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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.

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    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.2011-07-21
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    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.2011-07-21
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    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?2011-07-21
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    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 $\mathbb{C}[[x,y]]^{alg}[1/xy]\rightarrow \mathbb{C}$).2011-07-21
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    Qiauchu, I wonder what you realized that made you remove your last comment. I am now quite confused myself by this...2011-07-21
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    Well, I thought I had an argument that any morphism out of $\mathbb{C}[[x]]$ is automatically determined by its restriction to $\mathbb{C}[x]$, and I thought this applied to the $2$-dimensional case, but now I'm not confident about either of these thoughts...2011-07-21
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    I actually think you're right, now. There are no sections... There would be sections for things like $\mathbb{Z}[[x,y]]^{alg}[1/xy] \otimes \mathbb{C}$ which I confused with $\mathbb{C}[[x,y]]^{alg}[1/xy]$.2011-07-21

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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

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    Oh, I see. We don't even need to know whether a morphism out of $\mathbb{C}[[x, y]]$ is automatically continuous or not.2011-07-22