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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. | |||
== Statement == | |||
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 \pmod{p^k}</math> and <math>f'(r) \not\equiv 0 \pmod{p}</math> | |||
then there exists an integer ''s'' such that | |||
:<math>f(s) \equiv 0 \pmod{p^{k+m}}</math> and <math>r \equiv s \pmod{p^{k}}.</math> | |||
Furthermore, this ''s'' is unique modulo ''p''<sup>''k''+m</sup>, and can be computed explicitly as | |||
:<math>s = r + tp^k</math> where <math>t = - \frac{f(r)}{p^k} \cdot (f'(r)^{-1}).</math> | |||
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>. | |||
As an aside, if <math>f'(r) \equiv 0 \pmod{p}</math>, then 0, 1, or several ''s'' may exist (see Hensel Lifting below). | |||
=== Derivation === | |||
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 | |||
:<math>f(r + tp^k) = f(r) + tp^k\cdot f'(r) + O(p^{2k}).</math> | |||
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 | |||
:<math>0 \equiv f(r + tp^k) \equiv f(r) + tp^k \cdot f'(r)\pmod{p^{k+m}}</math> | |||
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 | |||
:<math>0 \equiv (z + tf'(r))p^k \pmod{p^{k+m}}</math>, | |||
which is to say | |||
:<math>0 \equiv z + tf'(r) \pmod{p^m}.</math> | |||
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>. | |||
== Hensel Lifting == | |||
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. | |||
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: | |||
::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>. | |||
That is, <math>f(r + tp^{k}) \equiv 0\,\bmod{p^{k+1}}\, </math> for all integers ''t''. | |||
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>. | |||
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. | |||
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>. | |||
== Hensel's Lemma for ''p''-adic Numbers == | |||
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 <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 | |||
::<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 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: | |||
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. | |||
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. | |||
== Examples == | |||
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 | |||
: <math>f(r) = r^2 - a \equiv 0 \,\bmod{p}</math> and <math>f'(r) = 2r \not\equiv 0 \,\bmod{p}</math>, | |||
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 | |||
: <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'' 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). | |||
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>f(r_1)=3^2-2=7</math> | |||
:<math>f(r_1)/p^1=7/7=1</math> | |||
:<math>f'(r_1)=2r_1=6</math> | |||
:<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> | |||
:<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 ''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>.) | |||
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 | |||
::<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 '''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). | |||
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. | |||
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: | |||
:: 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 | |||
::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>... | |||
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.) | |||
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>. | |||
(This is false when ''p'' is 2.) | |||
==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 | |||
:<math> f(a) \equiv 0 \,\bmod{f'(a)^2\mathfrak m}</math> | |||
then there is an exact root ''b'' ∈ ''A'' of ''f'' "close to" ''a''; that is, | |||
:<math>f(b) = 0</math> | |||
and | |||
:<math>b \equiv a \,\bmod{f'(a)\mathfrak m}.</math> | |||
Further, if ''f'' ′(''a'') is not a zero-divisor then ''b'' is unique. | |||
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> | |||
This result can be generalized to several variables as follows: | |||
'''Theorem''': Let ''A'' be a commutative ring that is complete with respect to an ideal '''m''' ⊂ ''A'' and | |||
''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 | |||
:''f''<sub>''i''</sub>('''a''') ≡ 0 mod (det J<sub>'''f'''</sub>('''a'''))<sup>2</sup>'''m''' | |||
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., | |||
:''f''<sub>''i''</sub>('''b''') = 0 for all ''i'', | |||
and furthermore this solution is "close" to '''a''' in the sense that | |||
:''b''<sub>''i''</sub> ≡ ''a''<sub>''i''</sub> mod ''J''<sub>'''f'''</sub>('''a''')'''m''' | |||
for 1 ≤ ''i'' ≤ ''n''. | |||
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''. | |||
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. | |||
==Related concepts== | |||
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]]'''. | |||
[[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]]. | |||
== See also == | |||
*[[Hasse–Minkowski theorem]] | |||
*[[Newton polygon]] | |||
==References== | |||
* {{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}} | |||
* {{Citation | last=Milne | first=J. G. | title=Étale cohomology | publisher=[[Princeton University Press]] | isbn=978-0-691-08238-7 | year=1980}} | |||
[[Category:Modular arithmetic]] | |||
[[Category:Commutative algebra]] | |||
[[Category:Lemmas]] | |||
Revision as of 22:13, 3 January 2014
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Another method by way of which you could find out whether the agent is sweet is by checking the feedback, of the shoppers, on the website. There are various individuals would publish their comments on the web site of the Singapore property agent. You can take a look at these feedback and the see whether it will be clever to hire that specific Singapore property agent. You may even get in contact with the developer immediately. Many Singapore property brokers know the developers and you may confirm the goodwill of the agent by asking the developer., then this root corresponds to a unique root of the same equation modulo any higher power of Buying, selling and renting HDB and personal residential properties in Singapore are simple and transparent transactions. Although you are not required to engage a real property salesperson (generally often known as a "public listed property developers In singapore agent") to complete these property transactions, chances are you'll think about partaking one if you are not accustomed to the processes concerned.
Professional agents are readily available once you need to discover an condominium for hire in singapore In some cases, landlords will take into account you more favourably in case your agent comes to them than for those who tried to method them by yourself. You need to be careful, nevertheless, as you resolve in your agent. Ensure that the agent you are contemplating working with is registered with the IEA – Institute of Estate Brokers. Whereas it might sound a hassle to you, will probably be worth it in the end. The IEA works by an ordinary algorithm and regulations, so you'll protect yourself in opposition to probably going with a rogue agent who prices you more than they should for his or her service in finding you an residence for lease in singapore.
There isn't any deal too small. Property agents who are keen to find time for any deal even if the commission is small are the ones you want on your aspect. Additionally they present humbleness and might relate with the typical Singaporean higher. Relentlessly pursuing any deal, calling prospects even without being prompted. Even if they get rejected a hundred times, they still come again for more. These are the property brokers who will find consumers what they need eventually, and who would be the most successful in what they do. 4. Honesty and Integrity
This feature is suitable for you who need to get the tax deductions out of your PIC scheme to your property agency firm. It's endorsed that you visit the correct site for filling this tax return software. This utility must be submitted at the very least yearly to report your whole tax and tax return that you're going to receive in the current accounting 12 months. There may be an official website for this tax filling procedure. Filling this tax return software shouldn't be a tough thing to do for all business homeowners in Singapore.
A wholly owned subsidiary of SLP Worldwide, SLP Realty houses 900 associates to service SLP's fast rising portfolio of residential tasks. Real estate is a human-centric trade. Apart from offering comprehensive coaching applications for our associates, SLP Realty puts equal emphasis on creating human capabilities and creating sturdy teamwork throughout all ranges of our organisational hierarchy. Worldwide Presence At SLP International, our staff of execs is pushed to make sure our shoppers meet their enterprise and investment targets. Under is an inventory of some notable shoppers from completely different industries and markets, who've entrusted their real estate must the expertise of SLP Worldwide.
If you're looking for a real estate or Singapore property agent online, you merely need to belief your instinct. It is because you don't know which agent is sweet and which agent will not be. Carry out research on a number of brokers by looking out the internet. As soon as if you find yourself certain that a selected agent is dependable and trustworthy, you'll be able to choose to utilize his partnerise find you a house in Singapore. More often than not, a property agent is considered to be good if she or he places the contact data on his web site. This is able to imply that the agent does not thoughts you calling them and asking them any questions regarding properties in Singapore. After chatting with them you too can see them of their office after taking an appointment.
Another method by way of which you could find out whether the agent is sweet is by checking the feedback, of the shoppers, on the website. There are various individuals would publish their comments on the web site of the Singapore property agent. You can take a look at these feedback and the see whether it will be clever to hire that specific Singapore property agent. You may even get in contact with the developer immediately. Many Singapore property brokers know the developers and you may confirm the goodwill of the agent by asking the developer., which can be found by iteratively "lifting" the solution modulo successive powers of Buying, selling and renting HDB and personal residential properties in Singapore are simple and transparent transactions. Although you are not required to engage a real property salesperson (generally often known as a "public listed property developers In singapore agent") to complete these property transactions, chances are you'll think about partaking one if you are not accustomed to the processes concerned.
Professional agents are readily available once you need to discover an condominium for hire in singapore In some cases, landlords will take into account you more favourably in case your agent comes to them than for those who tried to method them by yourself. You need to be careful, nevertheless, as you resolve in your agent. Ensure that the agent you are contemplating working with is registered with the IEA – Institute of Estate Brokers. Whereas it might sound a hassle to you, will probably be worth it in the end. The IEA works by an ordinary algorithm and regulations, so you'll protect yourself in opposition to probably going with a rogue agent who prices you more than they should for his or her service in finding you an residence for lease in singapore.
There isn't any deal too small. Property agents who are keen to find time for any deal even if the commission is small are the ones you want on your aspect. Additionally they present humbleness and might relate with the typical Singaporean higher. Relentlessly pursuing any deal, calling prospects even without being prompted. Even if they get rejected a hundred times, they still come again for more. These are the property brokers who will find consumers what they need eventually, and who would be the most successful in what they do. 4. Honesty and Integrity
This feature is suitable for you who need to get the tax deductions out of your PIC scheme to your property agency firm. It's endorsed that you visit the correct site for filling this tax return software. This utility must be submitted at the very least yearly to report your whole tax and tax return that you're going to receive in the current accounting 12 months. There may be an official website for this tax filling procedure. Filling this tax return software shouldn't be a tough thing to do for all business homeowners in Singapore.
A wholly owned subsidiary of SLP Worldwide, SLP Realty houses 900 associates to service SLP's fast rising portfolio of residential tasks. Real estate is a human-centric trade. Apart from offering comprehensive coaching applications for our associates, SLP Realty puts equal emphasis on creating human capabilities and creating sturdy teamwork throughout all ranges of our organisational hierarchy. Worldwide Presence At SLP International, our staff of execs is pushed to make sure our shoppers meet their enterprise and investment targets. Under is an inventory of some notable shoppers from completely different industries and markets, who've entrusted their real estate must the expertise of SLP Worldwide.
If you're looking for a real estate or Singapore property agent online, you merely need to belief your instinct. It is because you don't know which agent is sweet and which agent will not be. Carry out research on a number of brokers by looking out the internet. As soon as if you find yourself certain that a selected agent is dependable and trustworthy, you'll be able to choose to utilize his partnerise find you a house in Singapore. More often than not, a property agent is considered to be good if she or he places the contact data on his web site. This is able to imply that the agent does not thoughts you calling them and asking them any questions regarding properties in Singapore. After chatting with them you too can see them of their office after taking an appointment.
Another method by way of which you could find out whether the agent is sweet is by checking the feedback, of the shoppers, on the website. There are various individuals would publish their comments on the web site of the Singapore property agent. You can take a look at these feedback and the see whether it will be clever to hire that specific Singapore property agent. You may even get in contact with the developer immediately. Many Singapore property brokers know the developers and you may confirm the goodwill of the agent by asking the developer.. More generally it is used as a generic name for analogues for complete commutative rings (including p-adic fields in particular) of the Newton method for solving equations. Since p-adic analysis is in some ways simpler than real analysis, there are relatively neat criteria guaranteeing a root of a polynomial.
Statement
Let 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
then there exists an integer s such that
Furthermore, this s is unique modulo pk+m, and can be computed explicitly as
In this formula for t, the division by pk denotes ordinary integer division (where the remainder will be 0), while negation, multiplication, and multiplicative inversion are performed in .
As an aside, if , then 0, 1, or several s may exist (see Hensel Lifting below).
Derivation
The lemma derives from considering the Taylor expansion of f around r. From , we see that s has to be of the form s = r + tpk for some integer t. Expanding gives
Reducing both sides modulo pk+m, we see that for to hold, we need
where the O(p2k) terms vanish because k+m ≤ 2k. Then we note that for some integer z since r is a root of f mod pk, so
which is to say
Then substituting back f(r)/pk for z and solving for t in gives the explicit formula for t mentioned above. The assumption that is not divisible by p ensures that has an inverse mod which is necessarily unique. Hence a solution for t exists uniquely modulo , and s exists uniquely modulo .
Hensel Lifting
Using the lemma, one can "lift" a root r of the polynomial f mod pk to a new root s mod pk+1 such that r ≡ s mod pk (by taking m=1; taking larger m also works). In fact, a root mod pk+1 is also a root mod pk, so the roots mod pk+1 are precisely the liftings of roots mod pk. The new root s is congruent to r mod p, so the new root also satisfies . So the lifting can be repeated, and starting from a solution rk of we can derive a sequence of solutions rk+1, rk+2, ... of the same congruence for successively higher powers of p, provided for the initial root rk. This also shows that f has the same number of roots mod pk as mod pk+1, mod p k+2, or any other higher power of p, provided the roots of f mod pk are all simple.
What happens to this process if r is not a simple root mod p? If we have a root mod pk at which the derivative mod p is 0, then there is not a unique lifting of a root mod pk to a root mod pk+1: either there is no lifting to a root mod pk+1 or there are multiple choices:
That is, for all integers t. Therefore if then there is no lifting of r to a root of f(x) mod pk+1, while if then every lifting of r to modulus pk+1 is a root of f(x) mod pk+1.
To see the difficulty that can arise in a concrete example, take p = 2, f(x) = x2 + 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. On the other hand, if we take f(x) = x2 - 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 2k.
Hensel's Lemma for p-adic Numbers
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 rk (roots mod pk) to rk+1 (roots mod pk+1) can be expressed in a much more intuitive way. Instead of choosing t to be an(y) integer which solves the congruence , let t be the rational number (the pk here is not really a denominator since f(rk) is divisible by pk). Then set
This fraction may not be an integer, but it is a p-adic integer, and the sequence of numbers rk converges in the p-adic integers to a root of f(x) = 0. Moreover, the displayed recursive formula for the (new) number rk+1 in terms of rk 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 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: if there is an integer a which satisfies |f(a)|p < |f′(a)|p2, then there is a unique p-adic integer b such f(b) = 0 and |b-a|p < |f'(a)|p. 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|p < |f'(a)|p needs additional work.
The statement of Hensel's lemma given above (taking ) 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)|p < 1 and |f'(a)|p = 1.
Examples
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 Zp. Indeed, let f(x)=x2-a. Its derivative is 2x, so if r is a square root of a mod p we have
where the second condition depends on p not being 2. The basic version of Hensel's lemma tells us that starting from r1= r we can recursively construct a sequence of integers { rk } such that
This sequence converges to some p-adic integer b and b2=a. In fact, b is the unique square root of a in Zp congruent to r1 modulo p. Conversely, if a is a perfect square in Zp 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 ) in the 7-adic integers. Modulo 7 one solution is 3 (we could also take 4), so we set . Hensel's lemma then allows us to find as follows:
And sure enough, . (If we had used the Newton method recursion directly in the 7-adics, then r2 = r1 - f(r1)/f'(r1) = 3 - 7/6 = 11/6, and 11/6 ≡ 10 mod 72.)
We can continue and find . 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 Z7 which has initial 7-adic expansion
If we started with the initial choice then Hensel's lemma would produce a square root of 2 in Z7 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 f(x) = x2 - 17 and a = 1. Then f(a) = -16 and f'(a) = 2, so |f(a)|2 < |f′(a)|22, which implies there is a unique 2-adic integer b satisfying b2 = 17 and |b- a|2 < |f'(a)|2 = 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.
In terms of lifting roots of x2 - 17 from one modulus 2k to the next 2k+1, 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 x2 - 17 mod 2k, 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 + 23 and 25 = 1 + 23 + 24, 7 = 1 + 2 + 22 and 23 = 1 + 2 + 22 + 24.
The 2-adic square roots of 17 have expansions
- 1 + 23 + 25 + 26 + 27 + 29 + 210 + ..., 1 + 2 + 22 + 24 + 28 + 211...
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 Z3. Let f(x) = x3 - 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)|3 < |f'(1)|32, 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.)
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 p2 is a p-th power in Zp. (This is false when p is 2.)
Generalizations
Suppose A is a commutative ring, complete with respect to an ideal , and let be a polynomial with coefficients in A. Then if a ∈ A is an "approximate root" of f in the sense that it satisfies
then there is an exact root b ∈ A of f "close to" a; that is,
and
Further, if f ′(a) is not a zero-divisor then b is unique.
As a special case, if and f ′(a) is a unit in A then there is a unique solution to f(b) = 0 in A such that
This result can be generalized to several variables as follows:
Theorem: Let A be a commutative ring that is complete with respect to an ideal m ⊂ A and fi(x) ∈ A[x1, …, xn] for i = 1,...,n be a system of n polynomials in n variables over A. Let f = (f1,...,fn), viewed as a mapping from An to An, and let Jf(x) be the Jacobian matrix of f. Suppose some a = (a1, …, an) ∈ An is an approximate solution to f = 0 in the sense that
- fi(a) ≡ 0 mod (det Jf(a))2m
for 1 ≤ i ≤ n. Then there is some b = (b1, …, bn) in An satisfying f(b) = 0, i.e.,
- fi(b) = 0 for all i,
and furthermore this solution is "close" to a in the sense that
- bi ≡ ai mod Jf(a)m
for 1 ≤ i ≤ n.
As a special case, if fi(a) ≡ 0 mod m for all i and det Jf(a) is a unit in A then there is a solution to f(b) = 0 with bi ≡ ai mod m for all i.
When n = 1, a = a is an element of A and Jf(a) = Jf(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.
Related concepts
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.
Masayoshi Nagata proved in the 1950s that for any commutative local ring A with maximal ideal m there always exists a smallest ring Ah containing A such that Ah is Henselian with respect to mAh. This Ah is called the Henselization of A. If A is noetherian, Ah will also be noetherian, and Ah is manifestly algebraic as it is constructed as a limit of étale neighbourhoods. This means that Ah is usually much smaller than the completion  while still retaining the Henselian property and remaining in the same category.
See also
References
- Many property agents need to declare for the PIC grant in Singapore. However, not all of them know find out how to do the correct process for getting this PIC scheme from the IRAS. There are a number of steps that you need to do before your software can be approved.
Naturally, you will have to pay a safety deposit and that is usually one month rent for annually of the settlement. That is the place your good religion deposit will likely be taken into account and will kind part or all of your security deposit. Anticipate to have a proportionate amount deducted out of your deposit if something is discovered to be damaged if you move out. It's best to you'll want to test the inventory drawn up by the owner, which can detail all objects in the property and their condition. If you happen to fail to notice any harm not already mentioned within the inventory before transferring in, you danger having to pay for it yourself.
In case you are in search of an actual estate or Singapore property agent on-line, you simply should belief your intuition. It's because you do not know which agent is nice and which agent will not be. Carry out research on several brokers by looking out the internet. As soon as if you end up positive that a selected agent is dependable and reliable, you can choose to utilize his partnerise in finding you a home in Singapore. Most of the time, a property agent is taken into account to be good if he or she locations the contact data on his website. This may mean that the agent does not mind you calling them and asking them any questions relating to new properties in singapore in Singapore. After chatting with them you too can see them in their office after taking an appointment.
Have handed an trade examination i.e Widespread Examination for House Brokers (CEHA) or Actual Property Agency (REA) examination, or equal; Exclusive brokers are extra keen to share listing information thus making certain the widest doable coverage inside the real estate community via Multiple Listings and Networking. Accepting a severe provide is simpler since your agent is totally conscious of all advertising activity related with your property. This reduces your having to check with a number of agents for some other offers. Price control is easily achieved. Paint work in good restore-discuss with your Property Marketing consultant if main works are still to be done. Softening in residential property prices proceed, led by 2.8 per cent decline within the index for Remainder of Central Region
Once you place down the one per cent choice price to carry down a non-public property, it's important to accept its situation as it is whenever you move in – faulty air-con, choked rest room and all. Get round this by asking your agent to incorporate a ultimate inspection clause within the possibility-to-buy letter. HDB flat patrons routinely take pleasure in this security net. "There's a ultimate inspection of the property two days before the completion of all HDB transactions. If the air-con is defective, you can request the seller to repair it," says Kelvin.
15.6.1 As the agent is an intermediary, generally, as soon as the principal and third party are introduced right into a contractual relationship, the agent drops out of the image, subject to any problems with remuneration or indemnification that he could have against the principal, and extra exceptionally, against the third occasion. Generally, agents are entitled to be indemnified for all liabilities reasonably incurred within the execution of the brokers´ authority.
To achieve the very best outcomes, you must be always updated on market situations, including past transaction information and reliable projections. You could review and examine comparable homes that are currently available in the market, especially these which have been sold or not bought up to now six months. You'll be able to see a pattern of such report by clicking here It's essential to defend yourself in opposition to unscrupulous patrons. They are often very skilled in using highly unethical and manipulative techniques to try and lure you into a lure. That you must also protect your self, your loved ones, and personal belongings as you'll be serving many strangers in your home. Sign a listing itemizing of all of the objects provided by the proprietor, together with their situation. HSR Prime Recruiter 2010 - Many property agents need to declare for the PIC grant in Singapore. However, not all of them know find out how to do the correct process for getting this PIC scheme from the IRAS. There are a number of steps that you need to do before your software can be approved.
Naturally, you will have to pay a safety deposit and that is usually one month rent for annually of the settlement. That is the place your good religion deposit will likely be taken into account and will kind part or all of your security deposit. Anticipate to have a proportionate amount deducted out of your deposit if something is discovered to be damaged if you move out. It's best to you'll want to test the inventory drawn up by the owner, which can detail all objects in the property and their condition. If you happen to fail to notice any harm not already mentioned within the inventory before transferring in, you danger having to pay for it yourself.
In case you are in search of an actual estate or Singapore property agent on-line, you simply should belief your intuition. It's because you do not know which agent is nice and which agent will not be. Carry out research on several brokers by looking out the internet. As soon as if you end up positive that a selected agent is dependable and reliable, you can choose to utilize his partnerise in finding you a home in Singapore. Most of the time, a property agent is taken into account to be good if he or she locations the contact data on his website. This may mean that the agent does not mind you calling them and asking them any questions relating to new properties in singapore in Singapore. After chatting with them you too can see them in their office after taking an appointment.
Have handed an trade examination i.e Widespread Examination for House Brokers (CEHA) or Actual Property Agency (REA) examination, or equal; Exclusive brokers are extra keen to share listing information thus making certain the widest doable coverage inside the real estate community via Multiple Listings and Networking. Accepting a severe provide is simpler since your agent is totally conscious of all advertising activity related with your property. This reduces your having to check with a number of agents for some other offers. Price control is easily achieved. Paint work in good restore-discuss with your Property Marketing consultant if main works are still to be done. Softening in residential property prices proceed, led by 2.8 per cent decline within the index for Remainder of Central Region
Once you place down the one per cent choice price to carry down a non-public property, it's important to accept its situation as it is whenever you move in – faulty air-con, choked rest room and all. Get round this by asking your agent to incorporate a ultimate inspection clause within the possibility-to-buy letter. HDB flat patrons routinely take pleasure in this security net. "There's a ultimate inspection of the property two days before the completion of all HDB transactions. If the air-con is defective, you can request the seller to repair it," says Kelvin.
15.6.1 As the agent is an intermediary, generally, as soon as the principal and third party are introduced right into a contractual relationship, the agent drops out of the image, subject to any problems with remuneration or indemnification that he could have against the principal, and extra exceptionally, against the third occasion. Generally, agents are entitled to be indemnified for all liabilities reasonably incurred within the execution of the brokers´ authority.
To achieve the very best outcomes, you must be always updated on market situations, including past transaction information and reliable projections. You could review and examine comparable homes that are currently available in the market, especially these which have been sold or not bought up to now six months. You'll be able to see a pattern of such report by clicking here It's essential to defend yourself in opposition to unscrupulous patrons. They are often very skilled in using highly unethical and manipulative techniques to try and lure you into a lure. That you must also protect your self, your loved ones, and personal belongings as you'll be serving many strangers in your home. Sign a listing itemizing of all of the objects provided by the proprietor, together with their situation. HSR Prime Recruiter 2010