The question comes from my attempt to understand the following question. height of contracted prime ideals in power series rings

$\bullet$ My original question: Let $(R,m)$ be a Noetherian local ring and $\hat{R}$ the completion of $R$ with respect to $m$. Let $Q$ be a prime ideal in $\hat{R}$ and $P = Q \cap R$. Then is height (codimension) of $Q$ equal to height of $P$?

$\bullet$ Remark: The inequality ht $Q \ge$ ht $P$ follows from the going-down property.

In particular, I wanted to know an answer to the question in the following setting.
Setting: Let $R$ be the localization of a polynomial ring over a field at the origin, i.e., $R = k[x_1,\dots,x_d]_{(x_1,\dots,x_d)}$. Let $\hat{R}$ be the completion of $R$ with respect to its maximal ideal $m := (x_1,\dots,x_d)$. After observing that the question has a positive answer when $R = \mathbb{C}[x]_{(x)}$, I was optimistic about the question. However I found that Matsumura (1988) showed that the formal fiber is of dimension $d-1$. I was surprised by the result since I thought polynomial rings over a field is one of the nicest rings.

The result of Matsumura says that the following corollary in Hartshorne's Algebraic Geometry book does not hold true without the assumption that $X$ is of finite type over a field; let $R := \mathbb{C}[x_1,\dots,x_d]_{(x_1,\dots,x_d)}$. Let $X = \hbox{Spec} (\hat{R}), Y = \hbox{Spec} (R)$, and the morphism $f: X \rightarrow Y$ is induced by the natural ring homomorphism from $R$ to $\hat{R}$. Since $\hat{R}$ is a domain, $\hbox{Spec} (\hat{R})$ is reduced and irreducible. Hence it has only one irreducible component which is of dimension $d$. However the fiber $X_y$ has dimension $d-1$ where $y \in Y$ is the point corresponding to the zero ideal of $R$.

$\bullet$ Corollary 9.6 (Chapter 3 of Hartshorne's Algebraic Geometry book)
Let $f: X \rightarrow Y$ be a flat morphism of schemems of finite type over a field $k$, and assume that $Y$ is irreducible. Then the following conditions are equivalent:
(i) every irreducible component of $X$ has dimension equal to $\dim Y + n$;
(ii) for any point $y \in Y$ (closed or not), every irreducible component of the fibre $X_y$ has dimension $n$.

I would like to ask the following two questions.

$\bullet$ Question 1: is there an algebraic or geometric intuition (or explanation) why the dimension of $X_y$ is $d-1$?

$\bullet$ Question 2: how restrictive is the assumption of being of finite type over a field in statements in algebraic geometry? For instance, what are examples of "good" or "big" statements in Hartshorne's Algebraic Geometry book (or from other sources) fail if one removes the condition?


1 Answer 1


For question 1, my geometric explanation is this: you can find an irreducible "analytic curve" $C$, i.e. a 1-dimensional closed subscheme of $X$, which is "as transcendental as possible", meaning that it is not contained in any algebraic hypersurface (i.e. does not map to a hypersurface in $Y$). Now if $p$ is the generic point of $C$, this means that $p$ maps to $y$, hence it is a point of $X_y$ and so is any generization of $p$, which proves that $\dim X_y≥d-1$ because $p$ has codimension $d-1$ in $X$. And of course $\dim X_y<d=\dim X$, so $\dim X_y=d-1$.

To construct $C$, just note that $\mathbb{C}[[t]]$ has infinite transcendence degree over $\mathbb{C}$, so you can find $f_1,\dots,f_d\in \mathbb{C}[[t]]$ with zero constant term (and nonzero degree 1 term, if you wish) which are algebraically independent over $\mathbb{C}$. Now take for $C$ the image of the closed immersion $\mathrm{Spec}\,\mathbb{C}[[t]]\to X$ corresponding to $\varphi:\mathbb{C}[[x_1,\dots,x_d]]\to \mathbb{C}[[t]]$ sending $x_i$ to $f_i$. (Geometrically, it is the "formal curve parametrized by $f_1,\dots,f_d$").

Remark: I don't have Matsumura's book here, but I wouldn't be surprised if this were essentially his proof!

  • $\begingroup$ Hi Laurent, $\quad$ Thank you for such a nice answer. I tried to understand this well, but I was not able to answer the following question. How does one see that the Krull dimension of $\mathbb{C} [[f_1,\dots,f_d ]]$ is 1? I also was not able to see that $\mathbb{C} [[f_1,\dots,f_d ]] \subseteq \mathbb{C}$ is an integral extension. Thank you. $\endgroup$
    – Youngsu
    Apr 22, 2013 at 3:13
  • $\begingroup$ I meant to say "$\mathbb{C}[[f_1,\dots,f_d]] \subseteq \mathbb{C}[[t]]$ is an integral extension". $\endgroup$
    – Youngsu
    Apr 22, 2013 at 3:15
  • 1
    $\begingroup$ Standard fact in commutative algebra: Let $A$ be a noetherian ring (I am not sure about the minimal assumptions), $J$ an ideal such that $A$ is $J$-adically complete and separated. Let $M$ be an $A$-module. If $M$ is $J$-adically complete and separated and $M/JM$ is finitely generated, then $M$ is finitely generated. Apply this to $A=\mathbb{C}[[f_1,\dots,f_d]]$, $J=(f_1,\dots,f_d)$ and $M=\mathbb{C}[[t]]$. Things are even simpler if, say, $f_1$ is a uniformizer in $\mathbb{C}[[t]]$: then it is easy to see directly that $\mathbb{C}[[f_1]]=\mathbb{C}[[t]]$. $\endgroup$ Apr 22, 2013 at 19:16
  • $\begingroup$ Thank you very much. Your explanation is very clear to me. $\endgroup$
    – Youngsu
    Apr 23, 2013 at 21:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.