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 $\sqrt{I}$, is defined as

$\sqrt{I}=\{r\in R \mid r^n\in I\ \hbox{for some positive integer}\ n\}.$

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 $R/I$.[1] The latter shows $\sqrt{I}$ is an ideal itself, containing I.

If the radical of I is finitely generated, then some power of $\sqrt{I}$ 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.

1. The radical of the ideal 4Z of integer multiples of 4 is 2Z.
2. The radical of 5Z is 5Z.
3. The radical of 12Z is 6Z.
4. In general, the radical of mZ is rZ, 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, $\sqrt{I^n} = \sqrt{I}$. A prime ideal is a radical ideal. So $\sqrt{P} = P$ for any prime ideal P.

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

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

$\sqrt{\operatorname{ann}_R(M)} = \bigcap_{\mathfrak{p} \in \operatorname{supp}M} \mathfrak{p} = \bigcap_{\mathfrak{p} \in \operatorname{ass}M} \mathfrak{p}$[5]

where $\operatorname{supp}M$ is the support of M and $\operatorname{ass}M$ 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:

• 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 {rn|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.
• 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 $x_1, x_2, \ldots, x_n$ over the field k, one has

$\operatorname{I}(\operatorname{V}(J)) = \operatorname{Rad} (J)\,$

where

$\operatorname{V}(J) = \{x \in k^n \ |\ f(x)=0 \mbox{ for all } f\in J\}$

and

$\operatorname{I}(S) = \{f \in k[x_1,x_2,\ldots x_n] \ |\ f(x)=0 \mbox{ for all } x\in S \}.$

Another way of putting it: The composition $\operatorname{I}(\operatorname{V}(-))=\operatorname{Rad}(-)\,$ 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.

## Notes

1. ^ 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, an and bm 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):
$(a+b)^{n+m-1}=\sum_{i=0}^{n+m-1}{n+m-1\choose i}a^ib^{n+m-1-i}.$
For each i, exactly one of the following conditions will hold:
• in
• n + m − 1 − im.
This says that in each expression aibn+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 an in I and an arbitrary element rR. Then, (ra)n = rnan is in I, so ra is in the radical. Thus the radical is an ideal.
2. ^ Atiyah–MacDonald 1969, Proposition 7.14
3. ^ Atiyah–MacDonald 1969, Proposition 4.2
4. ^ Proof: $R = \sqrt{\sqrt{I} + \sqrt{J} } = \sqrt{I + J}$ implies $I + J = R$.
5. ^ Lang 2002, Ch X, Proposition 2.10