While I was searching about polynomial ring extensions over Prufer domains it came to me the following question:
Let $D$ be a Prufer domain and $M$ a maximal ideal of $D$. Then, is $D[X]_{M[X]}$ a valuation ring?
If we set $S=D\setminus M$, then clearly $M[X]\cap S=\emptyset$, so by known results about localizations we get $$D[X]_{M[X]}\cong(D[X]_M)_{(M[X]_M)}\cong(D_M[X])_{(MD_M[X])}.$$
By the hypothesis we have that $D_M$ is a valuation ring, so in particular $D_M$ is a GCD domain, then $(D_M[X])_{M[X]_M}$ is also a GCD domain, but from here I'm unable to conclude anything else about $(D_M[X])_{M[X]_M}$.
Any help is appreciated.
By the isomorphisms given in my question it's enough to show that $(D_M[X])_{(MD_M[X])}$ is a valuation ring. In order to prove this let's take two non-zero elements $F,G\in (D_M[X])_{(MD_M[X])}$.
By definition we can write $$F=\frac{\sum_{i=0}^n a_iX^i}{\sum_{j=0}^m b_jX^j}, \;\; G=\frac{\sum_{i=0}^r c_iX^i}{\sum_{j=0}^s d_jX^j},$$ where $a_i,b_j,c_i,d_j\in D_M$ and $\sum_{j=0}^m b_jX^j, \sum_{j=0}^s d_jX^j\notin MD_M[X]$. We then define $$h_1=\sum_{k=0}^{n+s} \alpha_kX^k=\Bigl(\sum_{i=0}^n a_iX^i\Bigr)\Bigl(\sum_{j=0}^s d_jX^j\Bigr),\;\; h_2=\sum_{l=0}^{r+m} \beta_lX^l=\Bigl(\sum_{i=0}^r c_iX^i\Bigr)\Bigl(\sum_{j=0}^m b_jX^j\Bigr)$$
As $D_M$ is a valuation ring, then one of the $\alpha_k$ is a gcd of $\alpha_0,\ldots, \alpha_{n+s}$. WLOG let's suppose that $\alpha_0=\gcd(\alpha_0,\ldots, \alpha_{n+s})$. Anagolously, let's suppose that $\beta_0=\gcd(\beta_0,\ldots, \beta_{r+m})$. Then we have that either $\alpha_0\mid \beta_0$ or $\beta_0\mid \alpha_0$.
Let's assume the later case is true, so there exists $\gamma\in D_M$ such that $\alpha_0=\gamma\beta_0$. Therefore $$H=\frac{h_1}{h_2}=\frac{\Bigl(\sum_{i=0}^n a_iX^i\Bigr)\Bigl(\sum_{j=0}^s d_jX^j\Bigr)}{\Bigl(\sum_{i=0}^r c_iX^i\Bigr)\Bigl(\sum_{j=0}^m b_jX^j\Bigr)}=\frac{\sum_{k=0}^{n+s} \alpha_kX^k}{\sum_{l=0}^{r+m} \beta_lX^l}=\frac{\alpha_0 \Bigl(\sum_{k=0}^{n+s} \alpha'_kX^k\Bigr)}{\beta_0 \Bigl(\sum_{l=0}^{r+m} \beta'_lX^l\Bigr)}=\frac{\gamma\Bigl(\sum_{k=0}^{n+s} \alpha'_kX^k\Bigr)}{\sum_{l=0}^{r+m} \beta'_lX^l}$$
satisfies that $H\in (D_M[X])_{(MD_M[X])}$ because $\beta'_0=1\notin MD_M$. Moreover it's easy to see that $F=GH$, then $G\mid F$ in $(D_M[X])_{(MD_M[X])}$. Hence, $(D_M[X])_{(MD_M[X])}$ is a valuation ring and we're done.