$\operatorname{Trdeg}(\operatorname{Frac}(R/(a))) =\operatorname{Trdeg}(\operatorname{Frac}(R))-1$

Question

Let $R$ be a commuative (noetherian?) integral domain and $Q=\operatorname{Frac}(R)$. Assume that $\operatorname{Trdeg}(Q) := n < \infty$. Let $a \in R$ a nonzero prime element of $R$. I want to understand if that's true that $\operatorname{Trdeg}(\operatorname{Frac}(R/(a)))= n-1$ without using PID and identification of Krull dimension of the ring with transcendence degree of accociated field of fractions. Is there a direct argument known with similar spirit as for $R= K[x_1,...,x_n]$ that quoting out a nonzero prime element of $R$ reduces the maximal number of algebraically independent elements of $\operatorname{Frac}(R/(a))$?

In case of $R= K[x_1,...,x_n]$ we deal with special case where the transcendental basis $x_1,x_2,...,x_n \in K[x_1,...,x_n]$ already generate the ring, so quoting every nonzero $f \in K[x_1,...,x_n]$ provides already a nontrivial algebraic relation between $x_1,x_2,...,x_n$ and $\operatorname{Trdeg}(\operatorname{Frac}(K[x_1,...,x_n]/(f))) =\operatorname{Trdeg}(K(x_1,...,x_n))-1$ is automatically satisfied for prime $f$. Can this argument can be adapted for arbitrary commuative noetherian domain?

Answer 1

Did you mean an integral domain? The transcendence degree over the prime field? $(a)$ not containing any non-zero integer (ie. not $a=2,R=\Bbb{Z}$) ? If so then $$Tr\deg(Frac(R/(a)))\le Tr\deg(Frac(R))-1$$ use that $R$ is algebraic over $\Bbb{Z}/(c)[x_1,\ldots,x_n]$ where $c$ is the characteristic and $x_1=a$.

Then try with $R=\bigcup_{k\ge 1}\Bbb{Q}[x_1,x_2/x_1^k],a=x_1$ to see that sometimes $$Tr\deg(Frac(R/(a)))=Tr\deg(Frac(R))-2$$ But this ring is not Noetherian as $I_k=(x_2/x_1^k)$ is a strictly increasing sequence of ideals.