Radical of an ideal
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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').
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Definition
The radical of an ideal I in a commutative ring R, denoted by Rad(I) or √I, is defined as
- <math>\hbox{Rad}(I)=\{r\in R|r^n\in I\ \hbox{for some positive integer}\ n\}.</math>
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.
Examples
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.
Proof that the radical is an ideal
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 I. Use the binomial theorem to expand (a+b)n+m−1:
- <math>(a+b)^{n+m-1}=\sum_{i=0}^{n+m-1}{n+m-1\choose i}a^ib^{n+m-1-i}.</math>
For each i, exactly one of the following conditions will hold:
- i ≥ n
- n+m-1-i ≥ m.
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 r∈R. Then, (ra)n = rnan is in I, so ra is in the radical. Thus 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)). As can be easily seen, the nilradical of R is just the radical of the zero ideal (0). This brings about 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.
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, rn is zero in R/I, and hence rn is in I. Second, if rn is in I for some n, then the image of rn in R/I is zero, and hence rn 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.
Jacobson radicals
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.
Properties
- 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 {rn|n is a nonnegative integer}. S is multiplicatively closed, so we can form the localization S-1R. Form the quotient S-1R/S-1I. By Zorn's lemma we can choose a maximal ideal P in this ring. The preimage of P under the maps R→S-1R→S-1R/S-1I 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.
Applications
The primary motivation in studying radicals is the celebrated Hilbert's Nullstellensatz in commutative algebra. A nice, easy to understand version of this theorem states that for an algebraically closed field k, and for any finitely generated polynomial ideal J of it, one has
- <math>I(\hbox{V}(J)) = \hbox{Rad} (J)\,</math>
where
- <math> \hbox{V}(J) = \{x \in k | f(x)=0 \mbox{ for all } f\in J\}</math>
and
- <math> I(S) = \{f \in k[x_1,x_2,\ldots x_n] | f(x)=0 \mbox{ for all } x\in S \}.</math>de:Radikal (Mathematik)