Purchase Solution

Ideals : Show that N is contained in P for each prime ideal, P of a commutative ring R.

Not what you're looking for?

Ask Custom Question

Show that N is contained in P for each prime ideal, P of a commutative ring R.

Where N is the set of all nilpotent elements. "a" is nilpotent if a^n=0 for some positive integer n. N itself is an ideal.

Purchase this Solution
Solution Summary

It is shown that N is contained in P for each prime ideal, P of a commutative ring R.

Solution Preview

Proof:

N is the set of all nilpotent elements. P is a prime ideal in a commutative ring R. We want to show ...

Purchase this Solution
Free BrainMass Quizzes
Multiplying Complex Numbers

This is a short quiz to check your understanding of multiplication of complex numbers in rectangular form.

Know Your Linear Equations

Each question is a choice-summary multiple choice question that will present you with a linear equation and then make 4 statements about that equation. You must determine which of the 4 statements are true (if any) in regards to the equation.

Solving quadratic inequalities

This quiz test you on how well you are familiar with solving quadratic inequalities.

Geometry - Real Life Application Problems

Understanding of how geometry applies to in real-world contexts

Graphs and Functions

This quiz helps you easily identify a function and test your understanding of ranges, domains , function inverses and transformations.

View More Free Quizzes
Related BrainMass Solutions

  • Homomorphism of Commutative Rings, Prime Ideals, Maximal Ideals

    Then if P is a prime ideal of S, we want to show that PR is either R or a prime ideal of R. We note that R is a commutative ring, then each element in PR has the form pr, where p is in P and r is in R.

  • Ideals and Factor Rings : Prime Ideal

    17287 Ideals and Factor Rings: Prime Ideal Problem: Let R be a commutative ring. Show that every maximal ideal of R is prime and if R is finite, show that every prime ideal is maximal. ALSO is every prime ideal of Z(integers) maximal? Why?

  • Problems Pertaining to Group Rings

    Let P be a minimal prime ideal of Z[G] such that P n Z = {0}. Suppose P is not contained in M. Then there exists an element r = sum_{g in G}(a_g G) of P such that sum_{g in G}(a_g) is odd. Now consider 1 + r = 1 + a_e + sum_{g in G{e}}(a_g).

  • Ring and ideal proofs

    231958 Ring and ideal proofs ** Please see the attached file for the complete problem description ** Let R be a ring....Prove that radical I is an ideal.....Prove that there is a prime ideal P containing I....

  • Principal and Prime Ideals

    227852 Principal and Prime Ideals Let R be a commutative ring with 1. Show that the principal ideal generated by x in the polynomial ring R[x] is a prime ideal if and only if R is an integral domain.

  • Ring theory

    show that for z in E, p(z)=0 implies p(q(z))=0 2 If R is a commutative ring of prime characteristic p, show the function f:R-->R, f(r)=r^p is a ring homomorphism 3.

  • maximal ideal

    I is a maximal ideal in a finite commutative ring R if and only if |R/I| is a prime number. Because I is also a normal subgroup of R.

  • Maximal ideals

    247559 Maximal ideals With respect to the ideal I=<2,x> in Z[x] I believe this ideal is maximal because one theorem I have read suggests to me that all maximal ideals of Z[x] are in the form <p, f(x)> where p is prime and f(x) is an element of

  • Proving Local Rings

    Therefore, M contains all the non-units of R. (⟸) Suppose that R is a ring in which all the non-units form an ideal M. Then every proper ideal of R is contained in M since proper ideals cannot contain units.

View More Related Solutions