Consider the curves following in $\mathbb{A}_\mathbb{C}^2$: \begin{align*} &L:&y=0\\ &H_1:&xy-1=0\\ &H_2:&x^2-y^2-1=0\\ &C:&x^2+y^2-1=0\\ \end{align*} Between $H_1$ and $H_2$: \begin{align*} H_1&\rightarrow H_2\\ (x,y)&\rightarrow (\frac{x+y}{2},\frac{y-x}{2})\\ H_2&\rightarrow H_1\\ (x,y)&\rightarrow (x-y,x+y)\\ \end{align*} Between $H_1$ and $L$: \begin{align*} H_1&\rightarrow L\\ (x,y)&\rightarrow (x,0)\\ L&\rightarrow H_1\\ (x,y)&\rightarrow(x,\frac{1}{x}) \end{align*} Between $H_1$ and $C$: \begin{align*} H_1&\rightarrow C\\ (x,y)&\rightarrow (\frac{x+y}{2},i\frac{y-x}{2})\\ C&\rightarrow H_1\\ (x,y)&\rightarrow (x+iy,x-iy)\\ \end{align*} By the definition in $\textit{Basic Algebraic Geometry}$ by $\textit{Igor R.Schaferevich}$, page 12 line 10, these curves are equivalent. But indeed the rational function field of the line and that of the circle and two hyperbolas are not isomorphic. Where did I misunderstand?

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    $\begingroup$ The function field of the line and of the hyperbola are indeed isomorphic. The line has function field $k(t)$ and the hyperbola has function field $\text{frac } k[x,y]/(xy-1)$, which is isomorphic to $\text{frac } k[x,1/x]$, which is $k(t)$. With more care, one can show your morphisms above induce an isomorphism of function fields. $\endgroup$ – hwong557 Dec 21 '16 at 16:10
  • $\begingroup$ @hwong557: If so, according to this math.stackexchange.com/questions/1057321/… (that I find quite believable), the line is birationally ismorphic with the circle? $\endgroup$ – quangtu123 Dec 21 '16 at 18:04
  • $\begingroup$ $H_1$, $H_2$, and $C$ are all isomorphic, not just birational. So the line is birational to the three conics you mentioned. $\endgroup$ – hwong557 Dec 21 '16 at 19:06
  • $\begingroup$ @hwong557 the the field of rational functions on the line k(x) is the same as the field of rational functions on the circle k(C)? $\endgroup$ – quangtu123 Dec 21 '16 at 20:16
  • $\begingroup$ Yes. I've already described how rational functions on the line are the same as rational functions on hyperbola. But the hyperbola is isomorphic to the circle, so rational functions on the hyperbola and on the circle are the same. $\endgroup$ – hwong557 Dec 21 '16 at 20:18

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