# Ring theory

1. If E, F are fields and F is a subring of E, show each q in Aut(E/F) permutes the roots in E of each nonzero p(x) in F[x]. Hint if p(x)=a0+a1x+a2x^2+. . . +anx^n then p(x) has at most n roots in E. 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. If a,b are positive integers with greatest common divisor d and least common multiple m, show aZ+bZ=dZ and aZ intersect bZ=mZ

4. For the cyclic group (Z,+) = <1> prove the ideal of Z is principal

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#### Solution Preview

1. Suppose, q is an automorphism of E/F, that is, q(f)=f for all f in F. Let p(x) be a polynomial in F[x], and z be an element of E. Then, p(z) is an element of E, and we can consider q(p(z)). Let p(x)=a0+a1x+a2x^2+. . . +anx^n. Then

q(p(z))=q(a0)+q(a1z)+q(a2z^2)+...+q(anz^n)=a0+a1q(z)+a2[q(z)]^2+...+an[q(z)]^n

Suppose further that p(z)=0. Then, we have

q(p(z))a0+a1q(z)+a2[q(z)]^2+...+an[q(z)]^n=q(0)=0

and so q(z) is again a root ot p(x). Since q is an automirphism, it must be one-to-one, and since p(x) has at most n roots, by the pigeonhole principle, q just permutes the roots of p(x).

2. In a ring of prime characteristic p we have ...

#### Solution Summary

This posting exemplifies Ring theory.