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| {{for|other radicals|radical of a ring}}
| | Deciding on the daily calorie consumption plays a main character inside how effectively we lose weight. If you eat too countless calories, you won't lose any fat. If you eat too few, we body's metabolism may slow down, plus as a result, try to hold onto more of the calories you ingest.<br><br>What is interesting here, men [http://safedietplans.com/bmr-calculator bmr calculator] could take their temp any time. For ladies that are inside their menstrual years, the number one reading usually be found on the 2nd or third day after the menstrual flow begins. Now comes the fun part.<br><br>Some would think it is needless to utilize a basal metabolic rate as a guide for controlling a daily calories. It's really because effortless to limit oneself to 1200 to 1500 calories a day. This will be fine when you were all the same. However, this 1 size fits all attitude of dieting may work for certain, but for most they need anything more that fits closer to their lifestyle.<br><br>Then the more we exercise, the better gets your metabolism. And with a high metabolism you'll burn calories more effectively. As for a individual with a low metabolism rate, he/she usually burn fewer calories from the same amount of food plus store the calories which were not 'burned' into fats.<br><br>People constantly ask how numerous calories must they be consuming in a day to keep there weight, youll only have to a small math. The initial step is calculating the bmr that this might be the amount of energy the body takes in and must function correctly. We use about 60% of the calories you consume everyday for the general bodily functions such as merely by being alive plus breathing the others which influence a BMR are height, weight, age and sex.<br><br>Women that are on a diet with their male partner usually complain that they can't lose fat as quickly as their man partner. Unfortunately, this does appear to become the case. Scientists think that the differences might be hormonal (as are many variations between females and guys!) Women have more estrogen, causing them to shop more body stamina because fat; whereas guys have more testosterone, helping them to convert body stamina to muscle faster.<br><br>Frankly, because a pharmacist, I am not convinced by strong health evidence which they on the contrary do what they claim to do. Many are stimulants that might be harmful to some people. Others simply work by decreasing the appetite temporarily. Occasionally they "work" simply because we merely invested $40.00 found on the bottle of medications...and we don't have enough income left to purchase junk food! |
| In [[commutative ring]] theory, a branch of [[mathematics]], the '''radical of an ideal''' ''I'' is an [[ideal (ring theory)|ideal]] such that an element ''x'' is in the radical if some power of ''x'' is in ''I''. A '''radical ideal''' (or '''semiprime ideal''') is an ideal that is its own radical (this can be phrased as being a [[Fixed point (mathematics)|fixed point]] of an operation on ideals called 'radicalization'). The radical of a [[primary ideal]] is prime.
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| Radical ideals defined here are generalized to noncommutative rings in the [[Semiprime ring]] article.
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| ==Definition==
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| The '''radical''' of an ideal ''I'' in a [[commutative ring]] ''R'', denoted by Rad(''I'') or <math>\sqrt{I}</math>, is defined as
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| :<math>\sqrt{I}=\{r\in R|r^n\in I\ \hbox{for some positive integer}\ n\}.</math> | |
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| Intuitively, one can think of the radical of ''I'' as obtained by taking all the possible roots of elements of ''I''. Equivalently, the radical of ''I'' is the pre-image of the ideal of nilpotent elements (called [[nilradical of a ring|nilradical]]) in <math>R/I</math>.<ref>A direct proof can be give as follows:
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| Let ''a'' and ''b'' be in the radical of an ideal ''I''. Then, for some positive integers ''m'' and ''n'', ''a''<sup>''n''</sup> and ''b''<sup>''m''</sup> are in ''I''. We will show that ''a'' + ''b'' is in the radical of ''I''. Use the [[binomial theorem]] to expand (''a''+''b'')<sup>''n''+''m''−1</sup> (with commutativity assumed):
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| :<math>(a+b)^{n+m-1}=\sum_{i=0}^{n+m-1}{n+m-1\choose i}a^ib^{n+m-1-i}.</math>
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| For each ''i'', exactly one of the following conditions will hold:
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| *''i'' ≥ ''n''
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| *''n'' + ''m'' − 1 − ''i'' ≥ ''m''.
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| This says that in each expression ''a''<sup>''i''</sup>''b''<sup>''n''+''m''− 1 − ''i''</sup>, either the exponent of ''a'' will be large enough to make this power of ''a'' be in ''I'', or the exponent of ''b'' will be large enough to make this power of ''b'' be in ''I''. Since the product of an element in ''I'' with an element in ''R'' is in ''I'' (as ''I'' is an ideal), this product expression will be in ''I'', and then (''a''+''b'')<sup>''n''+''m''−1</sup> is in ''I'', therefore ''a''+''b'' is in the radical of ''I''. | |
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| To finish checking that the radical is an ideal, we take an element ''a'' in the radical, with ''a''<sup>''n''</sup> in ''I'' and an arbitrary element ''r''∈''R''. Then, (''ra'')<sup>''n''</sup> = ''r''<sup>''n''</sup>''a''<sup>''n''</sup> is in ''I'', so ''ra'' is in the radical. Thus the radical is an ideal.</ref> The latter shows <math>\sqrt{I}</math> is an ideal itself, containing ''I''.
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| If the radical of ''I'' is finitely generated, then some power of <math>\sqrt{I}</math> is contained in ''I''.<ref>{{harvnb|Atiyah–MacDonald|1969|loc=Proposition 7.14}}</ref> In particular, If ''I'' and ''J'' are ideals of a noetherian ring, then ''I'' and ''J'' have the same radical if and only if ''I'' contains some power of ''J'' and ''J'' contains some power of ''I''.
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| If an ideal ''I'' coincides with its own radical, then ''I'' is called a ''radical ideal'' or ''[[semiprime ideal]]''.
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| ==Examples==
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| Consider the ring '''Z''' of [[integer]]s.
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| # The radical of the ideal 4'''Z''' of integer multiples of 4 is 2'''Z'''.
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| # The radical of 5'''Z''' is 5'''Z'''.
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| # The radical of 12'''Z''' is 6'''Z'''.
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| # In general, the radical of ''m'''''Z''' is ''r'''''Z''', where ''r'' is the product of all prime factors of ''m'' (see [[radical of an integer]]). In fact, this generalizes to an arbitrary ideal; see the properties section.
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| The radical of a [[primary ideal]] is prime. If the radical of an ideal ''I'' is maximal, then ''I'' is primary.<ref>{{harvnb|Atiyah–MacDonald|1969|loc=Proposition 4.2}}</ref>
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| If ''I'' is an ideal, <math>\sqrt{I^n} = \sqrt{I}</math>. A prime ideal is a radical ideal. So <math>\sqrt{P^n} = P</math> for any prime ideal ''P''.
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| Let ''I'', ''J'' be ideals of a ring ''R''. If <math>\sqrt{I}, \sqrt{J}</math> are [[Ideal (ring theory)#Types of ideals|comaximal]], then <math>I, J</math> are comaximal.<ref>Proof: <math>R = \sqrt{\sqrt{I} + \sqrt{J} } = \sqrt{I + J}</math> implies <math>I + J = R</math>.</ref>
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| Let ''M'' be a finitely generated module over a noetherian ring ''R''. Then
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| :<math>\sqrt{\operatorname{ann}_R(M)} = \bigcap_{\mathfrak{p} \in \operatorname{supp}M} \mathfrak{p} = \bigcap_{\mathfrak{p} \in \operatorname{ass}M} \mathfrak{p}</math><ref>{{harvnb|Lang|2002|loc=Ch X, Proposition 2.10}}</ref>
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| where <math>\operatorname{supp}M</math> is the [[support of a module|support]] of ''M'' and <math>\operatorname{ass}M</math> is the set of [[associated prime]]s of ''M''.
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| ==Properties==
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| This section will continue the convention that ''I'' is an ideal of a commutative ring ''R'':
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| *It is always true that Rad(Rad(''I''))=Rad(''I''). Moreover, Rad(''I'') is the smallest radical ideal containing ''I''.
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| *Rad(''I'') is the intersection of all the prime ideals of ''R'' that contain ''I''. On one hand, every prime ideal is radical, and so this intersection contains Rad(''I''). Suppose ''r'' is an element of ''R'' which is not in Rad(''I''), and let ''S'' be the set {''r<sup>n</sup>''|''n'' is a nonnegative integer}. By the definition of Rad(''I''), ''S'' must be disjoint from ''I''. ''S'' is also [[multiplicatively closed subset|multiplicatively closed]]. Thus, by a variant of [[Krull's theorem]], there exists a prime ideal ''P'' that contains ''I'' and is still disjoint from ''S''. (see [[prime ideal]].) Since ''P'' contains ''I'', but not ''r'', this shows that ''r'' is not in the intersection of prime ideals containing ''I''. This finishes the proof. The statement may be strengthened a bit: the radical of ''I'' is the intersection of all prime ideals of ''R'' that are [[Minimal prime (commutative algebra)|minimal]] among those containing ''I''.
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| *Specializing the last point, the nilradical (the set of all nilpotent elements) is equal to the intersection of all prime ideals of ''R''.
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| *An ideal ''I'' in a ring ''R'' is radical if and only if the [[quotient ring]] ''R/I'' is [[reduced ring|reduced]].
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| *The radical of a homogeneous ideal is homogeneous.
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| ==Applications==
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| The primary motivation in studying ''radicals'' is the celebrated ''[[Hilbert's Nullstellensatz]]'' in [[commutative algebra]]. An easily understood version of this theorem states that for an [[algebraically closed field]] ''k'', and for any finitely generated polynomial ideal ''J'' in the ''n'' indeterminates <math>x_1, x_2, \ldots, x_n</math> over the field ''k'', one has
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| :<math>\operatorname{I}(\operatorname{V}(J)) = \operatorname{Rad} (J)\,</math>
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| where
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| :<math> \operatorname{V}(J) = \{x \in k^n \ |\ f(x)=0 \mbox{ for all } f\in J\}</math>
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| and
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| :<math> \operatorname{I}(S) = \{f \in k[x_1,x_2,\ldots x_n] \ |\ f(x)=0 \mbox{ for all } x\in S \}.</math>
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| Another way of putting it: The composition <math>\operatorname{I}(\operatorname{V}(-))=\operatorname{Rad}(-)\,</math> on the set of ideals of a ring is in fact a [[closure operator]]. From the definition of the radical, it is clear that taking the radical is an [[idempotent]] operation.
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| ==See also==
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| * [[Jacobson radical]]
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| * [[Nilradical of a ring]]
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| == Notes ==
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| {{reflist}}
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| == References ==
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| *[[Michael Atiyah|M. Atiyah]], [[Ian G. Macdonald|I.G. Macdonald]], ''Introduction to Commutative Algebra'', [[Addison–Wesley]], 1994. ISBN 0-201-40751-5
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| * [[David Eisenbud|Eisenbud, David]], ''Commutative Algebra with a View Toward Algebraic Geometry'', Graduate Texts in Mathematics, 150, Springer-Verlag, 1995, ISBN 0-387-94268-8.
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| * {{Lang Algebra|edition=3r}}
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| [[Category:Ideals]]
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Deciding on the daily calorie consumption plays a main character inside how effectively we lose weight. If you eat too countless calories, you won't lose any fat. If you eat too few, we body's metabolism may slow down, plus as a result, try to hold onto more of the calories you ingest.
What is interesting here, men bmr calculator could take their temp any time. For ladies that are inside their menstrual years, the number one reading usually be found on the 2nd or third day after the menstrual flow begins. Now comes the fun part.
Some would think it is needless to utilize a basal metabolic rate as a guide for controlling a daily calories. It's really because effortless to limit oneself to 1200 to 1500 calories a day. This will be fine when you were all the same. However, this 1 size fits all attitude of dieting may work for certain, but for most they need anything more that fits closer to their lifestyle.
Then the more we exercise, the better gets your metabolism. And with a high metabolism you'll burn calories more effectively. As for a individual with a low metabolism rate, he/she usually burn fewer calories from the same amount of food plus store the calories which were not 'burned' into fats.
People constantly ask how numerous calories must they be consuming in a day to keep there weight, youll only have to a small math. The initial step is calculating the bmr that this might be the amount of energy the body takes in and must function correctly. We use about 60% of the calories you consume everyday for the general bodily functions such as merely by being alive plus breathing the others which influence a BMR are height, weight, age and sex.
Women that are on a diet with their male partner usually complain that they can't lose fat as quickly as their man partner. Unfortunately, this does appear to become the case. Scientists think that the differences might be hormonal (as are many variations between females and guys!) Women have more estrogen, causing them to shop more body stamina because fat; whereas guys have more testosterone, helping them to convert body stamina to muscle faster.
Frankly, because a pharmacist, I am not convinced by strong health evidence which they on the contrary do what they claim to do. Many are stimulants that might be harmful to some people. Others simply work by decreasing the appetite temporarily. Occasionally they "work" simply because we merely invested $40.00 found on the bottle of medications...and we don't have enough income left to purchase junk food!