If R is commutative intergal domain. and K is a subring of R such that K is field. if R is considered as vector space over K, it is of finite dimension. prove that R is field.
can any one explain, please.
let a be in R, but not in K (if there is no such element, then R = K, and thus R is a field). suppose that dim_{K}(R) = n, and consider the n+1 elements:
1,a,a^{2},...,a^{n}. these must be linearly dependent over K so we have c_{0},c_{1},....,c_{n} not all 0 with:
c_{0} + c_{1}a +...+ c_{n}a^{n} = 0.
if p(x) = c_{n}x^{n} +...+ c_{1}x + c_{0}, we have p(a) = 0.
now p(x) is a non-zero polynomial with p(a) = 0, so let m(x) be a monic polynomial of least degree with m(a) = 0
(since there is at least one such polynomial with p(a) = 0, there must be one of minimal degree, which has non-zero leading coefficient, which we can therefore divide by to obtain a monic polynomial).
so me have m(a) = 0, for some polynomial m(x) = x^{r} + d_{r-1}x^{r-1} +...+ d_{1}x + d_{0}, with r ≤ n.
note that we cannot have d_{0} = 0, for if so, then m(x) factors as xq(x), and since a ≠ 0 (since a is not in K, and 0 is), then q(a) = 0, contradicting the minimality of the degree of m(x).
(this is where we use the fact that R is an integral domain. for if 0 = m(a) = aq(a), then either a = 0, or q(a) = 0, since an integral domain has no zero-divisors).
thus d_{0} = -d_{1}a - d_{2}a^{2}-...- d_{r-1}a^{r-1} - a^{r}, so
1 = (-d_{1}/d_{0} - (d_{2}/d_{0})a -...- (d_{r-1}/d_{0})a^{r-1} - (1/d_{0})a^{r-1})a
which shows that every non-zero a in R is a unit.