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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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    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