TITLE: Finitely generated R-module is a field iff R is a field?
QUESTION [7 upvotes]: Not sure how to do this one. If $S$ is a field, then I was considering that $\exists r_1,\ldots, r_n\in R$ s.t. $1 = r_1s_1+\cdots+r_ns_n$ so for $r = rr_1s_1+\cdots+rr_ns_n$. Maybe that is somehow useful for taking inverses of elements.  
The assumption that $S$ is an integral domain is necessary because otherwise we could have $S = \mathbb{Z}_p[x]/f(x)$ where $f(x)$ is not irreducible. This is still a finitely generated $\mathbb{Z}_p$-module, but its not a field. 
Any hints or solutions would be much appreciated. I feel like this isn't that hard and I'm missing something simple
Source: https://dornsife.usc.edu/assets/sites/363/docs/F17_510ab.pdf

REPLY [3 votes]: Hints:
$\Rightarrow$: If $R$ is a field, let $s\in S$, and consider multiplication by $s$ in $S$. Check this is an injective $R$-linear map. What can you conclude, knowing $S$ is a finite dimensional $R$-vector space?
$\Leftarrow$: If $S$ is a field, consider $r\in R$; you know $r^{-1}\in S$, hence it is a root of a monic polynomial in $R[X]$. Deduce from this polynomial equation that  $r^{-1}$ is a polynomial in $r$, hence it belongs to $R$.