I am reading through Stephen McAdam's Asymptotic Prime Divisors. I am stuck on Lemma 3.1, which states:
Lemma 3.1. Let $R$ be a Noetherian domain with integral closure $\bar{R}$. Let $(V,N)$ be a D. V. R. overring of $R$ and suppose that $V$ is a localization of an integral extension of a finitely generated extension of $R$. If the transcendence degree of $V/N$ over $R/N \cap R$ is 0, then $height(N \cap \bar{R}) = 1.$
Now to me, it seems like we need $N \cap R$ to be a maximal ideal of $R$ so that $R/N \cap R$ is a field. This would follow immediately if $R \rightarrow V$ was an integral extension, but I can't establish that.
According to our setup, we have: $$ R \xrightarrow{f.g.} A \xrightarrow{integral} T \xrightarrow{localization} V = S^{-1}T.$$
It would be a great start to show $N \cap T$ is maximal in $T$. But even this isn't necessarily the case. Can someone help out?
In general this is false. Consider $R=A=T=k[[x,y]]$ for some field $k$ and $V=R_{(x)}$, which is a DVR. The unique maximal ideal of $R$ is $(x,y)$, but the image of $y$ in $V$ is invertible and so it is not in the maximal ideal of $V$.
I think that in the lemma the transcendence degree of $V/N$ over $R/N \cap R$ means over the fraction field of $R/N \cap R$. It is not weird.