Let $A$ be a commutative ring with identity. If $f = a_0 + a_1 x + \cdots + a_n x^n \in A[x]$ is a polynomial, define $c(f) = A a_0 + A a_1 + \cdots + A a_n$ the ideal of $A$ generated by the coefficients of $f$. Consider $S$ the subset of $A[x]$ made up of primitive polynomials, i.e. polynomials $f \in A[x]$ such that $c(f) = A$. It is not difficult to prove that $S$ is a multiplicative subset of $A[x]$. Consider the ring $ A(x) = S^{-1} (A[x]). $ It is easy to show that $S$ does not contain zero-divisors of $A[x]$, hence $A \subseteq A[x] \subseteq A(x)$. If $I$ is an ideal of $A$ then $(I \cdot A(x)) \cap A = I$. If $\mathfrak{p}$ is a prime ideal of $A$ then $\mathfrak{p} \cdot A(x)$ is a prime ideal of $A(x)$. The map $\phi \colon \mathrm{Spec} A(x) \to \mathrm{Spec} A$ defined by $\phi(P) = P \cap A$ is surjective, because a right-inverse is $\mathfrak{p} \mapsto \mathfrak{p} \cdot A(x)$. It is clear that $\dim A \leq \dim A(x) \leq \dim A[x]$.
- If $M$ is a maximal ideal of $A(x)$, then does there exist a maximal ideal $\mathfrak{m}$ of $A$ such that $M = \mathfrak{m} \cdot A(x)$?
- Is the map $\phi \colon \mathrm{Spec} A(x) \to \mathrm{Spec} A$ injective? If $A$ is noetherian, then $\dim A(x) = \dim A$?
- If $A$ is a normal domain, then is $A(x)$ a normal domain?