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

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?

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.
• Yes I mean an integral domain. I think that also in spirit of your hint we can work with $R= k[x_1,..., x_n]/P$ where $k= \mathbb{Z}/(c)$ and $P$ prime ideal of $k[x_1,..., x_n]$. So the question stays why if we take a nonzero $a \in R$ then the transcendence degree of $R/(a)$ drops by $1$ from that one of $R$? Commented Feb 18, 2021 at 3:57
• For $R= k[x_1,..., x_n]$ that's easy but for arbitrary $P$ I don't know how to solve it without identification of trdeg with Krull dimension. Indeed, applying this identification it's easy to show to claim, but I'm curious if one can show it without this tool Commented Feb 18, 2021 at 4:02
• Because $x_1=a$. That's a general property of transcendence bases: we can take for $x_1$ any non-algebraic element, and the cardinality of the base won't depend on this choice. Commented Feb 18, 2021 at 4:54
• I'm not getting on with the calculation in your example $R=\bigcup_{k\ge 1}\Bbb{Q}[x_1,x_2/x_1^k]$ and $R/(a)$ with $a:=x_1$. The case $Tr\deg(Frac(R))$ looks not hard. For every $k \ge 1$ we have $\Bbb{Q}[x_1,x_2] \subset \Bbb{Q}[x_1,x_2/x_1^k]$ and $\Bbb{Q}[x_1,x_2/x_1^k] \subset \Bbb{Q}(x_1,x_2)$, so $Frac(R)= \Bbb{Q}(x_1,x_2)$ as I see so far, is this calculation correct? Then trgeg is $2$. Commented Feb 19, 2021 at 1:38
• $R/(a) = \Bbb{Q}$ Commented Feb 19, 2021 at 3:33