Let $k$ be a field. Then $k[[x,y]]$ is a complete local noetherian regular domain of dimension $2$. What are the prime ideals?

I've browsed through the paper "Prime ideals in power series rings" (Jimmy T. Arnold), but it does not give a satisfactory answer. Perhaps there is none. Of course you might think it is more natural to consider only certain prime ideals (for example open/closed ones w.r.t. the adic topology), but I'm interested in the whole spectrum.

A first approximation is the subring $k[[x]] \otimes_k k[[y]]$. If we know its spectrum, perhaps we can compute the fibers of $\text{Spec } k[[x,y]] \to \text{Spec } k[[x]] \otimes_k k[[y]]$. Now the spectrum of the tensor product consists of $(x),(y),(x,y)$ and $\text{Spec } k((x)) \otimes_k k((y))$. The latter one is still very complicated, I think. For example we have the kernel of $k((x)) \otimes_k k((y)) \to k((x))$. Also, for every $p \in k[[x]]$, we have the prime ideal $(y - p)$.


3 Answers 3


The ring $k[[x,y]]$ is a local UFD of dimension 2; so its prime ideals are the zero ideal, the maximal ideal, and all the ideals generated by an irreducible element. They are all closed (all ideals in a noetherian local ring are closed). Classifying them is an extremely complicated business, already when $k = \mathbb C$.

The ring $k[[x]] \otimes_k k[[y]]$ is truly nasty, it is not even noetherian, and I doubt it would help.

  • 3
    $\begingroup$ It is a regular local ring, hence UFD by Auslander and Buchsbaum. Am I missing something here? $\endgroup$
    – Olivier
    May 10, 2010 at 12:23
  • 6
    $\begingroup$ Almost any book that treat UDF's, for example, Zariski and Samuel, or Matsumura's "Commutative Ring Theory" (Theorem 20.8). Anyway, $k[[x,y]]$ is a regular local ring, and those are well known to be UFD's (the Auslander-Buchsbaum theorem, 20.3 in Matsumura). $\endgroup$
    – Angelo
    May 10, 2010 at 12:25
  • 2
    $\begingroup$ Sorry, you wrote "where can I find a proof". The answer is e.g Matsumura Commutative Ring Theory Theorem 20.3 page 163 (of the second edition) $\endgroup$
    – Olivier
    May 10, 2010 at 12:26
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    $\begingroup$ Maybe it offers some insight to note that there are two quite different types of prime ideals in $k[[x,y]]$. If you start with a non-maximal prime ideal $P$ of $k[x,y]$ which is contained in $\langle x,y\rangle$, then you can push it forward to get $Pk[[x,y]]$; this might not be prime, but it will be the intersection of finitely many prime ideals of $k[[x,y]]$. Think of $P$ as a curve in the affine plane which passes through the origin; then these primes of $k[[x,y]]$ are the local branches at $0$. ctd... $\endgroup$ May 10, 2010 at 12:38
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    $\begingroup$ This process gives all the prime ideals of $k[[x,y]]$ whose contraction to $k[x,y]$ is non-zero. Unfortunately, there are also LOTS of prime ideals $Q$ of $k[[x,y]]$ whose contraction to $k[x,y]$ is zero, and these are a bit weird, e.g. the localization $k[[x,y]]_Q$ contains $k(x,y)$. $\endgroup$ May 10, 2010 at 12:38

I really come long after the battle on this one, but I have something to add to Angelo's answer. To rephrase what Angelo said, there is only one prime ideal of height 2, namely the maximal ideal $(x,y)$, and one of height $0$. The problem is to classify the prime ideals of height one which (as in any UFD) are exactly the principal ideals generated by an irreducible element $f(x,y)$.

Now there are several possible meaning for "classify", but one reasonable sense is to describe all such prime ideal up to an automorphism of $k[[x,y]]$ (automorphism of $k$-algebra, I mean, of course.)

Theorem: if $k=\mathbb C$, then up to a an automorphism of $k[[x,y]]$ any prime ideal of height one is of the form $(f(x,y))$ where $f(x,y)$ is an irreducible polynomial in two variables.

A proof for this theorem (that is, a combination of two references that do the job) can be found in my answer in the following question. I think that the theorem is true for any field $k$, but one of the references used there use $k=\mathbb C$.

EDIT: I give a little bit more explanation as requested. Let start with $f(x,y)$ an irreducible element. One has $f(0,0)=0$ otherwise $f$ would be a unit. Then a theorem of Artin (references in the question linked above) tells that up to a change of variable $x \rightarrow x'$, $y \rightarrow y'$, one can assume that $f(x,y)=g(x',y')$ will be analytical, that is the power series $g$ converges on some neighborhood of $(0,0)$. In algebraic terms, the change of variables defines an automorphism $\psi$ if $k[[x,y]]$ sending $x$ to $x'$ and $y$ to $y'$, and $g = f \circ \psi^{-1}$. This theorem is actually much more general: it works in any dimension and over any field, with analytic replaced then by "in the Henselization of the subring of polynomials". In the case of dimension $2$ over $\mathbb C$, i would not be surprised if it was much older (Puiseux? Newton?).

The second step is proving that analytic $g(x,y)$ can be changed, up to a second change of variable to an $h(x,y)$ which is simply a polynomial. The reference given in the linked question is to a book of complex geometry, hence is over $\mathbb C$, but the result may be true over any field. However it is not true in three and more variable, as explained in Moret-Bailly's answer to this question.

So we have an automorphism $\psi$ such that $h = f \circ \psi^{-1}$. Since $(f)$ is prime, so is $h$, hence $h$ is irreducible as a power series, hence as a polynomial.

To make this more concrete, let me gives two examples. if $f(x,y) =a x + b y$ + higher terms, with $a$, or $b$, or both, non zeros (This condition is equivalent to $A/(f)$ being regular, or a discrete valuation ring.) Say $a \neq 0$, then one can take for new variables $x'=f(x,y)$, $y'=y$, and $h(x,y)$ is just $x$.

If on the contrary $a=b=0$, looking at the quadratic terms of $f$ gives a quadratic forms over $k$. If this quadratic form $h$ is non-degenerate, then by the theory of Morse function $f$ can be transformed to $h$ (I guess $\operatorname{char} k \neq 2$ may be necessary).

  • $\begingroup$ I think you mean the height of $(0)$ is 0 and the height of $(x,y)$ is 2. Or you use a definition of height which is not standard. $\endgroup$
    – YCor
    Oct 18, 2013 at 19:06
  • $\begingroup$ Also could you explain in a little further detail the last few lines of your answer? $\endgroup$
    – YCor
    Oct 18, 2013 at 19:17
  • $\begingroup$ I changed the heights 0 and 2, and added details. If there is a mistake in my reasoning, it would appear more clearly now. $\endgroup$
    – Joël
    Oct 19, 2013 at 17:09
  • $\begingroup$ I was wondering if you could adress the converse, how can I check if an irreducible polynomial in $\mathbb{C}[x,y]$ stays irreducible in the power series ring? Concretely, what is $Spec(\mathbb{C}[[x,y]]/y^2-x^2-x^3)$? I.e how do I find the minimal prime containing $y^2-x^2-x^3$? $\endgroup$
    – harajm
    Mar 21, 2017 at 0:14
  • $\begingroup$ "The second step is proving that analytic $g(x,y)$ can be changed, up to a second change of variable to an $h(x,y)$ which is simply a polynomial." Weierstrass preparation theorem popped into my mind when I read this. $\endgroup$ May 15, 2017 at 0:39

You will find the proof of the fact that $k[[x_1,\dots,x_n]]$ is a UFD in Shafarevish, Basic algebraic geometry, Chapter II, 3.1. The proof uses the Weierstrass preparation lemma. The same section contains the proof that a codimension 1 prime ideal is principal. The only codimension prime ideal of codimension 2 is $(x,y)$ of course, since it has to be contained in $(x,y)$. Miles Reid's Undergraduate commutative algebra contains the proof as well, I think.


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