Deming regression

28 year-old Painting Investments Worker Truman from Regina, usually spends time with pastimes for instance interior design, property developers in new launch ec Singapore and writing. Last month just traveled to City of the Renaissance. 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

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

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 $R/I$. 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. 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.

If I is an ideal, $\sqrt{I^n}=\sqrt{I}$. A prime ideal is a radical ideal. So $\sqrt{P^n}=P$ 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.

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

$\sqrt{{\mathrm{ann}}_R(M)}=\bigcap _{\mathfrak{p}\in \mathrm{supp}M}\mathfrak{p}=\bigcap _{\mathfrak{p}\in \mathrm{ass}M}\mathfrak{p}$

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

• 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 {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.

• 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 $x_1,x_2,\dots ,x_n$ over the field k, one has

$\mathrm{I}(\mathrm{V}(J))=\mathrm{Rad}(J)$

where

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

and

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

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

See also

• Jacobson radical
• Nilradical of a ring

Notes

43 year old Petroleum Engineer Harry from Deep River, usually spends time with hobbies and interests like renting movies, property developers in singapore new condominium and vehicle racing. Constantly enjoys going to destinations like Camino Real de Tierra Adentro.

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