We have a simple proposition saying that
Proposition: Suppose $R$ is a UFD and let $Q(R)$ be the fraction field of $R$. Let $f(x)=\sum_{i=0}^n a_i x_i \in R[x]$ be a polynomial of degree $n\ge 1$. Suppose $\alpha\in Q(R)$ is a root of $f(x)$ and write $\alpha=c/d$ with $c,d\in R$ coprime. Then $c$ divides $a_0$ and $d$ divides $a_n$.
Here is a simple proof from our lecture notes:
Proof: Note that $0=f(c/d)=a_n (c/d)^n+ \cdots + a_1(c/d)+a_0$.
$\color{darkred}{\text{Multiplying the above equation by $d^n$, we get}}$ $$0=a_nc^n+a_{n-1}c^{n-1}d+\cdots + a_1cd^{n-1}+a_0 d^n. \quad (*)$$
Since $c|$ LHS, we have $c|$ RHS, which implies $c|a_0d^n$. Since $c$ and $d$ are coprime, we have $c|a_0$. Similarly, $(*)$ implies that $d|a_n$. $\tag*{$\square$}$
I have two questions.
Question $1$: If $R$ is a subring of an integral domain $U$, we say $c\in U$ is a root of $f(x)\in R[x]$ if $f(c)=0$. In the proposition above, $R$ is not even a subring of $Q(R)$. Why can we say some element of $Q(R)$ is a root of a polynomial in $R[x]?$
Question $2$: Even though we can regard $Q(R)$ as a left $R$-module and calculate $(c^n/d^n)*d^n=c^nd^n/d^n=c^n/1$ in the line in $\color{darkred}{\text{dark red}}$ above, we still cannot have the equality $(*)$ because $(c^n/1)\neq c^n$, where $1$ is the multiplicative identity in $R$. How do we get $(*)$?
Remark: I guess my professor has abused notations a lot. I just want to understand the places where he abused notations, i.e. the two questions above. Thanks for help.
The key point is that the fraction field of an integral domain $R$ is a field with an injective homomorphism $i\colon R\to Q(R)$ given by $i(r)=\frac{r}{1}$. Therefore, $i(R)\subseteq Q(R)$ is a subring isomorphic to $R$, and for this reason we can actually talk of $Q(R)$ as a field containing $R$ or, more precisely, the smallest field containing (an isomorphic copy of) $R$. Hence, the only slight abuse of notation is due to the identification of $R$ with its isomorphic copy in $Q(R)$.
For example, in a very rigorous sense $\mathbb{Z}$ and $\mathbb{Q}$ do not have anything to do with each other (elements of $\mathbb{Q}$ are fractions, and from a very rigid point of view the integer $n$ and the fraction $\frac{n}{1}$ are different things). However, if you consider the subring of $\mathbb{Q}$ given by all the elements $\frac{n}{1}$ with $n\in\mathbb{Z}$, it is easily seen that it is isomorphic to $\mathbb{Z}$, and therefore we can harmlessly say that $\mathbb{Z}$ is a subring of $\mathbb{Q}$.