yes it is. it suffices to prove that any non-zero element in R is a unit, since R is already known to be a commutative ring with identity (unity). moreover, we only have to prove that non-zero elements of R-K are units, since K is known to be a field (and hence any non-zero element of K has a multiplicative inverse in K).

so pick a in R - K. and suppose that dim_{K}(R) = n.

consider the set {1,a,a^{2},...,a^{n}}. this has n+1 elements, so must be a linearly dependent set.

therefore there are c_{0},c_{1},...,c_{n}in K, not all 0, such that:

c_{0}+c_{1}a+...+c_{n}a^{n}= 0.

hence the ideal in K[x], J = {f in K[x]: f(a) = 0} is non-zero. since K is a field, K[x] is a principal ideal domain, thus:

J = (m(x)), for some polynomial m(x) in K[x].

i claim m(x) is irreducible over K. for suppose that m(x) = h(x)k(x) where deg(h),deg(k) < deg(m) (so that neither h(x), nor k(x) is a unit).

then we have 0 = m(a) = h(a)k(a), since R is an integral domain. suppose h(a) = 0 (the proof if k(a) = 0 is similar).

then (m(x)) is contained in (h(x)), and this containment is strict, because h(x) is not in (m(x)):

if h(x) = r(x)m(x) = r(x)h(x)k(x), then (since K[x] is also an integral domain, since K is a field, and thus an integral domain)

1 = r(x)k(x), contradicting the fact that k(x) is not a unit. but, by definition, h(x) is in J, since h(a) = 0.

since h(x) cannot be both in J and not in J, this contradiction shows m(x) must be irreducible.

write m(x) = d_{0}+ d_{1}x +...+ d_{r}x^{r}.

i claim d_{0}≠ 0, since m(x) is irreducible (otherwise x divides m(x)).

since m(a) = 0, we have d_{0}+ d_{1}a +...+ d_{r}a^{r}= 0.

thus 1 = -(d_{1}/d_{0}+ (d_{2}/d_{0})a +...+ (d_{r}/d_{0})a^{r-1})a,

which shows that a is a unit, hence R is a field.