I have come across a problem while reading Chevelley's "Introduction to the theory of algebraic functions of one variable", in which he says that the genus of the field $L=\mathbb{R}(x,y)$ is 0, where $x$ is transcendental over $\mathbb{R}$ and $y$ satisfies $x^2+y^2+1=0$ and he further states that every place of this field is of degree 2. Here $L$ is an algebraic function field (i.e a finite field extension of a purely transcendental extension of $\mathbb{R}$). Can anyone explain how is it possible?
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Is there a typo? For if both $(x,y) \in \mathbb{R}$, it is obvious that both $x²,y² \geq 0$, hence this equation has no solutions. – Martigan Nov 07 '14 at 08:31
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$x$ and $y$ are not the elements of $\mathbb{R}$. $\mathbb{R}(x,y)$ is an algebraic function field of one variable. $x$ is transcendental over $\mathbb{R}$ implies that it is not in $\mathbb{R}$ and $y$ is algebraic over $\mathbb{R}(x)$. – bharath Nov 07 '14 at 09:14
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OK! Sorry I did not understood it like that. You should perhaps add that $x \in L$, $L$ being an extension of $\mathbb{R}$ (presumably $\mathbb{C}$). – Martigan Nov 07 '14 at 09:17
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@Martigan I have edited the question accordingly. hopefully now you can give me an answer. thanks – bharath Nov 07 '14 at 09:31
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Are you familiar with the isomorphism $\mathbf{C}(u)[v]/(u^2 + v^2 - 1) \cong \mathbf{C}(t)$? It might be useful to compare with that. – Nov 07 '14 at 09:38