Setup: Let $k$ be a field, let $n$ be a positive integer, and let $R := k[[x_1,\ldots,x_n]]$ denote the commutative ring of formal power series over $k$ in $x_1,\ldots,x_n$. We know that there is exactly one maximal ideal for $R$, namely $\langle x_1,\ldots,x_n \rangle$.

By localizing at the multiples of the $x_1,\ldots,x_n$, we can construct a multivariable Laurent series ring $L$. In particular, $L$ is equal to the ring of series of the form $$\sum _{m_1,\ldots,m_n \in \mathbb{Z}} \quad \lambda _{(m_1,\ldots,m_n)} \; x^{m_1}\cdots x^{m_n},$$ for $\lambda_{(m_1,\ldots,m_n)} \in k$, but with $\lambda _{(m_1,\ldots,m_n)} = 0$ when the minimum of the $m_1,\ldots,m_n$ is $\ll 0$.

When $n = 1$ it is well known that $L$ is a field. However, for $n > 1$ the situation is more subtle. For example, when $n=2$, the ideals of the form $\langle x_1 - \mu x_2 \rangle$, for nonzero $\mu \in k$, are maximal ideals.

My general question: Does anyone know of an explicit description of the maximal ideals of $L$, for $n > 1$?

A more refined question: Suppose $k$ is algebraically closed. Consider the Laurent polynomial ring $P := k[x_1^{\pm 1},\ldots,x_n^{\pm 1}]$, and let $H$ denote the multiplicative $n$-torus $(k^\times)^n$.

There is a natural action of $H$ on $P$, obtained via $$(\alpha_1,\ldots,\alpha_n)\cdot x_i = \alpha_ix_i,$$ for $(\alpha_1,\ldots,\alpha_n) \in H$. The action of $H$ on $P$ induces an action of $H$ on the maximal spectrum of $P$, and it follows from Hilbert's Nullstellensatz that there is exactly one $H$-orbit of maximal ideals.

So the refined question is: Consider the analogous action of $H$ on $L$. Are there finitely or infinitely many $H$-orbits of maximal ideals of $L$?

My own motivation: These commutative Laurent series rings show up as central subalgebras of $q$-commutative Laurent series rings (i.e., formal Laurent series where $x_i x_j = q_{ij}x_j x_i$ for suitable scalars $q_{ij}$). In joint work with Linhong Wang, I have been studying $q$-commutative power and Laurent series rings. Properties of the prime and primitive ideals in these noncommutative algebras are strongly influenced by the behavior of these central subalgebras.

Thank you for your time! Any hints or references greatly appreciated.


2 Answers 2


Your ring $L$ is a localization of the power series rings $R= k[[x_1,\cdots,x_n]]$ at the multiplicative set $M$ of monomials in $R$.

So the prime ideals of $L$ correspond to prime ideals in $R$ which do not meet $M$. Clearly, the maximal primes are the set of biggest primes which do not contain any variable. Since $R$ is local and the maximal ideal contains all the variables, these ideals are of dimension one (this is the key difference to the Laurent polynomials case, since in that case these ideals would be maximal in the polynomial ring, so Nullstellensatz applies)

For example, when $n=2$, you get all ideals of height one which does not contains $x_1$ or $x_2$. So they are all principal ideal $(f)$ with $f$ irreducible and $f\neq x_1$, $f\neq x_2$.

It shows that the answer to your refined question is NO. The ideals $(x-y)$ and $(x-y^2)$ can not be in the same orbit.

In general one can only describe the set of maximal ideals in $L$ as follows: they are prime ideals of height $n-1$ in $S$, minus the set $(x_i, P)$, where $P$ is a prime of height $n-2$ in $k[[x_1,...x_{i-1}, x_{i+1},...,x_n]]$.

I doubt one can say more in general, since there are many open questions about affine curves.

  • $\begingroup$ OK...thanks very much for this! ..Ed $\endgroup$
    – Ed Letzter
    Mar 11, 2011 at 3:20
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    $\begingroup$ Dear Ed, you are welcome! You may want to wait a bit before accepting though, for two reasons: 1) I may be wrong, 2) even if I am right, there are many fantastic people on this site, so you may get much better answer. $\endgroup$ Mar 11, 2011 at 3:24
  • $\begingroup$ Hmm. OK. I'm new (today) to this "online thing" and I'll take your advice. But thanks again! Be well. ..Ed $\endgroup$
    – Ed Letzter
    Mar 11, 2011 at 3:49
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    $\begingroup$ Dear Ed, no problem. Of course, once you are sure that this answers your question, I would not mind if you accept it again (-:. $\endgroup$ Mar 11, 2011 at 5:27
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    $\begingroup$ @Ed: Hailong's description gives another information (which may or may not be useful to you): the residue fields at these maximal ideals are all $k$-isomorphic to finite extensions of $k((x))$. If $k$ is perfect, they are of the form $k'((x))$ for finite extensions $k'$ of $k$. $\endgroup$ Mar 11, 2011 at 8:12

I think that the question difficult as illustrated by Hailong's answer. I suspect that it will be hard to even find a nice parameterisation of the $H$-orbits of maximal ideals in your refined question. Certianly, as Hailong implies there will be infinitely many such orbits.

You might find in helpful to consider a valuation theoretic approach. Konstantin Ardakov has done some work on related --- although not quite similar --- questions that I believe he has nearly finished writing up. Maybe he will appear and say more.

Edit: For the record Konstantin's work has now appeared here http://arxiv.org/abs/1108.0371

  • $\begingroup$ Hi Simon. Thanks. Yes, I'm a bit embarrassed now about the "finiteness" part of the question. ..Ed $\endgroup$
    – Ed Letzter
    Mar 11, 2011 at 12:33
  • $\begingroup$ @Simon: your second paragraph sounds very interesting. If you know Konstantin, perhaps he can be made aware of this question (-: ? I would like to hear what he has to say. Cheers. $\endgroup$ Mar 11, 2011 at 17:31

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