# $S$ is a commutative integral domain and a finitely generated $R$-Module where $R$ is a subring of $S$. $R$ is a field iff $S$ is a field

Assume $$S$$ is a commutative integral domain, and $$R \subseteq S$$ is a subring. Assume $$S$$ is finitely generated as an $$R$$-module, i.e., there exist elements $$s_1, \ldots, s_n \in S$$ such that $$S = s_1R + s_2R + \cdots + s_nR$$.

(a) Show that $$R$$ is a field if and only if $$S$$ is a field.

(b) Is the statement true if the assumption that $$S$$ is an integral domain is dropped?

I haven't been able to do either direction.

Starting with the forward direction I assume $$R$$ is a field and consider a non-zero element $$s ∈ S$$ and write $$s = s_1r_1 + · · · + s_nr_n$$ then subtract the sum from both sides to get $$s - s_1r_1 + · · · + s_nr_n = 0$$. I want to try to use the integral domain fact by somehow writing the left side as a product like so: $$s(1 - s^{-1}(s_1r_1 + · · · + s_nr_n)) = 0$$ and since s is non-zero $$1 - s^{-1}(s_1r_1 + · · · + s_nr_n) = 0$$ and therefore $$s^{-1}s = 1$$ but there are obvious problems with that argument. Any help or hints would be appreciated. This is not a homework problem I am studying for a qual.