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 $\displaystyle a \epsilon R$ 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, $\displaystyle \forall a\epsilon K$ $\displaystyle \exists \{r_i\} \subset R$ and $\displaystyle n \epsilon Z_{+} $ such that:

$\displaystyle a^n+r_{n-1}a^{n-1}+...+r_0=0$

Now, since $\displaystyle a \epsilon K$ and K is a field $\displaystyle \exists 1/a \epsilon K$. We may multiply the above equation by $\displaystyle (1/a)^n$:

$\displaystyle 1+r_{n-1}(1/a)+...+r_0(1/a)^n=0$

or

$\displaystyle r_0(1/a)^n+r_1(1/a)^{n-1}+...+1=0$.

Again, since $\displaystyle a \epsilon K$ and K is a field $\displaystyle \exists 1/a \epsilon K$, by hypothesis:

$\displaystyle (1/a)^m+q_{m-1}(1/a)^{m-1}+...+q_0=0$; $\displaystyle \{q_i\} \subset R$

where m does not need to be equal to n. We may multiply this equation by $\displaystyle r_0$:

$\displaystyle r_0(1/a)^m+r_0q_{m-1}(1/a)^{m-1}+...+r_0q_0=0$.

Compare this with

$\displaystyle r_0(1/a)^n+r_1(1/a)^{n-1}+...+1=0$.

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 $\displaystyle r_0q_0=1$ where both $\displaystyle r_0,q_0 \epsilon R$. Thus, by definition, $\displaystyle q_0$ is the multiplicative inverse of $\displaystyle r_0$.

This is as far as I've gotten. I can show by this method thatsomeelements of R have inverses in R, but not all of them. Only those $\displaystyle r_0$ 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