Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I've been thinking about weird rings recently, and I couldn't answer the following question to myself:

What are the sections of the inclusion $\mathbb{C}\rightarrow \mathbb{C}[[x,y]]^{alg}[\frac{1}{xy}]$ (the $alg$ in the superscript means that I only take those formal power series that are algebraic over $\mathbb{C}(x,y)$; though if you have an answer offhand for the ring of all power series, I suppose that would be interesting too)?

In other words, what are the possible values that $x$ and $y$ can take so that it gives us a well-defined section?

P.S. I put this under algebraic geometry because I'm given to believe that this has something to do with something called the etale stalk.

share|cite|improve this question
To be honest I never understood what morphisms out of $\mathbb{C}[[x]]$ look like as an abstract $\mathbb{C}$-algebra. Presumably the only morphism $\mathbb{C}[[x]] \to \mathbb{C}$ is the trivial one, but I don't think I actually know how to prove this. – Qiaochu Yuan Jul 21 '11 at 21:09
The $\mathbb{C}[[x]]$ case is much easier: this is a $1$ dimensional ring with only two prime ideals. The geometric point $Spec(\mathbb{C})\rightarrow Spec(\mathbb{C}[[x]])$ must go to (possibly a base extension of) the generic point of some subscheme, and so it must either go to the maximal point (what you called the trivial section) or the generic one. But the latter thing cannot happen since that would mean that the section $Spec(\mathbb{C})\rightarrow Spec(\mathbb{C}[[x]])$ would have to factor through $Spec(\overline{\mathbb{C}((x))})$ which it does not. – Nicole Jul 21 '11 at 21:20
Oh, I was missing something silly. $\mathbb{C}[[x]]$ contains $\frac{1}{1 - ax}$ for every $a \in \mathbb{C}$, so $x$ can only be sent to $0$. Okay, but why doesn't this resolve your question then? – Qiaochu Yuan Jul 21 '11 at 21:23
What makes this question interesting is that $\mathbb{C}[[x,y]]^{alg}$ (as well as without $alg$) is a $2$-dimensional ring, and when you invert $xy$ it becomes $1$-dimensional (because there is only one maximal ideal, which we remove). So there may be many maximal ideals in $\mathbb{C}[[x,y]]^{alg}[1/xy]$ but I don't know how to tell what they are, and whether or not they induce a geometric point (a section $\mathbb{C}[[x,y]]^{alg}[1/xy]\rightarrow \mathbb{C}$). – Nicole Jul 21 '11 at 21:25
Qiauchu, I wonder what you realized that made you remove your last comment. I am now quite confused myself by this... – Nicole Jul 21 '11 at 21:48

It seems to me this question was basically answered in the comments by Qiaochu and Nicole, so I'm just putting down a CW answer in order to note this. Please let me know if I am getting something wrong:

There are no $\mathbb{C}$-algebra maps $\mathbb{C}[[x,y]]^{alg}[1/(xy)] \to \mathbb{C}$.

Proof: Suppose, for the sake of contradiction, that $\phi$ is such a map. Then $\phi(x)$ can't be to $0$, as $\phi(x) \phi(y) \phi(1/(xy))$ must be $1$.

Let $\phi(x) =a \neq 0$. Then $1/(1-a^{-1}x)$ is in our ring. We are supposed to have $$\phi(1-a^{-1} x) \phi(1/(1-a^{-1}x)) = 1.$$ But the LHS is $$(1-a^{-1} \cdot a) \phi(1/(1-a^{-1}x)) = 0 \cdot \phi(1/(1-a^{-1}x)) =0,$$ a contradiction. QED

share|cite|improve this answer
Oh, I see. We don't even need to know whether a morphism out of $\mathbb{C}[[x, y]]$ is automatically continuous or not. – Qiaochu Yuan Jul 22 '11 at 19:34

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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