Let $\phi:R\rightarrow R'$ be an injective, finite map of integral domains. Is it true that induced map $\phi_1:\operatorname{Frac}R \rightarrow \operatorname{Frac}R'$ is also finite?
Note:
Finite in the sense of $R$-modules.
$\operatorname{Frac}R:=(R\setminus\{0\})^{-1}R$
My attempt:
I was trying to show that if $s_1,...,s_n$ generate $R'$ then $s_1,...s_n,1/s_1,...,1/s_n$ generate $\operatorname{Frac}R'$, but I couldn't express $\dfrac{r_1s_1+\cdots+r_ns_n}{t_1s_1+\cdots+t_ns_n}$ s.t. $r_i,t_i \in R$ with the generators above. I also doubt that the statement is correct, but I don't have much examples in my mind of finite injective maps between integral domains where the $R,R'$ aren't fields, so I can't come up with an counterexample.
Let $\frac{r}{s} \in \operatorname{Frac}(R')$. Since $\varphi$ is finite, it is integral, and so $$ s^n + a_{n-1} s^{n-1} + \dots + a_1 s + a_0 = 0 $$ for some $a_0,\dots, a_{n-1} \in R$. Since $R'$ is an integral domain, we have $a_0 \neq 0$ and so in $\operatorname{Frac}(R')$ we have $$ s^{-1} = \frac{-1}{a_0} (s^{n-1}+ a_{n-1}s^{n-2} + \dots + a_1 ) $$ and consequently $$ \frac{r}{s} = \frac{-1}{a_0} (rs^{n-1}+ a_{n-1}rs^{n-2} + \dots + a_1 r ) $$ lies in the $\operatorname{Frac}(R)$-span of $R'$. Thus, if $s_1, \dots, s_m$ generate $R'$ as $R$-module, they generate $\operatorname{Frac}(R')$ as $\operatorname{Frac}(R)$-module.