Let $R$ be a Noetherian domain of dimension $\ge 1$. Let $\mathfrak{p}\_i$, $i = 1, 2, ...$ be prime ideals of height one. Let $T = R[[X]]$ with $X$ is a indeterminate. For each $i \ge 1$ we set $\mathfrak{q}\_i = \mathfrak{p}\_i[[X]] = \mathfrak{p}\_iT$ the extension of $\mathfrak{p}\_i$. It is clearly that the height of $\mathfrak{q}\_i$ is one for all $i$. Set $S = T \setminus \cup\_{i \ge 1}\mathfrak{q}\_i$. 


**Question:** Is it true that $\dim T\_S = 1$?



Dimension of a ring after localization
If you can use the *Countable Prime Avoidance Lemma* (**Edit:** that is, if you are in situations in which the Countable Prime Avoidance Lemma applies), I think the answer is yes. Because any prime ideal of $T\_S$ comes from a prime ideal of $T$ that is contained in the union of the $\mathfrak{q}\_i$'s. But then the *Countable Prime Avoidance Lemma* would imply that this prime ideal must be contained in one of the $\mathfrak{q}\_i$'s. Hence $\dim T\_S$ cannot be bigger than $1$. 


The *Countable Prime Avoidance Lemma* will work, if for instance $R$ contains an uncountable number of elements $\{\mu\_\lambda\}\_\lambda$ such that $\mu\_\lambda-\mu\_\gamma$ is a unit in $R$ for every $\lambda\neq\gamma$. This condition will hold, if for example $R$ is local with $R/\mathfrak{m}$ uncountable. 


A reference for this version of the *Countable Prime Avoidance Lemma* is page 242 of the book **Cohen-Macaulay Representations** by Graham Leuschke and Roger Wiegand, but I am sure you can find it at other places, too.


