# Radical of an ideal

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In commutative ring theory, a branch of mathematics, the **radical of an ideal** *I* is an 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 of an operation on ideals called 'radicalization'). The radical of a primary ideal is prime.

Radical ideals defined here are generalized to noncommutative rings in the Semiprime ring article.

## Definition

The **radical** of an ideal *I* in a commutative ring *R*, denoted by Rad(*I*) or , is defined as

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) in .^{[1]} The latter shows is an ideal itself, containing *I*.

If the radical of *I* is finitely generated, then some power of is contained in *I*.^{[2]} 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*.

If an ideal *I* coincides with its own radical, then *I* is called a *radical ideal* or *semiprime ideal*.

## Examples

Consider the ring **Z** of integers.

- The radical of the ideal 4
**Z**of integer multiples of 4 is 2**Z**. - The radical of 5
**Z**is 5**Z**. - The radical of 12
**Z**is 6**Z**. - 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.

The radical of a primary ideal is prime. If the radical of an ideal *I* is maximal, then *I* is primary.^{[3]}

If *I* is an ideal, . A prime ideal is a radical ideal. So for any prime ideal *P*.

Let *I*, *J* be ideals of a ring *R*. If are comaximal, then are comaximal.^{[4]}

Let *M* be a finitely generated module over a noetherian ring *R*. Then

where is the support of *M* and is the set of associated primes of *M*.

## Properties

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

- It is always true that Rad(Rad(
*I*))=Rad(*I*). Moreover, Rad(*I*) is the smallest radical ideal containing*I*.

- 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*|^{n}*n*is a nonnegative integer}. By the definition of Rad(*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*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 among those containing*I*.

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

- An ideal
*I*in a ring*R*is radical if and only if the quotient ring*R/I*is reduced.

- The radical of a homogeneous ideal is homogeneous.

## Applications

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 over the field *k*, one has

where

and

Another way of putting it: The composition 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.

## See also

## Notes

- ↑ A direct proof can be give as follows:
Let
*a*and*b*be in the radical of an ideal*I*. Then, for some positive integers*m*and*n*,*a*^{n}and*b*^{m}are in*I*. We will show that*a*+*b*is in the radical of*I*. Use the binomial theorem to expand (*a*+*b*)^{n+m−1}(with commutativity assumed): For each*i*, exactly one of the following conditions will hold:*i*≥*n**n*+*m*− 1 −*i*≥*m*.

*a*^{i}*b*^{n+m− 1 − i}, 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*)^{n+m−1}is in*I*, therefore*a*+*b*is in the radical of*I*. To finish checking that the radical is an ideal, we take an element*a*in the radical, with*a*^{n}in*I*and an arbitrary element*r*∈*R*. Then, (*ra*)^{n}=*r*^{n}*a*^{n}is in*I*, so*ra*is in the radical. Thus the radical is an ideal. - ↑ Template:Harvnb
- ↑ Template:Harvnb
- ↑ Proof: implies .
- ↑ Template:Harvnb

## References

- M. Atiyah, I.G. Macdonald,
*Introduction to Commutative Algebra*, Addison-Wesley, 1994. ISBN 0-201-40751-5 - Eisenbud, David,
*Commutative Algebra with a View Toward Algebraic Geometry*, Graduate Texts in Mathematics, 150, Springer-Verlag, 1995, ISBN 0-387-94268-8. - Template:Lang Algebra