Explore BrainMass

Ring proofs

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

a. show that every subfield of complex numbers contains rational numbers
b. show that the prime field of real numbers is rational numbers
c. show that the prime field of complex numbers is rational numbers

a. Let R be a domain. Prove that the polynomial f(x) is a unit in R[x] if and only if f(x) is a nonzero constant which is a unit in R.

b. Show that ([2]x + [1])^2 = [1] in (integers modulo 4)[x] Conclude that the statement in part (a) may be false for the commutative rings that are not domains. [ An element z element of R is called a nilpoint if z^m = 0 for some integer m greater than or equal to one. For any commutative ring R, it can be proved that a polynomial f(x) = a(sub 0) + a(sub 1)x +...+a(sub n)x^n element R[x] is a unit in R[x] if and only if a(sub 0) is a unit in R and a(sub i) is nilpoint for all I greater than or equal to 1.]

© BrainMass Inc. brainmass.com October 25, 2018, 12:09 am ad1c9bdddf

Solution Preview

(1) By definition, the prime subfield of a field F is smallest subfield of F containing 1; in other words, the prime subfield is contained in every subfield of F containing 1. Moreover, if F is a ...

Solution Summary

This provides examples of proofs regarding subfields and prime fields, polynomial in a domain, and commutative rings.

See Also This Related BrainMass Solution

Finite Ring Proofs

Please see the attached pdf. I need a detailed, rigorous proof of this with explanation of the steps so I can learn.

Let R be a finite ring.
a. Prove that there are positive integers m and n with m>n such that x^m for ever x E R.
b. Give a direct proof (i.e. without appealing to part c) that if R is an integral domain, then it is a field
c. Suppose that R has identity, prove that if x E R is not a zero divisor, then it is a unit

View Full Posting Details