



Divide the equation $(1)$ by $x^3$ both sides,



$ frac{y^6}{x^3}-7 frac{y^2}{x}+6=0$



or, $ Y^6-7Y^2+6=0$, where $ Y=frac{y}{sqrt x}$,



This can be written as



$ (Y^2-1)(Y^2-2)(Y^2+3)=0 \ Rightarrow Y=pm 1, pm sqrt 2, pm sqrt{-3} . $



i.e., $ y=pm sqrt x, y=pm sqrt {2x}, y=pm sqrt{-3x}$.



These are the solutions by using $ text{Newton Polygon}$.



But what would be the solution by using Hensel's Lemma?



What would be the difference?



Please help me using Hensel's lemma and pointing out the difference or relation in both method.






p-adic-number-theory hensels-lemma







share|cite|improve this question









$endgroup$












  • $begingroup$
    $f(x,y) in mathbb{Z}_p[x,y], g(x,y ) = partial_x f(x,y) in mathbb{Z}_p[x,y]$. Given $b in mathbb{Z}_p$ and $A bmod p$, if $f(A bmod p,b bmod p) = 0 bmod p$ and $g(A bmod p,b bmod p) ne 0 bmod p$ then you can find $a in A+pmathbb{Z}_p$ such that $f(a,b)=0$ and that $a$ is unique.
    $endgroup$
    – reuns
    Dec 10 '18 at 18:44












  • $begingroup$
    @reuns, Please give a short answer so that I will get you. As i am new to this topic, I will not get by hints.
    $endgroup$
    – M. A. SARKAR
    Dec 11 '18 at 6:15










  • $begingroup$
    Well look at en.wikipedia.org/wiki/Hensel%27s_lemma
    $endgroup$
    – reuns
    Dec 11 '18 at 14:35














1












1








1


3



$begingroup$


Point out the main difference or relation between Newton's polygon and Hensel's lemma when it comes to find solution of the two variable polynomial $ f(x,y)=y^6-5xy^5+x^3y^4-7x^2y^2+6x^3+x^4=0$.



$ text{Using Newton's polygon}:$



Plot the points (exponents $(a,b)$) as given below:



$(6,0), (5,1), (4,3), (2,2), (0,3), (0,4)$.



The plot is given below:



enter image description here



Thus the vertices of the Newton polygon are $ (0,3), ( 2,2), (6,0)$.



The line segment joining these points is $ x+2y=6$.



Now the monomial s corresponding to the vertices $ (0,3), ( 2,2), (6,0)$ are



$y^6-7x^2y^2+6x^3=0, .......(1)$.



This can be solved easily.



Divide the equation $(1)$ by $x^3$ both sides,



$ frac{y^6}{x^3}-7 frac{y^2}{x}+6=0$



or, $ Y^6-7Y^2+6=0$, where $ Y=frac{y}{sqrt x}$,



This can be written as



$ (Y^2-1)(Y^2-2)(Y^2+3)=0 \ Rightarrow Y=pm 1, pm sqrt 2, pm sqrt{-3} . $



i.e., $ y=pm sqrt x, y=pm sqrt {2x}, y=pm sqrt{-3x}$.



These are the solutions by using $ text{Newton Polygon}$.



But what would be the solution by using Hensel's Lemma?



What would be the difference?



Please help me using Hensel's lemma and pointing out the difference or relation in both method.






p-adic-number-theory hensels-lemma







share|cite|improve this question









$endgroup$




Point out the main difference or relation between Newton's polygon and Hensel's lemma when it comes to find solution of the two variable polynomial $ f(x,y)=y^6-5xy^5+x^3y^4-7x^2y^2+6x^3+x^4=0$.



$ text{Using Newton's polygon}:$



Plot the points (exponents $(a,b)$) as given below:



$(6,0), (5,1), (4,3), (2,2), (0,3), (0,4)$.



The plot is given below:



enter image description here



Thus the vertices of the Newton polygon are $ (0,3), ( 2,2), (6,0)$.



The line segment joining these points is $ x+2y=6$.



Now the monomial s corresponding to the vertices $ (0,3), ( 2,2), (6,0)$ are



$y^6-7x^2y^2+6x^3=0, .......(1)$.



This can be solved easily.



Divide the equation $(1)$ by $x^3$ both sides,



$ frac{y^6}{x^3}-7 frac{y^2}{x}+6=0$



or, $ Y^6-7Y^2+6=0$, where $ Y=frac{y}{sqrt x}$,



This can be written as



$ (Y^2-1)(Y^2-2)(Y^2+3)=0 \ Rightarrow Y=pm 1, pm sqrt 2, pm sqrt{-3} . $



i.e., $ y=pm sqrt x, y=pm sqrt {2x}, y=pm sqrt{-3x}$.



These are the solutions by using $ text{Newton Polygon}$.



But what would be the solution by using Hensel's Lemma?



What would be the difference?



Please help me using Hensel's lemma and pointing out the difference or relation in both method.






p-adic-number-theory hensels-lemma




p-adic-number-theory hensels-lemma






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 10 '18 at 18:12









M. A. SARKARM. A. SARKAR

2,2381619




2,2381619












  • $begingroup$
    $f(x,y) in mathbb{Z}_p[x,y], g(x,y ) = partial_x f(x,y) in mathbb{Z}_p[x,y]$. Given $b in mathbb{Z}_p$ and $A bmod p$, if $f(A bmod p,b bmod p) = 0 bmod p$ and $g(A bmod p,b bmod p) ne 0 bmod p$ then you can find $a in A+pmathbb{Z}_p$ such that $f(a,b)=0$ and that $a$ is unique.
    $endgroup$
    – reuns
    Dec 10 '18 at 18:44












  • $begingroup$
    @reuns, Please give a short answer so that I will get you. As i am new to this topic, I will not get by hints.
    $endgroup$
    – M. A. SARKAR
    Dec 11 '18 at 6:15










  • $begingroup$
    Well look at en.wikipedia.org/wiki/Hensel%27s_lemma
    $endgroup$
    – reuns
    Dec 11 '18 at 14:35


















  • $begingroup$
    $f(x,y) in mathbb{Z}_p[x,y], g(x,y ) = partial_x f(x,y) in mathbb{Z}_p[x,y]$. Given $b in mathbb{Z}_p$ and $A bmod p$, if $f(A bmod p,b bmod p) = 0 bmod p$ and $g(A bmod p,b bmod p) ne 0 bmod p$ then you can find $a in A+pmathbb{Z}_p$ such that $f(a,b)=0$ and that $a$ is unique.
    $endgroup$
    – reuns
    Dec 10 '18 at 18:44












  • $begingroup$
    @reuns, Please give a short answer so that I will get you. As i am new to this topic, I will not get by hints.
    $endgroup$
    – M. A. SARKAR
    Dec 11 '18 at 6:15










  • $begingroup$
    Well look at en.wikipedia.org/wiki/Hensel%27s_lemma
    $endgroup$
    – reuns
    Dec 11 '18 at 14:35
















$begingroup$
$f(x,y) in mathbb{Z}_p[x,y], g(x,y ) = partial_x f(x,y) in mathbb{Z}_p[x,y]$. Given $b in mathbb{Z}_p$ and $A bmod p$, if $f(A bmod p,b bmod p) = 0 bmod p$ and $g(A bmod p,b bmod p) ne 0 bmod p$ then you can find $a in A+pmathbb{Z}_p$ such that $f(a,b)=0$ and that $a$ is unique.
$endgroup$
– reuns
Dec 10 '18 at 18:44






$begingroup$
$f(x,y) in mathbb{Z}_p[x,y], g(x,y ) = partial_x f(x,y) in mathbb{Z}_p[x,y]$. Given $b in mathbb{Z}_p$ and $A bmod p$, if $f(A bmod p,b bmod p) = 0 bmod p$ and $g(A bmod p,b bmod p) ne 0 bmod p$ then you can find $a in A+pmathbb{Z}_p$ such that $f(a,b)=0$ and that $a$ is unique.
$endgroup$
– reuns
Dec 10 '18 at 18:44














$begingroup$
@reuns, Please give a short answer so that I will get you. As i am new to this topic, I will not get by hints.
$endgroup$
– M. A. SARKAR
Dec 11 '18 at 6:15




$begingroup$
@reuns, Please give a short answer so that I will get you. As i am new to this topic, I will not get by hints.
$endgroup$
– M. A. SARKAR
Dec 11 '18 at 6:15












$begingroup$
Well look at en.wikipedia.org/wiki/Hensel%27s_lemma
$endgroup$
– reuns
Dec 11 '18 at 14:35




$begingroup$
Well look at en.wikipedia.org/wiki/Hensel%27s_lemma
$endgroup$
– reuns
Dec 11 '18 at 14:35










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