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Dalam [[matematika]], '''lema Hensel''' merupakan lema yang dinamai dari [[Kurt Hensel]]. Lema ini merupakan hasil dari [[aritmatika modular]] yang menyatakan bahwa jika [[persamaan polinomial]] memiliki [[Polinomial#Menyelesaikan persamaan polinomial|solusi akar-akar]] modulo [[bilangan prima]] {{math|''p''}}, maka akar-akarnya dapat ''terangkat'' dengan modulo akar tunggal dari setiap pangkat {{Math|''p''}} yang lebih tinggi. Lebih umumnya, jika polinomial memfaktor modulo {{Math|''p''}} menjadi dua polinomial koprima, maka faktorisasi tersebut dapat diangkat menjadi sebuah modulo faktorisasi dari setiap setiap pangkat {{Math|''p''}} yang lebih tinggi. Ini merupakan kasus dari akar yang berkorespondensi dengan kasus polinomial berderajat 1 untuk salah satu faktornya.
{{Dalam perbaikan}}
Dalam [[matematika]], '''Lemma Hensel''', atau dikenal juga sebagai '''Lemma pengangkat Hensel''', dinamai menurut [[Kurt Hensel]], adalah hasil dari [[aritmatika modular]], menyatakan bahwa jika [[persamaan polinomial]] memiliki [[Beberapa akar polinomial|akar sederhana]] modulo a [[bilangan prima]] {{math|''p''}}, maka akar ini sesuai dengan akar unik dari persamaan yang sama modulo pangkat apa pun yang lebih tinggi {{math|''p''}}, yang dapat ditemukan dengan secara berulang "[[angkat (matematika)|angkat]]" solusi modulo pangkat berurutan {{math|''p''}}. Lebih umum digunakan sebagai nama generik untuk analog untuk [[penyelesaian (teori gelanggang)|penyelesaian]] [[gelanggang komutatif]] (termasuk [[bidang p-adic|''p''-adic field]] secara khusus) dari [[Metode Newton]] untuk memecahkan persamaan. Karena [[analisis p-adic|''p''-analisis adic]] dalam beberapa hal lebih sederhana daripada [[analisis nyata]], ada kriteria yang relatif rapi yang menjamin root dari suatu polinomial.
== Statement ==
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<!--Many equivalent statements of Hensel's lemma exist. Arguably the most common statement is the following.
=== Pernyataan umum ===
Assume <math>K</math> is a field complete with respect to a normalised discrete [[valuation (algebra)|valuation]] <math>v</math>. Suppose, furthermore, that <math>\mathcal{O}_K</math> is the ring of integers of <math>K</math> (i.e. all elements of <math>K</math> with non-negative valuation), let <math>\pi\in K</math> be such that <math>v(\pi)=1</math> and let <math>k=\mathcal{O}_K/\pi\mathcal{O}_K</math> denote the [[residue field]]. Let <math>f(X)\in\mathcal{O}_K[X]</math> be a [[polynomial]] with coefficients in <math>\mathcal{O}_K</math>. If the reduction <math>\overline{f}(X)\in k[X]</math> has a simple root (i.e. there exists <math>k_0\in k</math> such that <math>\overline{f}(k_0)=0</math> and <math>\overline{f'}(k_0) \neq 0</math>), then there exists a unique <math>a\in\mathcal{O}_K</math> such that <math>f(a)=0</math> and the reduction <math>\overline{a}=k_0</math> in <math>k</math>.<ref>Serge Lang, ''Algebraic Number Theory'', Addison-Wesley Publishing Company, 1970, p. 43</ref>
=== Alternative statement ===
Another way of stating this (in less generality) is: let <math>f(x)</math> be a [[polynomial]] with [[integer]] (or ''p''-adic integer) coefficients, and let ''m'',''k'' be positive integers such that ''m'' ≤ ''k''. If ''r'' is an integer such that
:<math>f(r) \equiv 0 \bmod p^k \quad \text{and} \quad f'(r) \not\equiv 0 \bmod p</math>
then there exists an integer ''s'' such that
:<math>f(s) \equiv 0 \bmod p^{k+m} \quad \text{and} \quad r \equiv s \bmod p^k.</math>
Furthermore, this ''s'' is unique modulo ''p''<sup>''k''+''m''</sup>, and can be computed explicitly as the integer such that
:<math>s = r - f(r)\cdot a,</math>
where <math>a</math> is an integer satisfying
:<math>a \equiv [f'(r)]^{-1} \bmod p^m.</math>
Note that <math>f(r) \equiv 0 \bmod p^k </math> so that the condition <math>s \equiv r \bmod p^k </math> is met. As an aside, if <math>f'(r) \equiv 0 \bmod p</math>, then 0, 1, or several ''s'' may exist (see Hensel Lifting below).
=== Derivation ===
We use the Taylor expansion of ''f'' around ''r'' to write:
:<math>f(s) = \sum_{n=0}^N c_n (s-r)^n, \qquad c_n = f^{(n)}(r)/n!.</math>
From <math>r \equiv s \bmod p^k,</math> we see that ''s'' − ''r'' = ''tp<sup>k</sup>'' for some integer ''t''. Let
:<math>\begin{align}
f(s) &= \sum_{n=0}^N c_n \left(tp^k\right)^n \\
&= f(r) + t p^k f'(r) + \sum_{n=2}^N c_n t^n p^{kn} \\
&= f(r) + t p^k f'(r) + p^{2k}t^2g(t) && g(t) \in \Z[t] \\
&= zp^k + t p^k f'(r) + p^{2k}t^2g(t) && f(r) \equiv 0 \bmod p^k \\
&= (z+tf'(r)) p^k + p^{2k}t^2g(t)
\end{align}</math>
For <math>m \leqslant k,</math> we have:
:<math>\begin{align}
f(s) \equiv 0 \bmod p^{k+m} &\Longleftrightarrow (z + tf'(r))p^k \equiv 0 \bmod p^{k+m} \\
&\Longleftrightarrow z + tf'(r) \equiv 0 \bmod p^m \\
&\Longleftrightarrow tf'(r) \equiv -z \bmod p^m \\
&\Longleftrightarrow t \equiv -z [f'(r)]^{-1} \bmod p^m && p \nmid f'(r)
\end{align}</math>
The assumption that <math>f'(r)</math> is not divisible by ''p'' ensures that <math>f'(r)</math> has an inverse mod <math>p^m</math> which is necessarily unique. Hence a solution for ''t'' exists uniquely modulo <math>p^m,</math> and ''s'' exists uniquely modulo <math>p^{k+m}.</math>
=== Simple statement ===
For <math>f \in \mathbb{Z}_p[x]</math>, if there is a solution <math>a_0</math> of <math>f(a_0) \equiv 0 \bmod p</math> and <math>f'(x) \equiv 0 \bmod p</math> has no solutions, then there exists a unique lift <math>\alpha \in \mathbb{Z}_p</math> such that <math>f(\alpha) = 0</math>. Note that given a solution <math>\alpha \in \mathbb{Z}_p</math> where <math>f(\alpha) = 0</math>, its projection to <math>\mathbb{Z}/p</math> gives a solution to <math>f(\alpha) \equiv 0 \bmod p </math>, so Hensel's lemma gives a way to take solutions <math>\bmod p</math> and give a solution in <math>\mathbb{Z}_p</math>.-->
== Pengamatan ==
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<!--=== Frobenius ===
Note that given an <math>a \in \mathbb{F}_p</math> the [[Frobenius endomorphism]] <math>(-) \mapsto (-)^p</math> gives a polynomial <math>x^p - a</math> which always has zero derivative<blockquote><math>\begin{align}
\frac{d}{dx}x^p - a &= p\cdot x^{p-1} \\
&\equiv 0\cdot x^{p-1} \bmod p \\
& \equiv 0 \bmod p
\end{align}</math></blockquote>hence the ''p''-th roots of <math>a</math> do not exist in <math>\mathbb{Z}_p</math>. For <math>a = 1</math>, this implies <math>\mathbb{Z}_p</math> cannot contain the [[root of unity]] <math>\mu_p</math>.
=== Roots of unity ===
Although the <math>p</math>-th roots of unity are not contained in <math>\mathbb{F}_p</math>, there are solutions of <math>x^p - x = x(x^{p-1} - 1)</math>. Note<blockquote><math>\begin{align}
\frac{d}{dx} x^p - x &= px - 1 \\
&\equiv -1 \bmod p
\end{align}</math></blockquote>is never zero, so if there exists a solution, it necessarily lifts to <math>\mathbb{Z}_p</math>. Because the Frobenius gives <math>a^p = a</math>, all of the non-zero elements <math>\mathbb{F}_p^\times</math> are solutions. In fact, these are the only roots of unity contained in <math>\mathbb{Q}_p</math><ref>{{Cite web|title=Hensel's Lemma|url=https://kconrad.math.uconn.edu/blurbs/gradnumthy/hensel.pdf|last=Conrad|first=Keith|date=|website=|page=4|url-status=live|archive-url=|archive-date=|access-date=}}</ref>.-->
== Angkat Hensel ==
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<!--Using the lemma, one can "lift" a root ''r'' of the polynomial ''f'' modulo ''p<sup>k</sup>'' to a new root ''s'' modulo ''p''<sup>''k''+1</sup> such that ''r'' ≡ ''s'' mod ''p<sup>k</sup>'' (by taking ''m''=1; taking larger ''m'' follows by induction). In fact, a root modulo ''p''<sup>''k''+1</sup> is also a root modulo ''p<sup>k</sup>'', so the roots modulo ''p''<sup>''k''+1</sup> are precisely the liftings of roots modulo ''p<sup>k</sup>''. The new root ''s'' is congruent to ''r'' modulo ''p'', so the new root also satisfies <math>f'(s) \equiv f'(r) \not\equiv 0 \bmod p.</math> So the lifting can be repeated, and starting from a solution ''r<sub>k</sub>'' of <math>f(x) \equiv 0 \bmod p^k</math> we can derive a sequence of solutions ''r''<sub>''k''+1</sub>, ''r''<sub>''k''+2</sub>, ... of the same congruence for successively higher powers of ''p'', provided <math>f'(r_k) \not\equiv 0 \bmod p</math> for the initial root ''r<sub>k</sub>''. This also shows that ''f'' has the same number of roots mod ''p<sup>k</sup>'' as mod ''p''<sup>''k''+1</sup>, mod ''p'' <sup>''k''+2</sup>, or any other higher power of ''p'', provided the roots of ''f'' mod ''p<sup>k</sup>'' are all simple.
What happens to this process if ''r'' is not a simple root mod ''p''? Suppose
:<math>f(r) \equiv 0 \bmod p^k \quad \text{and} \quad f'(r) \equiv 0 \bmod p.</math>
Then <math>s \equiv r \bmod p^k </math> implies <math>f(s) \equiv f(r) \bmod p^{k+1}.</math> That is, <math>f(r + tp^k) \equiv f(r)\bmod p^{k+1} </math> for all integers ''t''. Therefore, we have two cases:
*If <math> f(r) \not\equiv 0 \bmod p^{k+1} </math> then there is no lifting of ''r'' to a root of ''f''(''x'') modulo ''p''<sup>''k''+1</sup>.
*If <math>f(r) \equiv 0 \bmod p^{k+1} </math> then every lifting of ''r'' to modulus ''p''<sup>''k''+1</sup> is a root of ''f''(''x'') modulo ''p''<sup>''k''+1</sup>.
'''Example.''' To see both cases we examine two different polynomials with ''p'' = 2:
<math>f(x) = x^2 +1</math> and ''r'' = 1. Then <math>f(1)\equiv 0 \bmod 2</math> and <math>f'(1) \equiv 0 \bmod 2.</math> We have <math>f(1) \not\equiv 0 \bmod 4</math> which means that no lifting of 1 to modulus 4 is a root of ''f''(''x'') modulo 4.
<math>g(x) = x^2 -17</math> and ''r'' = 1. Then <math>g(1)\equiv 0 \bmod 2</math> and <math>g'(1) \equiv 0 \bmod 2.</math> However, since <math>g(1) \equiv 0 \bmod 4,</math> we can lift our solution to modulus 4 and both lifts (i.e. 1, 3) are solutions. The derivative is still 0 modulo 2, so ''a priori'' we don't know whether we can lift them to modulo 8, but in fact we can, since ''g''(1) is 0 mod 8 and ''g''(3) is 0 mod 8, giving solutions at 1, 3, 5, and 7 mod 8. Since of these only ''g''(1) and ''g''(7) are 0 mod 16 we can lift only 1 and 7 to modulo 16, giving 1, 7, 9, and 15 mod 16. Of these, only 7 and 9 give ''g''(''x'') = 0 mod 32, so these can be raised giving 7, 9, 23, and 25 mod 32. It turns out that for every integer ''k'' ≥ 3, there are four liftings of 1 mod 2 to a root of ''g''(''x'') mod 2<sup>''k''</sup>.-->
== Lemma Hensel untuk ''p'' - bilangan adic ==
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<!--In the ''p''-adic numbers, where we can make sense of rational numbers modulo powers of ''p'' as long as the denominator is not a multiple of ''p'', the recursion from ''r<sub>k</sub>'' (roots mod ''p<sup>k</sup>'') to ''r''<sub>''k''+1</sub> (roots mod ''p''<sup>''k''+1</sup>) can be expressed in a much more intuitive way. Instead of choosing ''t'' to be an(y) integer which solves the congruence
:<math>tf'(r_k) \equiv -(f(r_k)/p^{k})\bmod p^m,</math>
let ''t'' be the rational number (the ''p<sup>k</sup>'' here is not really a denominator since ''f''(''r<sub>k</sub>'') is divisible by ''p<sup>k</sup>''):
:<math>-(f(r_k)/p^{k})/f'(r_k).</math>
Then set
:<math>r_{k+1} = r_k + tp^k = r_k - \frac{f(r_k)}{f'(r_k)}.</math>
This fraction may not be an integer, but it is a ''p''-adic integer, and the sequence of numbers ''r<sub>k</sub>'' converges in the ''p''-adic integers to a root of ''f''(''x'') = 0. Moreover, the displayed recursive formula for the (new) number ''r''<sub>''k''+1</sub> in terms of ''r<sub>k</sub>'' is precisely [[Newton's method]] for finding roots to equations in the real numbers.
By working directly in the ''p''-adics and using the [[P-adic order#p-adic absolute value|''p''-adic absolute value]], there is a version of Hensel's lemma which can be applied even if we start with a solution of ''f''(''a'') ≡ 0 mod ''p'' such that <math>f'(a)\equiv 0 \bmod p.</math> We just need to make sure the number <math>f'(a)</math> is not exactly 0. This more general version is as follows: if there is an integer ''a'' which satisfies:
:<math>|f(a)|_p < |f'(a)|_p^2,</math>
then there is a unique ''p''-adic integer ''b'' such ''f''(''b'') = 0 and <math>|b-a|_p <|f'(a)|_p.</math> The construction of ''b'' amounts to showing that the recursion from Newton's method with initial value ''a'' converges in the ''p''-adics and we let ''b'' be the limit. The uniqueness of ''b'' as a root fitting the condition <math>|b-a|_p <|f'(a)|_p</math> needs additional work.
The statement of Hensel's lemma given above (taking <math>m=1</math>) is a special case of this more general version, since the conditions that ''f''(''a'') ≡ 0 mod ''p'' and <math>f'(a)\not\equiv 0 \bmod p</math> say that <math>|f(a)|_p < 1</math> and <math>|f'(a)|_p = 1.</math>-->
== Contoh ==
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<!--Suppose that ''p'' is an odd prime and ''a'' is a non-zero [[quadratic residue]] modulo ''p''. Then Hensel's lemma implies that ''a'' has a square root in the ring of ''p''-adic integers <math>\Z_p.</math> Indeed, let <math>f(x)=x^2-a.</math> If ''r'' is a square root of ''a'' modulo ''p'' then:
: <math>f(r) = r^2 - a \equiv 0 \bmod p \quad \text{and} \quad f'(r) = 2r \not\equiv 0 \bmod p,</math>
where the second condition is dependent on the fact that ''p'' is odd. The basic version of Hensel's lemma tells us that starting from ''r''<sub>1</sub> = ''r'' we can recursively construct a sequence of integers <math>\{r_k\}</math> such that:
: <math>r_{k+1} \equiv r_k \bmod p^k, \quad r_k^2 \equiv a \bmod p^k. </math>
This sequence converges to some ''p''-adic integer ''b'' which satisfies ''b''<sup>2</sup> = ''a''. In fact, ''b'' is the unique square root of ''a'' in <math>\Z_p</math> congruent to ''r''<sub>1</sub> modulo ''p''. Conversely, if ''a'' is a perfect square in <math>\Z_p</math> and it is not divisible by ''p'' then it is a nonzero quadratic residue mod ''p''. Note that the [[quadratic reciprocity law]] allows one to easily test whether ''a'' is a nonzero quadratic residue mod ''p'', thus we get a practical way to determine which ''p''-adic numbers (for ''p'' odd) have a ''p''-adic square root, and it can be extended to cover the case ''p'' = 2 using the more general version of Hensel's lemma (an example with 2-adic square roots of 17 is given later).
To make the discussion above more explicit, let us find a "square root of 2" (the solution to <math>x^2-2=0</math>) in the 7-adic integers. Modulo 7 one solution is 3 (we could also take 4), so we set <math>r_1 = 3</math>. Hensel's lemma then allows us to find <math>r_2</math> as follows:
:<math>\begin{align}
f(r_1) &= 3^2-2=7 \\
f(r_1)/p^1 &=7/7=1 \\
f'(r_1) &=2r_1=6
\end{align}</math>
Based on which the expression
:<math>tf'(r_1) \equiv -(f(r_1)/p^k)\bmod p,</math>
turns into:
:<math>t\cdot 6 \equiv -1\bmod 7</math>
which implies <math>t = 1.</math> Now:
:<math>r_2 = r_1 + tp^1 = 3+1 \cdot 7 = 10 = 13_7.</math>
And sure enough, <math>10^2\equiv 2\bmod 7^2.</math> (If we had used the Newton method recursion directly in the 7-adics, then <math>r_2 = r_1 - f(r_1)/f'(r_1) = 3 - 7/6 = 11/6,</math> and <math>11/6 \equiv 10 \bmod 7^2.</math>)
We can continue and find <math>r_3 = 108 = 3 + 7 + 2\cdot 7^2 = 213_7</math>. Each time we carry out the calculation (that is, for each successive value of ''k''), one more base 7 digit is added for the next higher power of 7. In the 7-adic integers this sequence converges, and the limit is a square root of 2 in <math>\Z_7</math> which has initial 7-adic expansion
:<math>3 + 7 + 2\cdot7^2 + 6\cdot 7^3 + 7^4 + 2\cdot 7^5 + 7^6 + 2\cdot 7^7 + 4\cdot 7^8 + \cdots.</math>
If we started with the initial choice <math>r_1 = 4</math> then Hensel's lemma would produce a square root of 2 in <math>\Z_7</math> which is congruent to 4 (mod 7) instead of 3 (mod 7) and in fact this second square root would be the negative of the first square root (which is consistent with 4 = −3 mod 7).
As an example where the original version of Hensel's lemma is not valid but the more general one is, let <math>f(x) = x^2-17</math> and <math>a=1.</math> Then <math>f(a) =-16</math> and <math>f'(a) = 2,</math> so
:<math>|f(a)|_2 < |f'(a)|_2^2,</math>
which implies there is a unique 2-adic integer ''b'' satisfying
:<math>b^2 = 17 \quad \text{and} \quad |b-a|_2 < |f'(a)|_2 = \frac{1}{2},</math>
i.e., ''b'' ≡ 1 mod 4. There are two square roots of 17 in the 2-adic integers, differing by a sign, and although they are congruent mod 2 they are not congruent mod 4. This is consistent with the general version of Hensel's lemma only giving us a unique 2-adic square root of 17 that is congruent to 1 mod 4 rather than mod 2. If we had started with the initial approximate root ''a'' = 3 then we could apply the more general Hensel's lemma again to find a unique 2-adic square root of 17 which is congruent to 3 mod 4. This is the other 2-adic square root of 17.
In terms of lifting the roots of <math>x^2-17</math> from modulus 2<sup>''k''</sup> to 2<sup>''k''+1</sup>, the lifts starting with the root 1 mod 2 are as follows:-->
<!--:1 mod 2 > 1, 3 mod 4
:1 mod 4 > 1, 5 mod 8 and 3 mod 4 -> 3, 7 mod 8
:1 mod 8 > 1, 9 mod 16 and 7 mod 8 -> 7, 15 mod 16, while 3 mod 8 and 5 mod 8 don't lift to roots mod 16
:9 mod 16 -> 9, 25 mod 32 and 7 mod 16 -> 7, 23 mod 16, while 1 mod 16 and 15 mod 16 don't lift to roots mod 32.
For every ''k'' at least 3, there are ''four'' roots of ''x''<sup>2</sup> − 17 mod 2<sup>''k''</sup>, but if we look at their 2-adic expansions we can see that in pairs they are converging to just ''two'' 2-adic limits. For instance, the four roots mod 32 break up into two pairs of roots which each look the same mod 16:
:9 = 1 + 2<sup>3</sup> and 25 = 1 + 2<sup>3</sup> + 2<sup>4</sup>.
:7 = 1 + 2 + 2<sup>2</sup> and 23 = 1 + 2 + 2<sup>2</sup> + 2<sup>4</sup>.
The 2-adic square roots of 17 have expansions
:<math>1 + 2^3 +2^5 +2^6 +2^7 +2^9 + 2^{10} + \cdots </math>
:<math>1 + 2 + 2^2 + 2^4 + 2^8 + 2^{11} + \cdots </math>
Another example where we can use the more general version of Hensel's lemma but not the basic version is a proof that any 3-adic integer ''c'' ≡ 1 mod 9 is a cube in <math>\Z_3.</math> Let <math>f(x) =x^3-c</math> and take initial approximation ''a'' = 1. The basic Hensel's lemma cannot be used to find roots of ''f''(''x'') since <math>f'(r)\equiv 0 \bmod 3</math> for every ''r''. To apply the general version of Hensel's lemma we want <math>|f(1)|_3 <|f'(1)|_3^2,</math> which means <math>c\equiv 1 \bmod 27.</math> That is, if ''c'' ≡ 1 mod 27 then the general Hensel's lemma tells us ''f''(''x'') has a 3-adic root, so ''c'' is a 3-adic cube. However, we wanted to have this result under the weaker condition that ''c'' ≡ 1 mod 9. If ''c'' ≡ 1 mod 9 then ''c'' ≡ 1, 10, or 19 mod 27. We can apply the general Hensel's lemma three times depending on the value of ''c'' mod 27: if ''c'' ≡ 1 mod 27 then use ''a'' = 1, if ''c'' ≡ 10 mod 27 then use ''a'' = 4 (since 4 is a root of ''f''(''x'') mod 27), and if ''c'' ≡ 19 mod 27 then use ''a'' = 7. (It is not true that every ''c'' ≡ 1 mod 3 is a 3-adic cube, e.g., 4 is not a 3-adic cube since it is not a cube mod 9.)
In a similar way, after some preliminary work, Hensel's lemma can be used to show that for any ''odd'' prime number ''p'', any ''p''-adic integer ''c'' congruent to 1 modulo ''p''<sup>2</sup> is a ''p''-th power in <math>\Z_p.</math> (This is false for ''p'' = 2.)
== Generalisasi ==
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<!--Suppose ''A'' is a [[commutative ring]], [[completion of a ring|complete]] with respect to an [[ideal (ring theory)|ideal]] <math>\mathfrak{m},</math> and let <math>f(x) \in A[x].</math> ''a'' ∈ ''A'' is called an "approximate root" of ''f'', if
:<math> f(a) \equiv 0 \bmod f'(a)^2 \mathfrak{m}.</math>
If ''f'' has an approximate root then it has an exact root ''b'' ∈ ''A'' "close to" ''a''; that is,
:<math>f(b) = 0 \quad \text{and} \quad b \equiv a \bmod{\mathfrak m}.</math>
Furthermore, if <math>f'(a)</math> is not a zero-divisor then ''b'' is unique.
This result can be generalized to several variables as follows:
:'''Theorem.''' Suppose ''A'' be a commutative ring that is complete with respect to ideal <math>\mathfrak{m} \subset A.</math> Let <math>f_1, \ldots, f_n \in A[x_1, \ldots, x_n]</math> be a system of ''n'' polynomials in ''n'' variables over ''A''. View <math>\mathbf{f} = (f_1, \ldots, f_n),</math> as a mapping from ''A<sup>n</sup>'' to itself, and let <math>J_{\mathbf{f}}(\mathbf{x})</math> denote its [[Jacobian matrix]]. Suppose '''a''' = (''a''<sub>1</sub>, ..., ''a''<sub>''n''</sub>) ∈ ''A<sup>n</sup>'' is an approximate solution to '''f''' = '''0''' in the sense that
::<math>f_i(\mathbf{a}) \equiv 0 \bmod (\det J_{\mathbf{f}}(a))^2 \mathfrak{m}, \qquad 1 \leqslant i \leqslant n.</math>
:Then there is some '''b''' = (''b''<sub>1</sub>, ..., ''b''<sub>''n''</sub>) ∈ ''A<sup>n</sup>'' satisfying '''f'''('''b''') = '''0''', i.e.,
::<math>f_i(\mathbf{b}) =0, \qquad 1 \leqslant i \leqslant n.</math>
:Furthermore this solution is "close" to '''a''' in the sense that
::<math>b_i \equiv a_i \bmod \det J_{\mathbf{f}}(a) \mathfrak{m}, \qquad 1 \leqslant i \leqslant n.</math>
As a special case, if <math>f_i(\mathbf{a}) \equiv 0 \bmod \mathfrak{m}</math> for all ''i'' and <math>\det J_{\mathbf{f}}(\mathbf{a})</math> is a unit in ''A'' then there is a solution to '''f'''('''b''') = '''0''' with <math>b_i \equiv a_i \bmod \mathfrak{m}</math> for all ''i''.
When ''n'' = 1, '''a''' = ''a'' is an element of ''A'' and <math>J_{\mathbf{f}}(\mathbf{a}) = J_f(a)=f'(a).</math> The hypotheses of this multivariable Hensel's lemma reduce to the ones which were stated in the one-variable Hensel's lemma.-->
== Lihat pula ==
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