Explore BrainMass

Principal ideal proof

An ideal I of a commutative ring R is said to be principal if it is generated by a single element, that is, if I is of the form
{ra|r element of R} for some fixed a element of R. Notice that Corollary 6.6 (below) shows that ideals of F[x] of the form I_F,a are principal. Now prove that every ideal of F[x], where F is any field, is principal. (Hint: If I = {0}, then we are done - why? Otherwise, there must be some nonzero element in I of least degree, call it q. Given an arbitrary element f is a member of I, write f = bq +r (division algorithm). Use the definition of an ideal to show that r must be in I. Conclude from the choice of q and r that r must be zero. This shows that I is a subset of{rq|r is an element of F[x]}. Why does the reverse inclusion also hold?)

Corollary 6.6 - If p(x) is any nonzero element of I_F,a of least degree,then I_F,a = {f(x)p(x) |f(x) is an element F[x]}.

Solution Preview

Let I be an ideal of F[x].

If I = 0, then I = <0> is certainly principal.

Suppose then I =/= {0} and let g(x) be a nonzero element of I of minimal degree. If the degree of g(x) is 0, then ...

Solution Summary

This is a proof regarding principal ideals.