Extensions between integral domains give extensions of fields of the same degree. Assume that $S \subset R$ is a ring extension where, both $S$, $R$ are integral domains. Furthermore, assume that $R$ is a free $S$-module of rank $n$. Is it true that the extension of fields $\mathrm{Quot}(S) \subset \mathrm{Quot}(R)$ is of degree $n$? If yes, how do we prove that?

_Hint:_ Show that $\text{Quot}(R) \xleftarrow{\cong} R\otimes_S \text{Quot}(S)$; in words, it suffices to have all elements of $S\setminus\\{0\\}$ invertible to conclude that all elements of $R\setminus\\{0\\}$ are invertible. Use that $R$ is integral over $S$ for that. This reduces the question to the preservation of degree under base extension. Let me know if you need more details!