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 $\dim T_S = 1$?


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.

  • $\begingroup$ Another reference for countable prime avoidance is Baire's category theorem and prime avoidance in complete local rings by Sharp and Vamos (Arch. Math. 44 (1985), 243-248), as mentioned in mathoverflow.net/questions/102546. $\endgroup$ – Fred Rohrer Feb 26 '14 at 20:23
  • $\begingroup$ I know Countable Prime Avoidance Lemma. But I do not know how to apply it for my question. Could you give a detail answer. $\endgroup$ – Pham Hung Quy Feb 27 '14 at 7:27
  • $\begingroup$ Recall that an element of $T$ (a formal power series) is a unit if and only if its constant term (degree zero term) is a unit. Now if $R$ has an uncountable number of elements $\{\mu_\lambda\}$ such that $\mu_\lambda-\mu_\gamma$ is a unit in $R$ for $\lambda\neq\gamma$, then you can produce an uncountable number of elements of $T$ with the same property, just consider formal power series whose constant term (degree zero term) equals one of the $\mu_\lambda$'s. Fred Roher has kindly provided more details below. $\endgroup$ – Mahdi Majidi-Zolbanin Feb 27 '14 at 13:05
  • $\begingroup$ I only known Countable Prime Avoidance Lemma for complete local ring before. Thanks you very much for the new form. But the ring in my question do not have this assumption. $\endgroup$ – Pham Hung Quy Feb 27 '14 at 15:55

I will prove that $T_S$ satisfy Countable Prime Advoidance Lemma in the form mentioned in Mandi's answer (See the book of Graham Leuschke and Roger Wiegand)

  1. First note that an element of $T = R[[X]]$ is a unit if its constant term is a unit in $R$. After localization such element is also a unit in $T_S$.

  2. Since $X \notin \mathfrak{q}_i = \mathfrak{p}_i[[X]]$ we have $X$ contained in the multiplicative closed set $S$. Therefore $X$ is a unit in $T_S$.

  3. Consider the following set $$B = \{u_{\mu} = b_0 + b_1X + \cdots + b_iX^i + \cdots | b_i = 0 \text{ or } 1, \text{ and } u_{\mu} \neq 0\}.$$ It is clearly that $B$ in a uncountable set of units in $T_S$.

  4. Moreover, if $u$ and $v$ in $B$ we have $u - v$ (edit: $u-v$ nay not in $B$ but it is a unit) is a unit. Thus the assumption of Countable Prime Advoidance Lemma is satisfying.

The following application is a generalization of my question.

Corollary. Let $\{p_i\}_{i \ge 1}$ is a countable set of prime ideals with not containment relation. Let $T = R[[X]]$ and $S = T \setminus \cup_{i\ge 1}\mathfrak{p}_iT$. Then $R \to T_S$ is a flat extension and $$\text{Max}(T_S) = \{\mathfrak{p}_iT_S\}_{i \ge 1}.$$

  • $\begingroup$ Can you explain why $u-v$ is a unit? What if both $u$ and $v$ start with $1$? Then $u-v$ will start with $0$ and is not a unit.(?) $\endgroup$ – Mahdi Majidi-Zolbanin Feb 28 '14 at 14:12
  • $\begingroup$ Example $u = 1 + X$ and $v = 1 + X^2$ we have $u - v = X - X^2 = X(1 -X)$ is unit in $T_S$ since both $X$ and $1-X$ are units in $T_S$. $\endgroup$ – Pham Hung Quy Feb 28 '14 at 14:40
  • $\begingroup$ I see! I forgot $X$ was a unit in $T_S$. Very interesting. $\endgroup$ – Mahdi Majidi-Zolbanin Mar 1 '14 at 2:03

Let me give a detailed account of Mahdi's argument.

First, we have to suppose (which is not totally clear in the question), that we consider only countably many primes $\mathfrak{p}_i$ of height $1$.

Second, we will suppose that $R/\mathfrak{m}$ is uncountable for every maximal ideal $\mathfrak{m}$ of $R$.

Then, $\dim(S^{-1}T)=1$.

Indeed, replacing $R$ by its localisation at a maximal ideal we can suppose that $R$ is in addition local. Then, in order to show that $\dim(S^{-1}T)=1$ we will show that every non-zero prime of $S^{-1}T$ has height $1$. So, let $\mathfrak{p}$ be a non-zero prime of $S^{-1}T$. Then, there exists a non-zero prime $\mathfrak{p}'$ of $T$ with $S\cap\mathfrak{p}'=\emptyset$ such that $\mathfrak{p}=S^{-1}\mathfrak{p}'$, and moreover $\mathfrak{p}$ and $\mathfrak{p}'$ have the same height. Thus it suffices to show that $\mathfrak{p}'$ has height $1$. The condition $S\cap\mathfrak{p}'=\emptyset$ is equivalent to $\mathfrak{p}'\subseteq\bigcup_{i\in\mathbb{N}}\mathfrak{q}_i$. Since $R$ is local with uncountable residue field, the same holds for $T$. Hence, countable prime avoidance holds in $T$, and so we see that there exists $i\in\mathbb{N}$ such that $\mathfrak{p}'\subseteq\mathfrak{q}_i$. As $\mathfrak{p}'$ is non-zero, $\mathfrak{q}_i$ has height $1$ and $T$ is a domain, it follows that $\mathfrak{p}'=\mathfrak{q}_i$ has height $1$ as desired.

(Using another variant of countable prime avoidance, also proven by Sharp and Vamos, gives the same conclusion under the hypothesis that $R$ is a complete local ring.)

  • $\begingroup$ Thanks you Fred. But in my question we do not assume that $R/\mathfrak{m}$ is uncountable. $\endgroup$ – Pham Hung Quy Feb 27 '14 at 15:24
  • $\begingroup$ Em Quý, have you tried to use Example 2.4 in the article by Sharp and Vámos that I mentioned above to construct a counterexample? $\endgroup$ – Fred Rohrer Feb 27 '14 at 18:33
  • $\begingroup$ Dear Fred, I do not have that paper. Could you please sent it for me. I think I can prove it. $\endgroup$ – Pham Hung Quy Feb 28 '14 at 3:52

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