Mixing
Hensel's lemma for the completion of $mathbb{F}_q(t)$
$begingroup$
If we want to find the roots of a polynomial $f(x)$ modulo a prime $p$ to the power of $n$, we can use Hensel's lemma. Let's say we want to find all roots of $x^3+x^2+4x+1$ mod $49$. Then we can use Hensel's lemma for p-adic integers and compute the roots only mod $7$.
Is there also a version of Hensel's lemma, not only for the completion of $mathbb{Q}$, but for the completion of $mathbb{F}_q(t)$? Or exists any other root lifting technique in this case? I didn't found anything so far...
number-theory function-fields hensels-lemma
asked Jan 29 at 21:44
SqyuliSqyuli
344111
$endgroup$
add a comment |
$begingroup$
If we want to find the roots of a polynomial $f(x)$ modulo a prime $p$ to the power of $n$, we can use Hensel's lemma. Let's say we want to find all roots of $x^3+x^2+4x+1$ mod $49$. Then we can use Hensel's lemma for p-adic integers and compute the roots only mod $7$.
Is there also a version of Hensel's lemma, not only for the completion of $mathbb{Q}$, but for the completion of $mathbb{F}_q(t)$? Or exists any other root lifting technique in this case? I didn't found anything so far...
number-theory function-fields hensels-lemma
asked Jan 29 at 21:44
SqyuliSqyuli
344111
$endgroup$
2
$begingroup$
Sure. Let $f(X) in mathbb{F}_q[[t]][X]$, $q$ an odd prime power. Hensel lemma is just saying that if for $a_k in mathbb{F}_q[[t]]$, $f(a_k) equiv 0 bmod t^k mathbb{F}_q[[t]]$ then $f(a_k+bt^{k}) = f(a_k)+b t^{k} f'(a_k) +O(t^{2k})$ so if $f'(a_k) equiv f'(a_1) notequiv 0 bmod t mathbb{F}_q[[t]]$ then you can always choose $a_{k+1} = a_k + b t_{k}, b equiv - frac{f(a_k)}{t^{k} f'(a_1)} bmod t mathbb{F}_q[[t]]$ such that $f(a_{k+1}) equiv 0 bmod t^{k+1}mathbb{F}_q[[t]]$, moreover the obtained limit $lim_{k to infty} a_k$ is the unique lift of $a_1$.
$endgroup$
– reuns
Jan 29 at 22:41
$begingroup$
Sounds good, but what do you mean with mod $...mathbb{F}_q[[t]]$? In the case of the p-adic integers I only look at mod $p$ and not $p mathbb{Q}_p$.
$endgroup$
– Sqyuli
Jan 30 at 8:03
$begingroup$
With $mathbb{Z}_p$ it works the same way replacing $t, mathbb{F}_q[[t]]$ by $p, mathbb{Z}_p$. My comment applies to any complete DVR whose uniformizer satisfies $(a+bpi^k)^n = a^n+bpi^k n a^{n-1}+O(pi^{k+1})$
$endgroup$
– reuns
Jan 30 at 10:25
add a comment |
$begingroup$
If we want to find the roots of a polynomial $f(x)$ modulo a prime $p$ to the power of $n$, we can use Hensel's lemma. Let's say we want to find all roots of $x^3+x^2+4x+1$ mod $49$. Then we can use Hensel's lemma for p-adic integers and compute the roots only mod $7$.
Is there also a version of Hensel's lemma, not only for the completion of $mathbb{Q}$, but for the completion of $mathbb{F}_q(t)$? Or exists any other root lifting technique in this case? I didn't found anything so far...
number-theory function-fields hensels-lemma
asked Jan 29 at 21:44
SqyuliSqyuli
344111
$endgroup$
If we want to find the roots of a polynomial $f(x)$ modulo a prime $p$ to the power of $n$, we can use Hensel's lemma. Let's say we want to find all roots of $x^3+x^2+4x+1$ mod $49$. Then we can use Hensel's lemma for p-adic integers and compute the roots only mod $7$.
Is there also a version of Hensel's lemma, not only for the completion of $mathbb{Q}$, but for the completion of $mathbb{F}_q(t)$? Or exists any other root lifting technique in this case? I didn't found anything so far...
number-theory function-fields hensels-lemma
number-theory function-fields hensels-lemma
asked Jan 29 at 21:44
SqyuliSqyuli
344111
asked Jan 29 at 21:44
SqyuliSqyuli
344111
asked Jan 29 at 21:44
SqyuliSqyuli
344111
asked Jan 29 at 21:44
SqyuliSqyuli
344111
asked Jan 29 at 21:44
SqyuliSqyuli
344111
344111
2
$begingroup$
Sure. Let $f(X) in mathbb{F}_q[[t]][X]$, $q$ an odd prime power. Hensel lemma is just saying that if for $a_k in mathbb{F}_q[[t]]$, $f(a_k) equiv 0 bmod t^k mathbb{F}_q[[t]]$ then $f(a_k+bt^{k}) = f(a_k)+b t^{k} f'(a_k) +O(t^{2k})$ so if $f'(a_k) equiv f'(a_1) notequiv 0 bmod t mathbb{F}_q[[t]]$ then you can always choose $a_{k+1} = a_k + b t_{k}, b equiv - frac{f(a_k)}{t^{k} f'(a_1)} bmod t mathbb{F}_q[[t]]$ such that $f(a_{k+1}) equiv 0 bmod t^{k+1}mathbb{F}_q[[t]]$, moreover the obtained limit $lim_{k to infty} a_k$ is the unique lift of $a_1$.
$endgroup$
– reuns
Jan 29 at 22:41
$begingroup$
Sounds good, but what do you mean with mod $...mathbb{F}_q[[t]]$? In the case of the p-adic integers I only look at mod $p$ and not $p mathbb{Q}_p$.
$endgroup$
– Sqyuli
Jan 30 at 8:03
$begingroup$
With $mathbb{Z}_p$ it works the same way replacing $t, mathbb{F}_q[[t]]$ by $p, mathbb{Z}_p$. My comment applies to any complete DVR whose uniformizer satisfies $(a+bpi^k)^n = a^n+bpi^k n a^{n-1}+O(pi^{k+1})$
$endgroup$
– reuns
Jan 30 at 10:25
add a comment |
2
$begingroup$
Sure. Let $f(X) in mathbb{F}_q[[t]][X]$, $q$ an odd prime power. Hensel lemma is just saying that if for $a_k in mathbb{F}_q[[t]]$, $f(a_k) equiv 0 bmod t^k mathbb{F}_q[[t]]$ then $f(a_k+bt^{k}) = f(a_k)+b t^{k} f'(a_k) +O(t^{2k})$ so if $f'(a_k) equiv f'(a_1) notequiv 0 bmod t mathbb{F}_q[[t]]$ then you can always choose $a_{k+1} = a_k + b t_{k}, b equiv - frac{f(a_k)}{t^{k} f'(a_1)} bmod t mathbb{F}_q[[t]]$ such that $f(a_{k+1}) equiv 0 bmod t^{k+1}mathbb{F}_q[[t]]$, moreover the obtained limit $lim_{k to infty} a_k$ is the unique lift of $a_1$.
$endgroup$
– reuns
Jan 29 at 22:41
$begingroup$
Sounds good, but what do you mean with mod $...mathbb{F}_q[[t]]$? In the case of the p-adic integers I only look at mod $p$ and not $p mathbb{Q}_p$.
$endgroup$
– Sqyuli
Jan 30 at 8:03
$begingroup$
With $mathbb{Z}_p$ it works the same way replacing $t, mathbb{F}_q[[t]]$ by $p, mathbb{Z}_p$. My comment applies to any complete DVR whose uniformizer satisfies $(a+bpi^k)^n = a^n+bpi^k n a^{n-1}+O(pi^{k+1})$
$endgroup$
– reuns
Jan 30 at 10:25
2
2
$begingroup$
Sure. Let $f(X) in mathbb{F}_q[[t]][X]$, $q$ an odd prime power. Hensel lemma is just saying that if for $a_k in mathbb{F}_q[[t]]$, $f(a_k) equiv 0 bmod t^k mathbb{F}_q[[t]]$ then $f(a_k+bt^{k}) = f(a_k)+b t^{k} f'(a_k) +O(t^{2k})$ so if $f'(a_k) equiv f'(a_1) notequiv 0 bmod t mathbb{F}_q[[t]]$ then you can always choose $a_{k+1} = a_k + b t_{k}, b equiv - frac{f(a_k)}{t^{k} f'(a_1)} bmod t mathbb{F}_q[[t]]$ such that $f(a_{k+1}) equiv 0 bmod t^{k+1}mathbb{F}_q[[t]]$, moreover the obtained limit $lim_{k to infty} a_k$ is the unique lift of $a_1$.
$endgroup$
– reuns
Jan 29 at 22:41
$begingroup$
Sure. Let $f(X) in mathbb{F}_q[[t]][X]$, $q$ an odd prime power. Hensel lemma is just saying that if for $a_k in mathbb{F}_q[[t]]$, $f(a_k) equiv 0 bmod t^k mathbb{F}_q[[t]]$ then $f(a_k+bt^{k}) = f(a_k)+b t^{k} f'(a_k) +O(t^{2k})$ so if $f'(a_k) equiv f'(a_1) notequiv 0 bmod t mathbb{F}_q[[t]]$ then you can always choose $a_{k+1} = a_k + b t_{k}, b equiv - frac{f(a_k)}{t^{k} f'(a_1)} bmod t mathbb{F}_q[[t]]$ such that $f(a_{k+1}) equiv 0 bmod t^{k+1}mathbb{F}_q[[t]]$, moreover the obtained limit $lim_{k to infty} a_k$ is the unique lift of $a_1$.
$endgroup$
– reuns
Jan 29 at 22:41
$begingroup$
Sounds good, but what do you mean with mod $...mathbb{F}_q[[t]]$? In the case of the p-adic integers I only look at mod $p$ and not $p mathbb{Q}_p$.
$endgroup$
– Sqyuli
Jan 30 at 8:03
$begingroup$
Sounds good, but what do you mean with mod $...mathbb{F}_q[[t]]$? In the case of the p-adic integers I only look at mod $p$ and not $p mathbb{Q}_p$.
$endgroup$
– Sqyuli
Jan 30 at 8:03
$begingroup$
With $mathbb{Z}_p$ it works the same way replacing $t, mathbb{F}_q[[t]]$ by $p, mathbb{Z}_p$. My comment applies to any complete DVR whose uniformizer satisfies $(a+bpi^k)^n = a^n+bpi^k n a^{n-1}+O(pi^{k+1})$
$endgroup$
– reuns
Jan 30 at 10:25
$begingroup$
With $mathbb{Z}_p$ it works the same way replacing $t, mathbb{F}_q[[t]]$ by $p, mathbb{Z}_p$. My comment applies to any complete DVR whose uniformizer satisfies $(a+bpi^k)^n = a^n+bpi^k n a^{n-1}+O(pi^{k+1})$
$endgroup$
– reuns
Jan 30 at 10:25
add a comment |
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$begingroup$
Sure. Let $f(X) in mathbb{F}_q[[t]][X]$, $q$ an odd prime power. Hensel lemma is just saying that if for $a_k in mathbb{F}_q[[t]]$, $f(a_k) equiv 0 bmod t^k mathbb{F}_q[[t]]$ then $f(a_k+bt^{k}) = f(a_k)+b t^{k} f'(a_k) +O(t^{2k})$ so if $f'(a_k) equiv f'(a_1) notequiv 0 bmod t mathbb{F}_q[[t]]$ then you can always choose $a_{k+1} = a_k + b t_{k}, b equiv - frac{f(a_k)}{t^{k} f'(a_1)} bmod t mathbb{F}_q[[t]]$ such that $f(a_{k+1}) equiv 0 bmod t^{k+1}mathbb{F}_q[[t]]$, moreover the obtained limit $lim_{k to infty} a_k$ is the unique lift of $a_1$.
$endgroup$
– reuns
Jan 29 at 22:41
$begingroup$
Sounds good, but what do you mean with mod $...mathbb{F}_q[[t]]$? In the case of the p-adic integers I only look at mod $p$ and not $p mathbb{Q}_p$.
$endgroup$
– Sqyuli
Jan 30 at 8:03
$begingroup$
With $mathbb{Z}_p$ it works the same way replacing $t, mathbb{F}_q[[t]]$ by $p, mathbb{Z}_p$. My comment applies to any complete DVR whose uniformizer satisfies $(a+bpi^k)^n = a^n+bpi^k n a^{n-1}+O(pi^{k+1})$
$endgroup$
– reuns
Jan 30 at 10:25