imported>Cutx64 at 03:27, 18 October 2025

2025-10-18T03:27:16Z

New page

{{Short description|Concept in algebra}}{{for|other radicals|radical of a ring}}
In [[ring theory]], a branch of [[mathematics]], the '''radical''' of an [[ideal (ring theory)|ideal]] <math>I</math> of a [[commutative ring]] is another ideal defined by the property that an element <math>x</math> is in the radical [[if and only if]] some power of <math>x</math> is in <math>I</math>. Taking the radical of an ideal is called ''radicalization''. A '''radical ideal''' (or '''semiprime ideal''') is an ideal that is equal to its radical. The radical of a [[primary ideal]] is a [[prime ideal]].

This concept is generalized to [[non-commutative ring]]s in the [[semiprime ring]] article.

==Definition==

The '''radical''' of an ideal <math>I</math> in a [[commutative ring]] <math>R</math>, denoted by <math>\operatorname{rad}(I)</math> or <math>\sqrt{I}</math>, is defined as
:<math>\sqrt{I} = \left\{r\in R \mid r^n\in I\ \hbox{for some}\ n \in \Z^{+}\!\right\},</math>

(note that <math>I \subseteq \sqrt{I}</math>).
Intuitively, <math>\sqrt{I}</math> is obtained by taking all roots of elements of <math>I</math> within the [[ring (mathematics)|ring]] <math>R</math>. Equivalently, <math>\sqrt{I}</math> is the [[preimage]] of the ideal of [[nilpotent]] elements (the [[nilradical of a ring|nilradical]]) of the [[quotient ring]] <math>R/I</math> (via the natural map <math>\pi\colon R\to R/I</math>). The latter proves that <math>\sqrt{I}</math> is an ideal.<ref group="Note">Here is a direct proof that <math>\sqrt{I}</math> is an ideal. Start with <math>a,b\in\sqrt{I}</math> with some powers <math>a^n,b^m \in I</math>. To show that <math>a+b\in\sqrt{I}</math>, we use the [[binomial theorem]] (which holds for any commutative ring):

:<math>\textstyle (a+b)^{n+m-1}=\sum_{i=0}^{n+m-1}\binom{n+m-1}{i}a^ib^{n+m-1-i}.</math>

For each <math>i</math>, we have either <math>i\geq n</math> or <math>n+m-1-i\geq m</math>. Thus, in each term <math>a^i b^{n+m-1-i}</math>, one of the exponents will be large enough to make that factor lie in <math>I</math>. Since any element of <math>I</math> times an element of <math>R</math> lies in <math>I</math> (as <math>I</math> is an ideal), this term lies in <math>I</math>. Hence <math>(a+b)^{n+m-1} \in I</math>, and so <math>a+b\in\sqrt{I}</math>.

To finish checking that the radical is an ideal, take <math>a\in\sqrt{I}</math> with <math>a^n\in I</math>, and any <math>r \in R</math>. Then <math>(ra)^n=r^na^n\in I</math>, so <math>ra\in\sqrt{I}</math>. Thus the radical is an ideal.</ref>

If the radical of <math>I</math> is [[Ideal_(ring_theory)#Types_of_ideals|finitely generated]], then some power of <math>\sqrt{I}</math> is contained in <math>I</math>.<ref>{{harvnb|Atiyah|Macdonald|1994|loc=Proposition 7.14}}</ref> In particular, if <math>I</math> and <math>J</math> are ideals of a [[Noetherian ring]], then <math>I</math> and <math>J</math> have the same radical if and only if <math>I</math> contains some power of <math>J</math> and <math>J</math> contains some power of <math>I</math>.

If an ideal <math>I</math> coincides with its own radical, then <math>I</math> is called a ''radical ideal'' or ''[[semiprime ideal]]''.

==Examples==
* Consider the ring <math>\Z</math> of [[Integer#Algebraic_properties|integers]].
*# The radical of the ideal <math>4\Z</math> of integer multiples of <math>4</math> is <math>2\Z</math> (the [[Parity_(mathematics)|evens]]).
*# The radical of <math>5\Z</math> is <math>5\Z</math>.
*# The radical of <math>12\Z</math> is <math>6\Z</math>.
*# In general, the radical of <math>m\Z</math> is <math>r\Z</math>, where <math>r</math> is the product of all distinct [[prime factor]]s of <math>m</math>, the largest [[square-free integer|square-free]] factor of <math>m</math> (see [[Radical of an integer]]). In fact, this generalizes to an arbitrary ideal (see the [[#Properties|Properties]] section).
* Consider the ideal <math>I = \left(y^4\right) \subseteq \Complex[x,y]</math>. It is trivial to show <math>\sqrt{I}=(y)</math> (using the basic property {{awrap|<math>\sqrt{I^n} = \sqrt{I}</math>),}} but we give some alternative methods:{{clarify|The following, which is nice, seems to be using the [[Hilbert nullstellensaz]] implicitly.|date=January 2019}} The radical <math>\sqrt{I}</math> corresponds to the [[nilradical of a ring|nilradical]] <math>\sqrt{0}</math> of the quotient ring <math>R = \Complex[x,y]/\!\left(y^4\right)</math>, which is the [[intersection (set theory)|intersection]] of all prime ideals of the quotient ring. This is contained in the [[Jacobson radical]], which is the intersection of all [[maximal ideal]]s, which are the [[kernel (algebra)|kernels]] of [[ring homomorphism|homomorphisms]] to [[field (mathematics)|fields]]. Any ring homomorphism <math>R \to \Complex</math> must have <math>y</math> in the kernel in order to have a well-defined homomorphism (if we said, for example, that the kernel should be <math>(x,y-1)</math> the composition of <math>\Complex[x,y] \to R \to \Complex</math> would be <math>\left(x, y^4, y-1\right)</math>, which is the same as trying to force <math>1=0</math>). Since <math>\Complex</math> is [[algebraically closed]], every homomorphism <math>R \to \mathbb{F}</math> must factor through <math>\Complex</math>, so we only have to compute the intersection of <math>\{\ker(\Phi) : \Phi \in \operatorname{Hom}(R,\Complex) \}</math> to compute the radical of <math>(0).</math> We then find that <math>\sqrt{0} = (y) \subseteq R.</math>

==Properties==
This section will continue the convention that <math>I</math> is an ideal of a commutative ring <math>R</math>:

*It is always true that <math display="inline">\sqrt{\sqrt{I}} = \sqrt{I}</math>, i.e. radicalization is an [[idempotent]] operation. Moreover, <math>\sqrt{I}</math> is the smallest radical ideal containing <math>I</math>.
*<math>\sqrt{I}</math> is the intersection of all the [[prime ideal|prime ideals]] of <math>R</math> that contain <math>I</math><math display="block">\sqrt{I}=\bigcap_{\stackrel{\mathfrak{p}\text{ prime}}{R\supset\mathfrak{p}\supseteq I}}\mathfrak{p},</math>and thus the radical of a prime ideal is equal to itself. Proof: ''On one hand, every prime ideal is radical, and so this intersection contains <math>\sqrt{I}</math>. Suppose <math>r</math> is an element of <math>R</math> that is not in <math>\sqrt{I}</math>, and let <math>S</math> be the set <math>\left\{r^n \mid n = 0, 1, 2, \ldots \right\}</math>. By the definition of <math>\sqrt{I}</math>, <math>S</math> must be [[disjoint sets|disjoint]] from <math>I</math>. <math>S</math> is also [[multiplicatively closed subset|multiplicatively closed]]. Thus, by a variant of [[Krull's theorem]], there exists a prime ideal <math>\mathfrak{p}</math> that contains <math>I</math> and is still disjoint from <math>S</math> (see [[Prime ideal]]). Since <math>\mathfrak{p}</math> contains <math>I</math>, but not <math>r</math>, this shows that <math>r</math> is not in the intersection of prime ideals containing <math>I</math>. This finishes the proof.'' The statement may be strengthened a bit: the radical of <math>I</math> is the intersection of all prime ideals of <math>R</math> that are [[Minimal prime (commutative algebra)|minimal]] among those containing <math>I</math>.
*Specializing the last point, the [[nilradical of a ring|nilradical]] (the set of all nilpotent elements) is equal to the intersection of all prime ideals of <math>R</math><ref group="Note">For a direct proof, see also the [[Nilradical of a ring#Commutative rings|characterisation of the nilradical of a ring]].</ref> <math display="block">\sqrt{0} = \mathfrak{N}_R = \bigcap_{\mathfrak{p}\subsetneq R\text{ prime}}\mathfrak{p}.</math>This property is seen to be equivalent to the former via the natural map <math>\pi\colon R\to R/I</math>, which yields a [[bijection]] <math>u</math>: <math display="block">\left\lbrace\text{ideals }J\mid R\supseteq J\supseteq I\right\rbrace
\quad {\overset{u}{\rightleftharpoons}} \quad
\left\lbrace\text{ideals }J\mid J\subseteq R/I\right\rbrace,</math> defined by <math>u \colon J\mapsto J/I=\lbrace r+I\mid r\in J\rbrace.</math><ref>{{Cite book| url=https://bookstore.ams.org/gsm-104| title=Algebra: Chapter 0| last=Aluffi| first=Paolo| publisher=AMS| year=2009| isbn=978-0-8218-4781-7| pages=142}}</ref><ref group=Note>This fact is also known as [[Isomorphism theorems#Third isomorphism theorem 2|fourth isomorphism theorem]].</ref>
*An ideal <math>I</math> in a ring <math>R</math> is radical if and only if the [[quotient ring]] <math>R/I</math> is [[reduced ring|reduced]].
*The radical of a [[homogeneous ideal]] is homogeneous.
*The radical of an intersection of ideals is equal to the intersection of their radicals: <math> \sqrt{I \cap J} = \sqrt{I} \cap \sqrt{J}</math>.
*The radical of a [[primary ideal]] is prime. If the radical of an ideal <math>I</math> is maximal, then <math>I</math> is primary.<ref>{{harvnb|Atiyah|Macdonald|1994|loc=Proposition 4.2}}</ref>
*If <math>I</math> is an ideal, <math>\sqrt{I^n} = \sqrt{I}</math>. Since prime ideals are radical ideals, <math>\sqrt{\mathfrak{p}^n} = \mathfrak{p}</math> for any prime ideal <math>\mathfrak{p}</math>.
*Let <math>I,J</math> be ideals of a ring <math>R</math>. If <math>\sqrt{I}, \sqrt{J}</math> are [[Ideal (ring theory)#Types of ideals|comaximal]], then <math>I, J</math> are comaximal.<ref group="Note">Proof: <math display="inline">R = \sqrt{\sqrt{I} + \sqrt{J} } = \sqrt{I + J}</math> implies <math>I + J = R</math>.</ref>
*Let <math>M</math> be a [[finitely generated module|finitely generated]] [[module (mathematics)|module]] over a [[Noetherian ring]] <math>R</math>. Then<ref>{{harvnb|Lang|2002|loc=Ch X, Proposition 2.10}}</ref><math display="block">\sqrt{\operatorname{ann}_R(M)} = \bigcap_{\mathfrak{p} \,\in\, \operatorname{supp}M} \mathfrak{p} = \bigcap_{\mathfrak{p} \,\in\, \operatorname{ass}M} \mathfrak{p}</math> where <math>\operatorname{supp}M</math> is the [[support of a module|support]] of <math>M</math> and <math>\operatorname{ass}M</math> is the set of [[associated prime]]s of <math>M</math>.

==Applications==
The primary motivation in studying radicals is [[Hilbert's Nullstellensatz]] in [[commutative algebra]]. One version of this celebrated theorem states that for any ideal <math>J</math> in the [[polynomial ring]] <math>\mathbb{k}[x_1, x_2, \ldots, x_n]</math> over an [[algebraically closed field]] <math>\mathbb{k}</math>, one has
:<math>\operatorname{I}(\operatorname{V}(J)) = \sqrt{J}</math>
where
:<math>\operatorname{V}(J) = \left\{x \in \mathbb{k}^n \mid f(x)=0 \mbox{ for all } f \in J\right\}</math>
and
:<math>\operatorname{I}(V) = \{f \in \mathbb{k}[x_1, x_2,\ldots x_n] \mid f(x)=0 \mbox{ for all } x \in V \}.</math>
Geometrically, this says that if a [[algebraic variety|variety]] <math>V</math> is cut out by the [[polynomial equation]]s <math>f_1=0,\ldots,f_r=0</math>, then the only other polynomials that vanish on <math>V</math> are those in the radical of the ideal <math>(f_1,\ldots,f_r)</math>.

Another way of putting it: the composition <math>\operatorname{I}(\operatorname{V}(-))=\sqrt{-}</math> is a [[closure operator]] on the set of ideals of a ring.

==See also==
* [[Jacobson radical]]
* [[Nilradical of a ring]]
* [[Real radical]]

==Notes==
{{reflist|group=Note}}

==Citations==
{{reflist}}

== References ==
*{{Cite book|last1=Atiyah|first1=Michael Francis|author1-link=Michael Atiyah|year=1994|publisher=[[Addison-Wesley]]|first2=Ian G. |last2=Macdonald|author2-link=Ian G. Macdonald|title=Introduction to Commutative Algebra|isbn=0-201-40751-5|location=Reading, MA}}
*{{Cite book | last=Eisenbud|first= David |author-link=David Eisenbud |title=Commutative algebra with a view toward algebraic geometry | location=New York | publisher=[[Springer Science+Business Media|Springer-Verlag]] | series=[[Graduate Texts in Mathematics]] | volume=150 | year=1995 | mr=1322960 | isbn=0-387-94268-8 |ref=none}}
* {{Lang Algebra|edition=3r}}

[[Category:Ideals (ring theory)]]
[[Category:Closure operators]]

Radical of an ideal - Revision history

imported>Cutx64 at 03:27, 18 October 2025