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## Radical Math Definition Example Essays

Is Tess in ‘Tess of the d'Urbervilles' portrayed as being responsible for her own demise? [pdf 40 KB]

Yours is a beautifully clear essay. You write very well, and your prose is delightful to read. You've also done your research and it shows. There is a remarkable lack of vagary about society or feminism in your piece, and you've picked canny quotes from your secondary sources that elucidate and situate your arguments.

You've also located some wonderfully specific quotations from your primary source to support your argument that Hardy's narrator sympathises with Tess. Some of your close readings are wonderfully astute, as when you point out that Tess implores Angel, rather than commanding him. Slightly less persuasive is your assertion that Tess is the victim of Alec's eyes; I suspect you might have found better quotations, descriptions, or incidents denouncing Alec's gaze.

You are clearly very good at pursuing and proving an argument. I encourage you to be a bit more experimental in your next essay; perhaps choose a less straightforward topic and see where it takes you.

Please see penciled notes throughout on shortening sentences and watching for comma splices (please look this term up in a style manual if it is unfamiliar).

For other radicals, see radical of a ring.

In commutative ring theory, a branch of mathematics, the radical of an idealI 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 ringR, 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 . The latter shows is an ideal itself, containing I.[Note 1]

If the radical of I is finitely generated, then some power of 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 quotient ring . Notice that any morphism must have in the kernel in order to have a well-defined morphism (if we said, for example, that the kernel should be the composition of would be which is the same as trying to force ). Since is algebraically closed, every morphism must factor through , so we only have the compute the intersection of to compute the radical of . We then find that .

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 distinct prime factors of m (each prime factor of m occurs exactly once as a factor of the product r) (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, . 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.

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. Proof: 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 ringR/I is reduced.
• The radical of a homogeneous ideal is homogeneous.

## Applications

The primary motivation in studying radicals is Hilbert's Nullstellensatz in commutative algebra. One version of this celebrated theorem states that for an algebraically closed fieldk, 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.

## 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.
• Lang, Serge (2002), Algebra, Graduate Texts in Mathematics, 211 (Revised third ed.), New York: Springer-Verlag, ISBN 978-0-387-95385-4, MR 1878556, Zbl 0984.00001
1. ^A direct proof can be given 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): For each i, exactly one of the following conditions will hold: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.