# Field

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• May 13th 2012, 06:12 AM
saravananbs
Field
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 PM
Deveno
Re: 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 dimK(R) = n, and consider the n+1 elements:

1,a,a2,...,an. these must be linearly dependent over K so we have c0,c1,....,cn not all 0 with:

c0 + c1a +...+ cnan = 0.

if p(x) = cnxn +...+ c1x + c0, 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) = xr + dr-1xr-1 +...+ d1x + d0, with r ≤ n.

note that we cannot have d0 = 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 d0 = -d1a - d2a2-...- dr-1ar-1 - ar, so

1 = (-d1/d0 - (d2/d0)a -...- (dr-1/d0)ar-1 - (1/d0)ar-1)a

which shows that every non-zero a in R is a unit.
• May 13th 2012, 02:27 PM
saravananbs
Re: Field
oh! very thanks