Radical of an ideal explained

of a commutative ring is another ideal defined by the property that an element

is in the radical if and only if some power of

is in

. 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 rings in the semiprime ring article.

Definition

The radical of an ideal

in a commutative ring

, denoted by

\operatorname{rad}(I)

\sqrt{I}

, is defined as

\sqrt{I}=\left\{r\inR\midr^n\inI \hbox{forsome} n\in\Z⁺\right\},

(note that

I\subseteq\sqrt{I}

).Intuitively,

\sqrt{I}

is obtained by taking all roots of elements of

within the ring

. Equivalently,

\sqrt{I}

is the preimage of the ideal of nilpotent elements (the nilradical) of the quotient ring

R/I

(via the natural map

\pi\colonR\toR/I

). The latter proves that

\sqrt{I}

is an ideal.^[1]

If the radical of

is finitely generated, then some power of

\sqrt{I}

is contained in

. In particular, if

and

are ideals of a Noetherian ring, then

and

have the same radical if and only if

contains some power of

and

contains some power of

If an ideal

coincides with its own radical, then

is called a radical ideal or semiprime ideal.

Examples

Consider the ring

of integers.

1. The radical of the ideal

4\Z

of integer multiples of

2\Z

(the evens).

1. The radical of

5\Z

1. The radical of

12\Z

6\Z

1. In general, the radical of

m\Z

r\Z

, where

is the product of all distinct prime factors of

, the largest square-free factor of

(see Radical of an integer). In fact, this generalizes to an arbitrary ideal (see the Properties section).

Consider the ideal

I=\left(y^4\right)\subseteq\Complex[x,y]

. It is trivial to show

\sqrt{I}=(y)

(using the basic property but we give some alternative methods: The radical

\sqrt{I}

corresponds to the nilradical

\sqrt{0}

of the quotient ring

R=\Complex[x,y]/\left(y^4\right)

, which is the intersection of all prime ideals of the quotient ring. This is contained in the Jacobson radical, which is the intersection of all maximal ideals, which are the kernels of homomorphisms to fields. Any ring homomorphism

R\to\Complex

must have

in the kernel in order to have a well-defined homomorphism (if we said, for example, that the kernel should be

(x,y-1)

the composition of

\Complex[x,y]\toR\to\Complex

would be

\left(x,y^4,y-1\right)

, which is the same as trying to force

1=0

). Since

\Complex

is algebraically closed, every homomorphism

R\toF

must factor through

\Complex

, so we only have to compute the intersection of

\{\ker(\Phi):\Phi\in\operatorname{Hom}(R,\Complex)\}

to compute the radical of

(0).

We then find that

\sqrt{0}=(y)\subseteqR.

Properties

This section will continue the convention that I is an ideal of a commutative ring

It is always true that $\sqrt = \sqrt$ , i.e. radicalization is an idempotent operation. Moreover,

\sqrt{I}

is the smallest radical ideal containing

\sqrt{I}

is the intersection of all the prime ideals of

that contain

\sqrt=\bigcap_\mathfrak,

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
\sqrt{I}

. Suppose
r

is an element of
R

that is not in
\sqrt{I}

, and let
S

be the set
\left\{rⁿ\midn=0,1,2,\ldots\right\}

. By the definition of
\sqrt{I}

,
S

must be disjoint from
I

.
S

is also multiplicatively closed. Thus, by a variant of Krull's theorem, there exists a prime ideal
ak{p}

that contains
I

and is still disjoint from
S

(see Prime ideal). Since
ak{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

is the intersection of all prime ideals of

that are minimal among those containing

Specializing the last point, the nilradical (the set of all nilpotent elements) is equal to the intersection of all prime ideals of

^[2]

\sqrt = \mathfrak_R = \bigcap_\mathfrak.

This property is seen to be equivalent to the former via the natural map

\pi\colonR\toR/I

, which yields a bijection

\left\lbrace\textJ\mid R\supseteq J\supseteq I\right\rbrace\quad \quad\left\lbrace\textJ\mid J\subseteq R/I\right\rbrace,

defined by

u\colonJ\mapstoJ/I=\lbracer+I\midr\inJ\rbrace.

^[3] ^[4]

An ideal

in a ring

is radical if and only if the quotient ring

R/I

is reduced.

The radical of a homogeneous ideal is homogeneous.
The radical of an intersection of ideals is equal to the intersection of their radicals:

\sqrt{I\capJ}=\sqrt{I}\cap\sqrt{J}

The radical of a primary ideal is prime. If the radical of an ideal

is maximal, then

is primary.

is an ideal,

\sqrt{I^n}=\sqrt{I}

. Since prime ideals are radical ideals,

\sqrt{ak{p}^n}=ak{p}

for any prime ideal

ak{p}

I,J

be ideals of a ring

. If

\sqrt{I},\sqrt{J}

are comaximal, then

I,J

are comaximal.^[5]

be a finitely generated module over a Noetherian ring

. Then

\sqrt = \bigcap_ \mathfrak = \bigcap_ \mathfrak

where

\operatorname{supp}M

is the support of

and

\operatorname{ass}M

is the set of associated primes of

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

in the polynomial ring

k[x_1,x_2,\ldots,x_n]

over an algebraically closed field

, one has

\operatorname{I}(\operatorname{V}(J))=\sqrt{J}

where

\operatorname{V}(J)=\left\{x\inkⁿ\midf(x)=0forallf\inJ\right\}

and

\operatorname{I}(V)=\{f\ink[x_1,x_2,\ldotsx_n]\midf(x)=0forallx\inV\}.

Geometrically, this says that if a variety

is cut out by the polynomial equations

f_1=0,\ldots,f_r=0

, then the only other polynomials that vanish on

are those in the radical of the ideal

(f_1,\ldots,f_r)

Another way of putting it: the composition

\operatorname{I}(\operatorname{V}(-))=\sqrt{-}

is a closure operator on the set of ideals of a ring.

References

Book: Atiyah. Michael Francis. Michael Atiyah. 1994. Addison-Wesley. Ian G. . Macdonald. Ian G. Macdonald. Introduction to Commutative Algebra. 0-201-40751-5. Reading, MA.
Book: Eisenbud, David . David Eisenbud . Commutative algebra with a view toward algebraic geometry . New York . . . 150 . 1995 . 1322960 . 0-387-94268-8 . none.

Notes and References

Here is a direct proof that
\sqrt{I}

is an ideal. Start with
a,b\in\sqrt{I}

with some powers
a^n,b^m\inI

. To show that
a+b\in\sqrt{I}

, we use the binomial theorem (which holds for any commutative ring):
style(a+b)^n+m-1

n+m-1
=\sum
i=0

\binom{n+m-1}{i}a^ib^n+m-1-i.

For each

i

, we have either
i\geqn

or
n+m-1-i\geqm

. Thus, in each term
aⁱb^n+m-1-i

, one of the exponents will be large enough to make that factor lie in
I

. Since any element of
I

times an element of
R

lies in
I

(as
I

is an ideal), this term lies in
I

. Hence
(a+b)^n+m-1\inI

, and so
a+b\in\sqrt{I}

.
To finish checking that the radical is an ideal, take

a\in\sqrt{I}

with
a^n\inI

, and any
r\inR

. Then
(ra)^n=r^na^n\inI

, so
ra\in\sqrt{I}

. Thus the radical is an ideal.
For a direct proof, see also the characterisation of the nilradical of a ring.
Book: Aluffi, Paolo. Algebra: Chapter 0. AMS. 2009. 978-0-8218-4781-7. 142.
This fact is also known as fourth isomorphism theorem.
Proof: $R = \sqrt = \sqrt$ implies
I+J=R

.

	n+m-1
=\sum
	i=0

Radical of an ideal explained

Definition

Examples

Properties

Applications

See also

References

Notes and References