Let \(f\) be the rational function on \(\PP^2\) given by \(f = x_1/x_0\). Find the set of points where \(f\) is defined, and describe the corresponding regular function.
Now regard this function as a rational map from \(\PP^2\) to \(\AA^1\). Embed \(\AA^1\) in \(\PP^1\), and let \(\varphi: \PP^2 \dashrightarrow \PP^1\) be the resulting rational map. Find the set of points where \(\varphi\) is defined, and describe the corresponding morphism.
Solution
1. On the standard affine chart\[ D_+(x_0)=\{[x_0:x_1:x_2]:x_0\ne0\}, \]the rational function \(f=x_1/x_0\) is regular.
The chart \(D_+(x_0)\) is isomorphic to \(\AA^2\) with affine coordinates \[ u_1=\frac{x_1}{x_0}, \qquad u_2=\frac{x_2}{x_0}. \] Under this identification, \(f\) is exactly the coordinate function \(u_1\).
2. The rational function \(f\) is not regular at any point of the hyperplane \(H_0=Z(x_0)\).
First let \[ Q\in H_0\cap D_+(x_1). \] On the affine chart \(D_+(x_1)\), the function \[ u=\frac{x_0}{x_1} \] is regular and vanishes at \(Q\). If \(f=x_1/x_0\) were regular near \(Q\), then on the nonempty open set where both are defined we would have \[ u f=1. \] Equality of regular functions would extend this identity to a neighborhood of \(Q\), but evaluation at \(Q\) would give \(0=1\), a contradiction.
The only point of \(H_0\) not covered by \(D_+(x_1)\) is \[ P=[0:0:1]. \] On the affine chart \(D_+(x_2)\) around \(P\), put \[ u=\frac{x_0}{x_2}, \qquad v=\frac{x_1}{x_2}. \] Then \[ \mco_{P,\PP^2}\cong k[u,v]_{(u,v)}, \qquad f=\frac vu. \] If \(v/u=a/s\) in this local ring with \(s\notin(u,v)\), then \[ v s=u a. \] Since \(k[u,v]\) is a UFD and \(u\) does not divide \(v\), it must divide \(s\), contradicting \(s\notin(u,v)\). Thus \(f\) is not regular at \(P\) either.
3. The maximal domain of definition of \(f\) is\[ \boxed{D(f)=D_+(x_0)\cong\AA^2}, \]and the corresponding regular function is the affine coordinate \(x_1/x_0\).
Step 1 gives regularity on \(D_+(x_0)\), while step 2 excludes every point of its complement. Maximality now follows from P-AGH41RATFNDOM. This proves (a).
4. After the standard embedding\[ \AA^1\hookrightarrow\PP^1, \qquad t\longmapsto[1:t], \]the induced rational map is\[ \varphi([x_0:x_1:x_2])=[x_0:x_1]. \]
On \(D_+(x_0)\), \[ [1:f]=\left[1:\frac{x_1}{x_0}\right]=[x_0:x_1]. \] Thus the displayed homogeneous pair represents the same rational map. The two linear forms \(x_0,x_1\) have no common zero away from \[ P=[0:0:1], \] so they define a morphism \[ \PP^2\setminus\{P\}\longrightarrow\PP^1. \]
5. The rational map \(\varphi\) is not defined at \(P=[0:0:1]\).
Suppose that \(\varphi\) extended to a morphism on a neighborhood \(W\) of \(P\). Write \[ \varphi(P)=[a:b]. \] At least one of \(a,b\) is nonzero.
If \(a\ne0\), then after shrinking \(W\) we may assume the image lies in the target chart \(D_+(y_0)\). Its affine coordinate \(y_1/y_0\) would pull back to a regular function near \(P\). On the dense open subset where \(x_0\ne0\), that pullback is \[ \frac{x_1}{x_0}, \] contradicting step 2.
If \(b\ne0\), use instead the target chart \(D_+(y_1)\). Then \(y_0/y_1\) would pull back to the rational function \[ \frac{x_0}{x_1}, \] which is likewise not regular at \(P\) by the same local-ring argument with \(u\) and \(v\) interchanged. Both possibilities are impossible, so no extension exists at \(P\).
6. Therefore\[ \boxed{D(\varphi)=\PP^2\setminus\{[0:0:1]\}}, \]and on this maximal domain\[ \boxed{\varphi([x_0:x_1:x_2])=[x_0:x_1]}. \]
Step 4 gives the morphism on the displayed open set, and step 5 proves that the only missing point cannot be added. By P-AGH42RATMAPDOM, this is the maximal domain of definition. This proves (b).
7. Q.E.D.
Steps <1>1–<1>3 prove (a), and steps <1>4–<1>6 prove (b).
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Dependencies
P-AGH41RATFNDOMGluing regular functions and the domain of definition of a rational functionusesP-AGH42RATMAPDOMLargest open set on which a rational map is a morphismuses