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In ring theory, a branch of mathematics, the radical of an ideal is a kind of completion of the ideal. There are several special radicals associated with the entire ring - such as the nilradical and the Jacobson radical, which isolate certain "bad" properties of the ring. A radical 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').
## Definition

## Examples

## Proof that the radical is an ideal

## The nilradical of a ring

Consider the set of all nilpotent elements of R, which will be called the nilradical of R (and will be denoted by N(R)). One can easily see that the nilradical of R is just the radical of the zero ideal (0). This permits an alternative definition for the (general) radical of an ideal I in R. Define Rad(I) as the preimage of N(R/I), the nilradical of R/I, under the projection map R→R/I. ## Jacobson radicals

## Properties

## Applications

## References

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

- $hbox\{Rad\}(I)=\{rin\; R|r^nin\; I\; hbox\{for\; some\; positive\; integer\}\; n\}.$

Intuitively, one can think of the radical of I as obtained by taking all the possible roots of elements of I. Rad(I) turns out to be an ideal itself, containing I.

Consider the ring Z of integers.

- The radical of the ideal 4Z of integers multiple of 4 is 2Z.
- The radical of 5Z is 5Z.
- The radical of 12Z is 6Z.

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):

- $(a+b)^\{n+m-1\}=sum\_\{i=0\}^\{n+m-1\}\{n+m-1choose\; i\}a^ib^\{n+m-1-i\}.$

For each i, exactly one of the following conditions will hold:

- i ≥ n
- n + m − 1 − i ≥ m.

This says that in each expression 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.

To see that the two definitions for the radical of I are equivalent, note first that if r is in the preimage of √(R/I), then for some n, r^{ n} is zero in R/I, and hence r^{ n} is in I. Second, if r^{ n} is in I for some n, then the image of r^{ n} in R/I is zero, and hence r^{ n} is in the preimage of √(R/I).

This alternative definition can be very useful, as we shall see right below. See #Properties below for another characterization of the nilradical.

Let R be any ring, not necessarily commutative. The Jacobson radical of R is the intersection of the annihilators of all simple right R-modules.

There are several equivalent characterizations of the Jacobson radical, such as:

- J(R) is the intersection of the regular maximal right (or left) ideals of R.
- J(R) is the intersection of all the right (or left) primitive ideals of R.
- J(R) is the maximal right (or left) quasi-regular right (resp. left) ideal of R.

As with the nilradical, we can extend this definition to arbitrary two-sided ideals I by defining J(I) to be the preimage of J(R/I) under the projection map R→R/I.

If R is commutative, the Jacobson radical always contains the nilradical. If the ring R is a finitely generated Z-algebra, then the nilradical is equal to the Jacobson radical, and more generally: the radical of any ideal I will always be equal to the intersection of all the maximal ideals of R that contain I. This says that R is a Jacobson ring.

- If P is a prime ideal, then R/P is an integral domain, so it cannot have zero divisors, and in particular it cannot have nonzero nilpotents. Hence, the nilradical of R/P is {0}, and its preimage, being P, is a radical ideal.
- By using localization, we can see that Rad(I) is the intersection of all the prime ideals of R that contain I: Every prime ideal is radical, so the intersection J of the prime ideals containing I contains Rad(I). If r is an element of R which is not in Rad(I), then we let S be the set {r
^{n}|n is a nonnegative integer}. S is multiplicatively closed, so we can form the localization S^{-1}R. Form the quotient S^{-1}R/S^{-1}I. By Zorn's lemma we can choose a maximal ideal P in this ring. The preimage of P under the maps R→S^{-1}R→S^{-1}R/S^{-1}I is a prime ideal which contains I and does not meet S; in particular, it does not meet r, so r is not in J. - In particular, the nilradical is equal to the intersection of all prime ideals containing the 0 ideal, but all ideals must contain 0 so the nilradical can alternatively be defined as the intersection of the prime ideals.

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

- $I(hbox\{V\}(J))\; =\; hbox\{Rad\}\; (J),$

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

- $I(S)\; =\; \{f\; in\; k[x\_1,x\_2,ldots\; x\_n]\; |\; f(x)=0\; mbox\{\; for\; all\; \}\; xin\; S\; \}.$

Another way of putting it: The composition I V is a closure operator, and by identifying the closure as the radical we have established that taking the radical is an idempotent operation.

- Eisenbud, David, Commutative Algebra with a View Toward Algebraic Geometry, Graduate Texts in Mathematics, 150, Springer-Verlag, 1995, ISBN 0-387-94268-8.

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Last updated on Monday September 29, 2008 at 20:21:37 PDT (GMT -0700)

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