$\phi(x,y)=\frac{y-x^2}{x^2}$ for $\phi:X\to\mathbb{A}^1(\mathbb{C})$
$X$ being a variety $X=V(\langle x^5-x^4+2x^2y-y^2\rangle) \subset \mathbb{A}^2(\mathbb{C})$
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Does $X=A$? Because you didn't define $X$. – Patrick Da Silva Apr 21 '16 at 20:31
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Sorry for the confusion, I fixed it – B DIll Apr 21 '16 at 20:34
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You can use "\to" to get $\to$. :) – Eman Yalpsid Apr 21 '16 at 20:37
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A rational map is an equivalence class of maps defined on open subsets of $X$. The map you defined above is defined on the open subset of $X \subseteq \mathbb A^2(\mathbb C)$ where $x \neq 0$, so it represents a rational map.
Hope that helps,
Patrick Da Silva
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can you explain how I show explicitly that the map is defined on the set X? – B DIll Apr 21 '16 at 20:46
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@B Dill : It's not "defined as a map" on $X$, as a map, it's only defined on an open subset. But this means it defines a rational map on $X$ by definition. – Patrick Da Silva Apr 21 '16 at 20:50
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Does it suffice for me to say, "x5−x4+2x2y−y2 are elements of the function field of X therefore φ is a rational map on X? – B DIll Apr 21 '16 at 20:58
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@B Dill : The map $x^5 - x^4 + 2x^2 - y^2$ is identically zero on $X$. Actually, all you need to show is that the set of points in $\mathbb A^2(\mathbb C)$ where $\phi$ is defined intersects $X$ in a non-empty set, so that the rational map is well-defined. I suggest you read up on your definitions, you sound confused! – Patrick Da Silva Apr 21 '16 at 21:02