Explore BrainMass

The Problems in Ring Theory

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

When f1, ....fn are elements in the polynomial ring R[x, y], we denote by (f1,....,fn) the set { ∑gifi | g ∈ R[x, y]} which is an ideal of R[x, y]

Let R be a commutative ring and let I ⊆ R be an ideal. Show that √I := { f ∈ R | there exists n ∈ N such that fn ∈ I } is an ideal of R.

Let V(I) be defined as the set of all points (p, q) ∈R2 such that f(p, q) = 0 for every f ∈ I. Set V = V(I). Let I(V) be defined as the ideal of all the polynomials g ∈R[x, y] such that g(p, q) = 0 for all the points (p, q)∈ V. Let I = { f1, ....fn} be an ideal in the ring of polynomials in two variables R[x,y]. Show that √I ⊆ I(V).

Prove that the ideal I = (2x2 + 3y2 - 11, x2 - y2 - 3) and the ideal J = (x2 - 4, y2 - 1) are the same. Use this fact to find V(I).

Let I be an ideal of a commutative ring R. Show that if I + a = I + a1 and I + b = I + b1 then I + ab = I + a1b1.

© BrainMass Inc. brainmass.com March 21, 2019, 9:02 pm ad1c9bdddf

Solution Preview

1. Let R be a commutative ring and let be an ideal. Show that is an ideal of R.

Proof: First, let's show that is a subring of R.
Since and
Suppose and Then and such that and for some positive integers m and n. It follows that and since I is an ideal of R. Then (if m is even) or (if m is odd). So and by the same argument.

Now let's show that Then since R is commutative, we can use the Binomial Theorem to expand Thus we have

Now if we have and so It follows
where and
Since I is the ideal of R,

Now if ...

Solution Summary

Problems in ring theory are noted.