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 :
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?