A question on localization

Question

Let $D$ be an integral domain , let $S$ be a multiplicatively closed subset of $D$ not containing zero , then w.l.o.g. by natural embeddings , we can think of $D$ as a subring of $S^{-1} D$ which is further a subring of the fraction field of $D$ . Now if for every $q \in S^{-1} D , \exists $ monic $P_q(X) \in D[X] $ such that $P_q(q)=0$ , then is it true that every element of $S$ is a unit of $D$ ?

Answer 1

Suppose $P_q(X)=X^n+d_{n-1}X^{n-1}+...+d_0$ with $d_j\in D$ and for any $s\in S$ let $q=s^{-1}\in S^{-1}D$, then $$P_q(q)=s^{-n}+d_{n-1}s^{1-n}+...+d_0=0$$ after multplying the whole equation by $s^n$ yields $$1+d_{n-1}s+...+d_0s^n=0$$ and thus $$1=s\underbrace{(-d_{n-1}-d_{n-2}s-...-d_0s^{n-1})}_{\in D}$$ shows that $S$ consists of units in $D$.

Answer 2

Yes, this is true and the proof is very standard. Assume some $\frac{1}{s}$ admits a monic equation over $D$ and after some intuitive algebraic manipulations you will see that $s$ was a unit in $D$.