Thread: 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.

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

c₀ + c₁a +...+ c_naⁿ = 0.

if p(x) = c_nxⁿ +...+ 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) = x^r + d_r-1x^r-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-...- d_r-1a^r-1 - a^r, so

1 = (-d₁/d₀ - (d₂/d₀)a -...- (d_r-1/d₀)a^r-1 - (1/d₀)a^r-1)a

which shows that every non-zero a in R is a unit.

oh! very thanks