Hensel Lifting with a non-simple root












2












$begingroup$


I was doing a problem and it involves lifting a root x= 55, from mod $2^{10}$ to a solution mod $2^{19}$ but the root is non simple, i.e. $$f'(x) equiv 0 (mod 2)$$



Here,
$f(x) = x^{3} - 9x + 8 equiv 0 $ $(mod 2^{10})$



Where x = 55 is a solution of f(x).



I am wondering how I go about trying to lift to this power modulus.






number-theory elementary-number-theory hensels-lemma







share|cite|improve this question











$endgroup$












  • $begingroup$
    Maybe substitute to find $(55+2^{10}x)^3-9(55+2^{10}x)+8equiv0pmod{2^{19}}$?
    $endgroup$
    – awllower
    Feb 14 '18 at 15:43






  • 1




    $begingroup$
    You could use the fact that $f(x)=(x-1)(x^2+x-8)$. You can do any kind of Hensel on the quadratic factor, and your desired number is the nonunit root. Alternatively, you can write $g(x)=f(x+1)=x(x^2 + 3x - 6)$ and work with the quadratic factor here. In addition, there are several ways of using Strong Hensel, which I guess you’re not aware of.
    $endgroup$
    – Lubin
    Feb 14 '18 at 22:28








  • 1




    $begingroup$
    @awllower Thank you! I got there, I found that the lift was $x = 55+83 cdot 2^{10}$ if you wanted the answer.
    $endgroup$
    – Bunneh
    Feb 15 '18 at 0:36


















2












$begingroup$


I was doing a problem and it involves lifting a root x= 55, from mod $2^{10}$ to a solution mod $2^{19}$ but the root is non simple, i.e. $$f'(x) equiv 0 (mod 2)$$



Here,
$f(x) = x^{3} - 9x + 8 equiv 0 $ $(mod 2^{10})$



Where x = 55 is a solution of f(x).



I am wondering how I go about trying to lift to this power modulus.






number-theory elementary-number-theory hensels-lemma







share|cite|improve this question











$endgroup$












  • $begingroup$
    Maybe substitute to find $(55+2^{10}x)^3-9(55+2^{10}x)+8equiv0pmod{2^{19}}$?
    $endgroup$
    – awllower
    Feb 14 '18 at 15:43






  • 1




    $begingroup$
    You could use the fact that $f(x)=(x-1)(x^2+x-8)$. You can do any kind of Hensel on the quadratic factor, and your desired number is the nonunit root. Alternatively, you can write $g(x)=f(x+1)=x(x^2 + 3x - 6)$ and work with the quadratic factor here. In addition, there are several ways of using Strong Hensel, which I guess you’re not aware of.
    $endgroup$
    – Lubin
    Feb 14 '18 at 22:28








  • 1




    $begingroup$
    @awllower Thank you! I got there, I found that the lift was $x = 55+83 cdot 2^{10}$ if you wanted the answer.
    $endgroup$
    – Bunneh
    Feb 15 '18 at 0:36
















2












2








2





$begingroup$


I was doing a problem and it involves lifting a root x= 55, from mod $2^{10}$ to a solution mod $2^{19}$ but the root is non simple, i.e. $$f'(x) equiv 0 (mod 2)$$



Here,
$f(x) = x^{3} - 9x + 8 equiv 0 $ $(mod 2^{10})$



Where x = 55 is a solution of f(x).



I am wondering how I go about trying to lift to this power modulus.






number-theory elementary-number-theory hensels-lemma







share|cite|improve this question











$endgroup$




I was doing a problem and it involves lifting a root x= 55, from mod $2^{10}$ to a solution mod $2^{19}$ but the root is non simple, i.e. $$f'(x) equiv 0 (mod 2)$$



Here,
$f(x) = x^{3} - 9x + 8 equiv 0 $ $(mod 2^{10})$



Where x = 55 is a solution of f(x).



I am wondering how I go about trying to lift to this power modulus.






number-theory elementary-number-theory hensels-lemma




number-theory elementary-number-theory hensels-lemma






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Feb 14 '18 at 20:08







Bunneh

















asked Feb 14 '18 at 15:13









BunnehBunneh

557




557












  • $begingroup$
    Maybe substitute to find $(55+2^{10}x)^3-9(55+2^{10}x)+8equiv0pmod{2^{19}}$?
    $endgroup$
    – awllower
    Feb 14 '18 at 15:43






  • 1




    $begingroup$
    You could use the fact that $f(x)=(x-1)(x^2+x-8)$. You can do any kind of Hensel on the quadratic factor, and your desired number is the nonunit root. Alternatively, you can write $g(x)=f(x+1)=x(x^2 + 3x - 6)$ and work with the quadratic factor here. In addition, there are several ways of using Strong Hensel, which I guess you’re not aware of.
    $endgroup$
    – Lubin
    Feb 14 '18 at 22:28








  • 1




    $begingroup$
    @awllower Thank you! I got there, I found that the lift was $x = 55+83 cdot 2^{10}$ if you wanted the answer.
    $endgroup$
    – Bunneh
    Feb 15 '18 at 0:36




















  • $begingroup$
    Maybe substitute to find $(55+2^{10}x)^3-9(55+2^{10}x)+8equiv0pmod{2^{19}}$?
    $endgroup$
    – awllower
    Feb 14 '18 at 15:43






  • 1




    $begingroup$
    You could use the fact that $f(x)=(x-1)(x^2+x-8)$. You can do any kind of Hensel on the quadratic factor, and your desired number is the nonunit root. Alternatively, you can write $g(x)=f(x+1)=x(x^2 + 3x - 6)$ and work with the quadratic factor here. In addition, there are several ways of using Strong Hensel, which I guess you’re not aware of.
    $endgroup$
    – Lubin
    Feb 14 '18 at 22:28








  • 1




    $begingroup$
    @awllower Thank you! I got there, I found that the lift was $x = 55+83 cdot 2^{10}$ if you wanted the answer.
    $endgroup$
    – Bunneh
    Feb 15 '18 at 0:36


















$begingroup$
Maybe substitute to find $(55+2^{10}x)^3-9(55+2^{10}x)+8equiv0pmod{2^{19}}$?
$endgroup$
– awllower
Feb 14 '18 at 15:43




$begingroup$
Maybe substitute to find $(55+2^{10}x)^3-9(55+2^{10}x)+8equiv0pmod{2^{19}}$?
$endgroup$
– awllower
Feb 14 '18 at 15:43




1




1




$begingroup$
You could use the fact that $f(x)=(x-1)(x^2+x-8)$. You can do any kind of Hensel on the quadratic factor, and your desired number is the nonunit root. Alternatively, you can write $g(x)=f(x+1)=x(x^2 + 3x - 6)$ and work with the quadratic factor here. In addition, there are several ways of using Strong Hensel, which I guess you’re not aware of.
$endgroup$
– Lubin
Feb 14 '18 at 22:28






$begingroup$
You could use the fact that $f(x)=(x-1)(x^2+x-8)$. You can do any kind of Hensel on the quadratic factor, and your desired number is the nonunit root. Alternatively, you can write $g(x)=f(x+1)=x(x^2 + 3x - 6)$ and work with the quadratic factor here. In addition, there are several ways of using Strong Hensel, which I guess you’re not aware of.
$endgroup$
– Lubin
Feb 14 '18 at 22:28






1




1




$begingroup$
@awllower Thank you! I got there, I found that the lift was $x = 55+83 cdot 2^{10}$ if you wanted the answer.
$endgroup$
– Bunneh
Feb 15 '18 at 0:36






$begingroup$
@awllower Thank you! I got there, I found that the lift was $x = 55+83 cdot 2^{10}$ if you wanted the answer.
$endgroup$
– Bunneh
Feb 15 '18 at 0:36












1 Answer
1






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0












$begingroup$

A general method is: If $x$ is a singular root $mod p^j$, then either it lifts to p roots $mod p^{j+1}$, or it lifts to no root, depending on if $f(x)$ is congruent to $0 mod p^{j+1}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    A remark. f(55) = 81*2^11, f'(55) = 4533*2^1. Since 11 >= 2*1 + 1, the solution 55 can be lifted to mod p^j with arbitrarily large j.
    $endgroup$
    – james0910
    Dec 6 '18 at 9:04











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









0












$begingroup$

A general method is: If $x$ is a singular root $mod p^j$, then either it lifts to p roots $mod p^{j+1}$, or it lifts to no root, depending on if $f(x)$ is congruent to $0 mod p^{j+1}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    A remark. f(55) = 81*2^11, f'(55) = 4533*2^1. Since 11 >= 2*1 + 1, the solution 55 can be lifted to mod p^j with arbitrarily large j.
    $endgroup$
    – james0910
    Dec 6 '18 at 9:04
















0












$begingroup$

A general method is: If $x$ is a singular root $mod p^j$, then either it lifts to p roots $mod p^{j+1}$, or it lifts to no root, depending on if $f(x)$ is congruent to $0 mod p^{j+1}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    A remark. f(55) = 81*2^11, f'(55) = 4533*2^1. Since 11 >= 2*1 + 1, the solution 55 can be lifted to mod p^j with arbitrarily large j.
    $endgroup$
    – james0910
    Dec 6 '18 at 9:04














0












0








0





$begingroup$

A general method is: If $x$ is a singular root $mod p^j$, then either it lifts to p roots $mod p^{j+1}$, or it lifts to no root, depending on if $f(x)$ is congruent to $0 mod p^{j+1}$.






share|cite|improve this answer











$endgroup$



A general method is: If $x$ is a singular root $mod p^j$, then either it lifts to p roots $mod p^{j+1}$, or it lifts to no root, depending on if $f(x)$ is congruent to $0 mod p^{j+1}$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 6 '18 at 5:12









Tianlalu

3,08621038




3,08621038










answered Dec 6 '18 at 4:50









james0910james0910

111




111












  • $begingroup$
    A remark. f(55) = 81*2^11, f'(55) = 4533*2^1. Since 11 >= 2*1 + 1, the solution 55 can be lifted to mod p^j with arbitrarily large j.
    $endgroup$
    – james0910
    Dec 6 '18 at 9:04


















  • $begingroup$
    A remark. f(55) = 81*2^11, f'(55) = 4533*2^1. Since 11 >= 2*1 + 1, the solution 55 can be lifted to mod p^j with arbitrarily large j.
    $endgroup$
    – james0910
    Dec 6 '18 at 9:04
















$begingroup$
A remark. f(55) = 81*2^11, f'(55) = 4533*2^1. Since 11 >= 2*1 + 1, the solution 55 can be lifted to mod p^j with arbitrarily large j.
$endgroup$
– james0910
Dec 6 '18 at 9:04




$begingroup$
A remark. f(55) = 81*2^11, f'(55) = 4533*2^1. Since 11 >= 2*1 + 1, the solution 55 can be lifted to mod p^j with arbitrarily large j.
$endgroup$
– james0910
Dec 6 '18 at 9:04


















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