Talk:Hensel's lemma

Fixed error in "generalizations" section

The generalizations section used to say that a is an "approximate root" of f(x) if f(a) = 0 mod m, and that approximate roots always lift to exact roots. This is not true, consider f(x) = x^2 +1 and a=1, in the 2-adic numbers. I changed the definition of "approximate root" to f(a) = 0 mod f'(a)^2 m; this statement is true, and is a special case of the multivariate result following it. I think this whole section would benefit from more citations, but I don't have sources immediately to hand. (David E Speyer)

Conversion

I was going to convert this stuff to HTML to make it easier to inline, but then I realized that I couldn't find a "not congruent" HTML symbol. It would be better if we could inline things somehow using HTML, but I haven't figured out a good way to do it yet. — Preceding unsigned comment added by Gauge (talk • contribs)

Right. I think we are stuck with PNG for the time being. Oleg Alexandrov 21:16, 19 July 2005 (UTC)

More general forms

Btw, Hensel's Lemma also has much more general forms pertaining to lifting coprime factorizations of polynomials over commutative rings. I might add this at some point if no one else does. - Gauge 18:23, 19 July 2005 (UTC)

Good luck! Oleg Alexandrov 21:16, 19 July 2005 (UTC)

I just implemented this stuff, although in more direct terms than the coprime factorizations mentioned above. My exposition may clash a bit with the relatively concrete number-theoretic introduction.

C'est la vie. - Gauge 20:07, 25 June 2006 (UTC)

Bar

What does the bar over the f'(r) signify, in the third formula? -- J. Shipman

The derivative of f(r) DrBurningBunny (talk) 03:58, 30 March 2018 (UTC)

Unclear

"Let every odd number greater than one thousand be prime." What on all earth is that supposed to mean? --holger —Preceding unsigned comment added by 134.99.156.25 (talk) 16:24, 27 April 2009 (UTC)

Deleted statement

I got rid of a statement about completeness with respect to an ideal being equivalent to the existence of an absolute value because it's not true. An absolute value should satisfy |a| = 0 iff a = 0. But you can have rings that are complete with respect to an ideal and that contain zero divisors. If ab = 0, a, b ≠ 0, you get a problem: 0 = |0| = |ab| = |a||b|, so either |a| = 0 or |b| = 0. No good. As an example, the ring Z_2 x Z_3 is isomorphic to the completion of Z with respect to the ideal 6Z. But Z_2 x Z_3 clearly has zero divisor: (1,0)*(0,1) = 0. 69.203.157.67 (talk) 01:41, 21 July 2010 (UTC)

Local fields version

The version of Hensel's Lemma for local fields doesn't seem to be covered in the generalisation -- what is the most general way to state the version for local fields? —Preceding unsigned comment added by 140.247.44.157 (talk) 17:06, 6 November 2010 (UTC)

Confusing

I find the "First form" section extremely cryptic. The reason is that it weaves several different statements of the lemma with their derivations, remarks and caveats in the derivation, a description of Hensel lifting, and intuitions relating to Newton's method, all into one passage. The reader has to guess which parts pertain to the statement, which parts to the derivation, and which parts are just remakrs, and then piece together what the general statement is. It's too much work and I for one can't tell with confidence what can be concluded when what premises are satisfied. I'm adding my understanding of what statement was supposed to be extracted from the "First form" section, but an expert needs to fix this. I'm sure it's either inaccurate, incomplete, or more likely both. Especially my handling of Taylor series is suspect. I've also separated the derivation from the description of lifting. I hope this helps address the comments by user 140.247.44.157 above. -- jun 194.47.12.68 (talk) 19:13, 31 August 2011 (UTC)

Suggestion

The text is p-adic orientated which may put low-math users like myself off. For example, the strange use of m and k in the first part, surely one would wish to lift a solution in p^m to one in p^(m+k)? But the reason that k is used as a basis becomes clear when we get to p-adics. And the only example is a p-adic one. I recommend a simple example immediately following the first part to clarify, in particular, the comment that division by p^k is integer division with remainder zero whilst the rest is z/p^mz. Such an example is to find the quadratic root of 3 mod 121. Begin with f(x) = x^2 - 3, f'(x) = 2x, present solution f(5) = 0 (mod 11), then take the s-formula term by term: -f(5)/p = -22/11 = -2; (2*5)^-1 = -12 mod 121 using Euclid's algorithm; s = 5 + (-2 * -12 )*11 = 27 mod 121. And verify 27^2 = 3 mod 121. There - we have convinced our audience and now they will go on to see (vaguely) that maybe p-adics look useful. A1jrj (talk) 10:24, 26 July 2014 (UTC)

Correction possibly needed in derivation

I'm not an algebraist, so I'm raising this issue here but not amending the article myself. Someone with expertise, please do so if I'm not mistaken.

The derivation in the article begins with the assertion that, if $f(r) = 0$, then $f(s) = \sum_{n=0}^N c_n (s-r)^n$, where $c_n = f^{(n)}(r)$. However, the correct formula (given by Taylor's theorem) should be that $c_n = f^{(n)}/n!$. The latter formula agrees with the one stated at the end of lecture 6 in https://ocw.mit.edu/courses/mathematics/18-781-theory-of-numbers-spring-2012/lecture-notes/ .

The author of those notes states that, because of the division of the nth derivative by $n!$, it must be proved that $c_n$ is an integer to complete the proof of Hensel's lemma. 71.58.74.4 (talk) 18:44, 7 January 2019 (UTC)

Linear lifting proof

Questions for User:D.Lazard. I may be missing something, or the proof in "Linear lifting" may be incomplete.

Doesn't $af+bg\equiv 1\mod I$ imply that $\delta _{1}$ can always be taken to be $0$ since it is defined by $\delta _{1}\equiv af+bg-1\mod I$? Is there a missing exponent?

Also, the proof claims that existence is verified by

$\delta _{f}=\beta \delta _{h}b,\qquad \delta _{g}=\beta \delta _{h}a.$

Does it necessarily follow that $\deg \delta _{f}<\deg f$ and $\deg \delta _{g}<\deg g$? Eric Rowland (talk) 23:28, 5 July 2021 (UTC)

First point. You are right: $\delta _{1}$ can always be taken to be $0,$ but it is not always the best choice. This was meant with "similarly", which was meant to apply to the sentence between parentheses. This has to be clarified.

Second point. You are right also. I'll fix this soon. D.Lazard (talk) 06:15, 6 July 2021 (UTC)

Upgrading the article

I have now finished to rewrite the beginning of the article in a way that explains the motivations, states Hensel's lemma in a way that is general enough for all applications that I know of, and provides a proof of this general statement that is totally elementary in the sense that it consists only of explicit modular computation.

This new beginning makes redundant a large part of the remaining of the article, which is also confusing (for example the explicit p-adic example is included in section § Statement over a valuated field. I'll now remove redundant and non-encyclopedic content, and upgrade/update the content that deserves to be kept. D.Lazard (talk) 10:05, 9 July 2021 (UTC)

All those algorithmic remarks that you are peppering around the article are irrelevant to Hensel's lemma itself. Also claims like "gives the most efficient known algorithm" are meaningless without precise indication of the measure(s) of efficiency, computational model, etc. details that wouldn't belong to this article. So, that claim has no place here either. Thatwhichislearnt (talk) 22:26, 5 February 2024 (UTC)