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| In [[mathematics]], '''Hensel's lemma''', also known as '''Hensel's lifting lemma''', named after [[Kurt Hensel]], is a result in [[modular arithmetic]], stating that if a [[polynomial equation]] has a [[Multiplicity (mathematics)#Multiplicity of a root of a polynomial|simple root]] modulo a [[prime number]] {{math|''p''}}, then this root corresponds to a unique root of the same equation modulo any higher power of {{math|''p''}}, which can be found by iteratively "[[lift (mathematics)|lift]]ing" the solution modulo successive powers of {{math|''p''}}. More generally it is used as a generic name for analogues for [[completion (ring theory)|complete]] [[commutative ring]]s (including [[p-adic field|''p''-adic field]]s in particular) of the [[Newton method]] for solving equations. Since [[p-adic analysis|''p''-adic analysis]] is in some ways simpler than [[real analysis]], there are relatively neat criteria guaranteeing a root of a polynomial.
| | == Leuca three are lying on the sofa == |
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| == Statement ==
| | Sent to friends. '<br><br>'You go.'<br><br>Rong Jun, Leuca three are lying on the sofa, the TV drama thunderstorms aftertaste of wine taste, simply do not want to get up.<br><br>practice is that desperate, and occasionally indulge the luxury of time to go and enjoy the nature.<br><br>......<br><br>'these two guys.' Luo Feng smiled and shook his head, then turned into a direct streamer fly directly out into the sky.<br><br>......<br><br>And now, another one in the beginning was the [http://www.nrcil.net/fancybox/lib/rakuten_LV_113.html 財布 ルイヴィトン メンズ] area within the manor.<br><br>'Today this course is [http://www.nrcil.net/fancybox/lib/rakuten_LV_64.html ルイヴィトン 新作 2014] to [http://www.nrcil.net/fancybox/lib/rakuten_LV_25.html ルイヴィトン ボストンバッグ] describe the space, golden rules of origin combination of these two methods, that Feng Luo will definitely go. [http://www.nrcil.net/fancybox/lib/rakuten_LV_42.html ルイヴィトン 帽子] Mo Yilin, Ash ...... you guys have to go to class, take the opportunity to find a [http://www.nrcil.net/fancybox/lib/rakuten_LV_31.html ルイヴィトン デザイナー] pretext, to find ways and Romania peak of a war, and feel this is the end of Feng [http://www.nrcil.net/fancybox/lib/rakuten_LV_99.html ルイヴィトン 値段] Luo. know some [http://www.nrcil.net/fancybox/lib/rakuten_LV_115.html ルイヴィトン 旅行カバン] of his trick, be prepared to fight ...... when eligibility is more relaxed! 'said the man wearing a silver robe.<br><br>tenth chapter of the original [http://www.nrcil.net/fancybox/lib/rakuten_LV_76.html ベルト ルイヴィトン] provocation Chapter IX<br><br>the beginning area, II floor lecture hall. |
| 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
| | 相关的主题文章： |
| | | <ul> |
| :<math>f(r) \equiv 0 \pmod{p^k}</math> and <math>f'(r) \not\equiv 0 \pmod{p}</math>
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| | | <li>[http://www.bjbbc.org/plus/feedback.php?aid=424 http://www.bjbbc.org/plus/feedback.php?aid=424]</li> |
| then there exists an integer ''s'' such that
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| | | <li>[http://www.zaiak.com/plus/feedback.php?aid=2 http://www.zaiak.com/plus/feedback.php?aid=2]</li> |
| :<math>f(s) \equiv 0 \pmod{p^{k+m}}</math> and <math>r \equiv s \pmod{p^{k}}.</math>
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| | | <li>[http://cgi.www5b.biglobe.ne.jp/~kaneda/cgi-bin/ita2/ita_honey_b_b_s5.cgi http://cgi.www5b.biglobe.ne.jp/~kaneda/cgi-bin/ita2/ita_honey_b_b_s5.cgi]</li> |
| Furthermore, this ''s'' is unique modulo ''p''<sup>''k''+m</sup>, and can be computed explicitly as
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| | | </ul> |
| :<math>s = r + tp^k</math> where <math>t = - \frac{f(r)}{p^k} \cdot (f'(r)^{-1}).</math>
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| In this formula for ''t'', the division by ''p''<sup>''k''</sup> denotes ordinary integer division (where the remainder will be 0), while negation, multiplication, and multiplicative inversion <math>f'(r)^{-1}</math> are performed in <math>\mathbb{Z}/p^m\mathbb{Z}</math>.
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| As an aside, if <math>f'(r) \equiv 0 \pmod{p}</math>, then 0, 1, or several ''s'' may exist (see Hensel Lifting below).
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| === Derivation ===
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| The lemma derives from considering the Taylor expansion of ''f'' around ''r''. From <math>r \equiv s \pmod{p^k}</math>, we see that ''s'' has to be of the form ''s = r + tp<sup>k</sup>'' for some integer ''t''. Expanding <math>f(r + tp^k)</math> gives
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| :<math>f(r + tp^k) = f(r) + tp^k\cdot f'(r) + O(p^{2k}).</math>
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| Reducing both sides modulo p<sup>k+m</sup>, we see that for <math>f(s) \equiv 0 \pmod{p^{k+m}}</math> to hold, we need
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| :<math>0 \equiv f(r + tp^k) \equiv f(r) + tp^k \cdot f'(r)\pmod{p^{k+m}}</math>
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| where the ''O''(''p''<sup>2''k''</sup>) terms vanish because ''k''+''m'' ≤ 2''k''. Then we note that <math>f(r) = zp^k</math> for some integer ''z'' since ''r'' is a root of ''f'' mod ''p''<sup>''k''</sup>, so
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| :<math>0 \equiv (z + tf'(r))p^k \pmod{p^{k+m}}</math>,
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| which is to say
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| :<math>0 \equiv z + tf'(r) \pmod{p^m}.</math>
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| Then substituting back ''f''(''r'')/''p''<sup>''k''</sup> for ''z'' and solving for ''t'' in <math>\mathbb{Z}/p^m\mathbb{Z}</math> gives the explicit formula for ''t'' mentioned above. 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>.
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| == Hensel Lifting ==
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| Using the lemma, one can "lift" a root ''r'' of the polynomial ''f'' mod ''p''<sup>''k''</sup> to a new root ''s'' mod ''p''<sup>''k''+1</sup> such that ''r'' ≡ ''s'' mod ''p''<sup>''k''</sup> (by taking ''m''=1; taking larger ''m'' also works). In fact, a root mod ''p''<sup>''k''+1</sup> is also a root mod ''p''<sup>''k''</sup>, so the roots mod ''p''<sup>''k''+1</sup> are precisely the liftings of roots mod ''p''<sup>''k''</sup>. The new root ''s'' is congruent to ''r'' mod ''p'', so the new root also satisfies <math>f'(s) \equiv f'(r) \not\equiv 0 \pmod{p}</math>. So the lifting can be repeated, and starting from a solution ''r''<sub>''k''</sub> of <math>f(x) \equiv 0 \pmod{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 \pmod{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.
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| What happens to this process if ''r'' is not a simple root mod ''p''? If we have a root mod ''p''<sup>''k''</sup> at which the derivative mod ''p'' is 0, then there is ''not'' a unique lifting of a root mod ''p''<sup>''k''</sup> to a root mod ''p''<sup>''k''+1</sup>: either there is no lifting to a root mod ''p''<sup>''k''+1</sup> or there are multiple choices:
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| ::if <math> f(r) \equiv 0 \,\bmod{p^k}</math> and <math> f'(r) \equiv 0 \,\bmod{p},</math> then <math> s \equiv r \,\bmod p^k \Rightarrow f(s) \equiv f(r) \,\bmod p^{k+1}</math>.
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| That is, <math>f(r + tp^{k}) \equiv 0\,\bmod{p^{k+1}}\, </math> for all integers ''t''.
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| Therefore 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'') mod ''p''<sup>''k''+1</sup>, while 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'') mod ''p''<sup>''k''+1</sup>.
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| To see the difficulty that can arise in a concrete example, take ''p'' = 2, ''f''(''x'') = ''x''<sup>2</sup> + 1, and ''r'' = 1. Then ''f''(1) ≡ 0 mod 2 and f'(1) ≡ 0 mod 2. We have ''f''(1) = 2 ≠ 0 mod 4 and no lifting of 1 to modulus 4 is a root of ''f''(''x'') mod 4.
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| On the other hand, if we take ''f''(''x'') = ''x''<sup>2</sup> - 17 and then 1 is a root of ''f''(''x'') mod 2 and for every positive integer ''k'' there is more than one lifting of 1 mod 2 to a root of ''f''(''x'') mod 2<sup>''k''</sup>.
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| == Hensel's Lemma for ''p''-adic Numbers ==
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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
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| <math>tf'(r_k) \equiv -(f(r_k)/p^{k})\,\bmod{p^m}\,</math>, let ''t'' be the rational number <math>\ -(f(r_k)/p^{k})/f'(r_k) </math> (the ''p''<sup>''k''</sup> here is not really a denominator since ''f''(''r''<sub>''k''</sub>) is divisible by ''p''<sup>''k''</sup>). Then set
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| ::<math>r_{k+1} = r_k + tp^k = r_k - \frac{f(r_k)}{f'(r_k)}.</math>
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| 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.
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| By working directly in the ''p''-adics and using the ''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 f'(''a'') ≡ 0 mod ''p''. We just need to make sure the number f'(''a'') is not exactly 0. This more general version is as follows:
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| if there is an integer ''a'' which satisfies |''f''(''a'')|<sub>''p''</sub> < |f′(''a'')|<sub>''p''</sub><sup>2</sup>, then there is a unique ''p''-adic integer ''b'' such ''f''(''b'') = 0 and |''b''-''a''|<sub>''p''</sub> < |f'(''a'')|<sub>''p''</sub>. 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 |''b''-''a''|<sub>''p''</sub> < |f'(''a'')|<sub>''p''</sub> needs additional work.
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| 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 f'(''a'') ≠ 0 mod ''p'' say that |''f''(''a'')|<sub>''p''</sub> < 1 and |f'(''a'')|<sub>''p''</sub> = 1.
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| == Examples ==
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| Suppose that ''p'' is an odd prime number and ''a'' is a [[quadratic residue]] modulo ''p'' that is nonzero mod ''p''. Then Hensel's lemma implies that ''a'' has a square root in the ring of ''p''-adic integers '''Z'''<sub>''p''</sub>. Indeed, let ''f''(''x'')=''x''<sup>2</sup>-''a''. Its derivative is 2''x'', so if ''r'' is a square root of ''a'' mod ''p'' we have
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| : <math>f(r) = r^2 - a \equiv 0 \,\bmod{p}</math> and <math>f'(r) = 2r \not\equiv 0 \,\bmod{p}</math>,
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| where the second condition depends on ''p'' not being 2. 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 { ''r''<sub>k</sub> } such that
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| : <math>r_{k+1} \equiv r_k \,\bmod{p^k}, \quad r_k^2 \equiv a \,\bmod{p^k}. </math> | |
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| This sequence converges to some ''p''-adic integer ''b'' and ''b''<sup>2</sup>=''a''. In fact, ''b'' is the unique square root of ''a'' in '''Z'''<sub>p</sub> congruent to ''r''<sub>1</sub> modulo ''p''. Conversely, if ''a'' is a perfect square in '''Z'''<sub>p</sub> 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).
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| 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:
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| :<math>f(r_1)=3^2-2=7</math>
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| :<math>f(r_1)/p^1=7/7=1</math>
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| :<math>f'(r_1)=2r_1=6</math>
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| :<math>tf'(r_1) \equiv -(f(r_1)/p^{k-1})\,\bmod{p},</math> that is, <math>t\cdot 6 \equiv -1\,\bmod{7}</math> | |
| :<math>\Rightarrow t = 1</math>
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| :<math>r_2 = r_1 + tp^1 = 3+1 \cdot 7 = 10 =13_7.</math>
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| 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 ''r''<sub>2</sub> = ''r''<sub>1</sub> - f(''r''<sub>1</sub>)/f'(''r''<sub>1</sub>) = 3 - 7/6 = 11/6, and 11/6 ≡ 10 mod 7<sup>2</sup>.)
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| 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 '''Z'''<sub>7</sub> which has initial 7-adic expansion
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| ::<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>
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| If we started with the initial choice <math>r_1 = 4</math> then Hensel's lemma would produce a square root of 2 in '''Z'''<sub>7</sub> 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).
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| As an example where the original version of Hensel's lemma is not valid but the more general one is, let ''f''(''x'') = ''x''<sup>2</sup> - 17 and ''a'' = 1. Then ''f''(''a'') = -16 and f'(''a'') = 2, so |''f''(''a'')|<sub>2</sub> < |f′(''a'')|<sub>''2''</sub><sup>2</sup>, which implies there is a unique 2-adic integer ''b'' satisfying ''b''<sup>2</sup> = 17 and |''b''- ''a''|<sub>2</sub> < |f'(''a'')|<sub>2</sub> = 1/2, 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.
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| In terms of lifting roots of ''x''<sup>2</sup> - 17 from one modulus 2<sup>''k''</sup> to the next 2<sup>''k''+1</sup>, the lifts starting with the root 1 mod 2 are as follows:
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| :: 1 mod 2 --> 1, 3 mod 4
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| :: 1 mod 4 --> 1, 5 mod 8 and 3 mod 4 ---> 3, 7 mod 8
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| :: 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
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| :: 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.
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| 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:
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| :: 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>.
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| The 2-adic square roots of 17 have expansions
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| ::1 + 2<sup>3</sup> + 2<sup>5</sup> + 2<sup>6</sup> + 2<sup>7</sup> + 2<sup>9</sup> + 2<sup>10</sup> + ..., 1 + 2 + 2<sup>2</sup> + 2<sup>4</sup> + 2<sup>8</sup> + 2<sup>11</sup>...
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| 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 '''Z'''<sub>3</sub>. Let ''f''(''x'') = ''x''<sup>3</sup> - c and take initial approximation ''a'' = 1. The basic Hensel's lemma can't be used to find roots of ''f''(''x'') since f'(''r'') ≡ 0 mod 3 for every ''r''. To apply the general version of Hensel's lemma we want |f(1)|<sub>3</sub> < |f'(1)|<sub>3</sub><sup>2</sup>, which means ''c'' ≡ 1 mod 27. 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.)
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| 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'' which is 1 mod ''p''<sup>2</sup> is a ''p''-th power in '''Z'''<sub>''p''</sub>.
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| (This is false when ''p'' is 2.)
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| ==Generalizations== | |
| Suppose ''A'' is a [[commutative ring]], complete with respect to an [[ideal (ring theory)|ideal]] <math>\mathfrak m_A</math>, and let <math>f(x) \in A[x]</math> be a [[polynomial]] with coefficients in ''A''. Then if ''a'' ∈ ''A'' is an "approximate root" of ''f'' in the sense that it satisfies
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| :<math> f(a) \equiv 0 \,\bmod{f'(a)^2\mathfrak m}</math>
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| then there is an exact root ''b'' ∈ ''A'' of ''f'' "close to" ''a''; that is,
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| :<math>f(b) = 0</math> | |
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| and
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| :<math>b \equiv a \,\bmod{f'(a)\mathfrak m}.</math>
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| Further, if ''f'' ′(''a'') is not a zero-divisor then ''b'' is unique.
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| As a special case, if <math>f(a) \equiv 0 \, \bmod{\mathfrak m}</math> and ''f'' ′(''a'') is a unit in ''A'' then there is a unique solution to ''f''(''b'') = 0 in ''A'' such that <math>b \equiv a \, \bmod{\mathfrak m}.</math>
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| This result can be generalized to several variables as follows:
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| '''Theorem''': Let ''A'' be a commutative ring that is complete with respect to an ideal '''m''' ⊂ ''A'' and
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| ''f''<sub>''i''</sub>('''x''') ∈ ''A''[''x''<sub>1</sub>, …, ''x''<sub>''n''</sub>] for ''i'' = 1,...,''n'' be a system of ''n'' polynomials in ''n'' variables over ''A''. Let '''f''' = (''f''<sub>1</sub>,...,''f''<sub>''n''</sub>), viewed as a mapping from ''A''<sup>''n''</sup> to ''A''<sup>''n''</sup>, and let ''J''<sub>'''f'''</sub>('''x''') be the [[Jacobian matrix]] of '''f'''. Suppose some '''a''' = (''a''<sub>1</sub>, …, ''a''<sub>''n''</sub>) ∈ ''A''<sup>''n''</sup> is an approximate solution to '''f''' = '''0''' in the sense that
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| :''f''<sub>''i''</sub>('''a''') ≡ 0 mod (det J<sub>'''f'''</sub>('''a'''))<sup>2</sup>'''m'''
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| for 1 ≤ ''i'' ≤ ''n''. Then there is some '''b''' = (''b''<sub>1</sub>, …, ''b''<sub>''n''</sub>) in ''A''<sup>''n''</sup> satisfying '''f'''('''b''') = '''0''', i.e.,
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| :''f''<sub>''i''</sub>('''b''') = 0 for all ''i'',
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| and furthermore this solution is "close" to '''a''' in the sense that
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| :''b''<sub>''i''</sub> ≡ ''a''<sub>''i''</sub> mod ''J''<sub>'''f'''</sub>('''a''')'''m'''
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| for 1 ≤ ''i'' ≤ ''n''.
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| As a special case, if ''f''<sub>''i''</sub>('''a''') ≡ 0 mod '''m''' for all ''i'' and det J<sub>'''f'''</sub>('''a''') is a unit in ''A'' then there is a solution to '''f'''('''b''') = '''0''' with ''b''<sub>''i''</sub> ≡ ''a''<sub>''i''</sub> mod '''m''' for all ''i''.
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| When ''n'' = 1, '''a''' = ''a'' is an element of ''A'' and ''J''<sub>'''f'''</sub>('''a''') = ''J''<sub>''f''</sub>(''a'') is ''f'' ′(''a''). The hypotheses of this multivariable Hensel's lemma reduce to the ones which were stated in the one-variable Hensel's lemma.
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| ==Related concepts==
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| Completeness of a ring is not a necessary condition for the ring to have the Henselian property: [[Goro Azumaya]] in 1950 defined a commutative [[local ring]] satisfying the Henselian property for the maximal ideal '''m''' to be a '''[[Henselian ring]]'''.
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| [[Masayoshi Nagata]] proved in the 1950s that for any commutative local ring ''A'' with maximal ideal '''m''' there always exists a smallest ring ''A''<sup>h</sup> containing ''A'' such that ''A''<sup>h</sup> is Henselian with respect to '''m'''''A''<sup>h</sup>. This ''A''<sup>h</sup> is called the '''[[Henselization]]''' of ''A''. If ''A'' is [[noetherian ring|noetherian]], ''A''<sup>h</sup> will also be noetherian, and ''A''<sup>h</sup> is manifestly algebraic as it is constructed as a limit of [[étale topology|étale neighbourhood]]s. This means that ''A''<sup>h</sup> is usually much smaller than the completion ''Â'' while still retaining the Henselian property and remaining in the same [[category theory|category]].
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| == See also ==
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| *[[Hasse–Minkowski theorem]]
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| *[[Newton polygon]]
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| ==References==
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| * {{Citation | last=Eisenbud | first=David | authorlink=David Eisenbud | title=Commutative algebra | publisher=[[Springer-Verlag]] | location=Berlin, New York | series=Graduate Texts in Mathematics | isbn=978-0-387-94269-8 | id={{MathSciNet | id = 1322960}} | year=1995 | volume=150}}
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| * {{Citation | last=Milne | first=J. G. | title=Étale cohomology | publisher=[[Princeton University Press]] | isbn=978-0-691-08238-7 | year=1980}}
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| [[Category:Modular arithmetic]]
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| [[Category:Commutative algebra]]
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| [[Category:Lemmas]]
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