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There is this notion of a Goldman domain which is a domain $A$ satisfying $Q(A)=A[u^{-1}]$ for some $u\in A$, where $Q(A)$ is the field of fractions of $A$. This is equivalent to saying $Q(A)$ is finitely generated as an $A$-algebra.

We can extend this notion to any commutative unital ring $A$, where we say $A$ is a Goldman ring if $Q(A)$, the total fraction ring of $A$, can be generated by the inverse of a single regular (meaning non-zero-divisor) element of $A$. (Of course, this generalization is a little silly from the perspective of the theory of Jacobson rings, where this notion is often used.)

One can show if $A$ is a domain (taking domains to be nontrivial) then $A[x]$ is never a Goldman domain. Does the same hold in general? That is, if $A$ is a nontrivial ring, can $A[x]$ ever be a Goldman ring? The proof I have seen when $A$ is a domain makes heavy use of this assumption.

I suspect it's not possible and tried to argue this by hand. Supposing $Q(A[x])=A[x][f^{-1}]$ for some (regular) $f\in A[x]$ I tried playing various games by writing things like $x^{-1}$ as polynomials in $f^{-1}$ (for example from this we can conclude the constant term $f(0)$ of $f$ is nilpotent, though this argument doesn't seem to go anywhere or e.g. extend to higher coefficients). I wanted to argue one can generate $Q(A[x])$ using a polynomial of smaller degree via these methods but didn't get anywhere.

P-addict
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This is indeed true and proven as Proposition 3.2. in Rings with a finitely generated total quotient ring by John Conway Adams. (You can access the paper online for free.) It follows easily from the observation* that $Q(A)=A[1/u]$ iff $u$ is contained in every regular prime ideal of $A$. (Here, an ideal is called regular if it contains regular element.)

*Since that observation is remarked without proof in the paper in section 2, let me give a proof here.

$\phantom{\iff}$ $u$ is contained in every regular prime ideal

$\iff$ for every regular $a \in A$ we have: every prime ideal containing $a$ also contains $u$

$\iff$ for every regular $a \in A$ we have $(A/\langle a \rangle)[1/u] = 0$

$\iff$ for every regular $a \in A$ we have $u^k = ab$ for some $b \in A$ and $k \geq 0$

$\iff$ for every regular $a \in A$ we have $\dfrac{1}{a} = \dfrac{b}{u^k}$ in $Q(A)$ for some $b \in A$ and $k \geq 0$

$\iff$ we have $Q(A) = A[1/u]$

Martin Brandenburg
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    Likely it is remarked without proof because the proof is exactly the same as the well-known classical proof for domains (which can be found in many places, e.g. Section 1-3 in Kaplansky's Commutative Rings). – Bill Dubuque Dec 30 '23 at 17:23