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.

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- May 13th 2012, 06:12 AMsaravananbsField
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. - May 13th 2012, 12:33 PMDevenoRe: Field
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. - May 13th 2012, 02:27 PMsaravananbsRe: Field
oh! very thanks