Got another one for y'all. I got part of the way through this, but I'm stuck for a way to continue.

Here's the question:

"Let K be a field, and let R be a subring of K. Suppose that every element of K satisfies a polynomial whose coefficients are in R and whose highest degree coefficient is 1. Show that R is actually a subfield of K."

The question goes on to give a couple of hints:

1) Show that R is an integral domain

2) Given a nonzero

we know that 1/a satisfies a suitable polynomial with coefficients in R. Manipulate this polynomial.

Here's what I've got so far:

Proof: R is an integral domain.

Assume R is a subring of the field K. Then R is an integral domain because any element of R is also an element of K, thus the subring R is commutative and contains no zero-divisors. (End of proof.)

For the second part, here's as far as I've gotten.

By hypothesis,

and

such that:

Now, since

and K is a field

. We may multiply the above equation by

:

or

.

Again, since

and K is a field

, by hypothesis:

;

where m does not need to be equal to n. We may multiply this equation by

:

.

Compare this with

.

I stated before that m was not necessarily equal to n, but these equations together show that the set of all possible values for m (given this a) contains the value n. So set m=n. Furthermore, R contains no zero-divisors, so the second equation is not trivial (that is, all of the coefficients are non-zero.) Finally, the leading coefficient is the same in both equations. Thus I conclude the polynomials are equal in the sense that ALL the coefficients of like terms are equal.

In particular, comparison of the constant term shows

where both

. Thus, by definition,

is the multiplicative inverse of

.

This is as far as I've gotten. I can show by this method that

*some* elements of R have inverses in R, but not all of them. Only those

values that come up in polynomials for given values of a will have inverses, not the whole set R.

Any thoughts to finish this, or am I beating down a dead end?

Thanks!

-Dan