A variety \(Y\) is rational if it is birationally equivalent to \(\PP^n\) for some \(n\); equivalently by (4.5), if \(K(Y)\) is a purely transcendental extension of \(k\).
Show that any conic in \(\PP^2\) is a rational curve.
Show that the cuspidal cubic \(y^2 = x^3\) is a rational curve.
Let \(Y\) be the nodal cubic curve \(y^2 z = x^2(x+z)\) in \(\PP^2\). Show that the projection \(\varphi\) from the point \(P = (0,0,1)\) to the line \(z = 0\) induces a birational map from \(Y\) to \(\PP^1\). Thus \(Y\) is a rational curve.
(The word nodal in characteristic two)
At \(P=[0:0:1]\), use the affine chart \(z=1\). The equation is \[ y^2=x^2(x+1), \] whose lowest-degree part is \[ y^2-x^2=(y-x)(y+x). \] If \(\operatorname{char}k\ne2\), these are two distinct tangent lines, so \(P\) is an ordinary node. If \(\operatorname{char}k=2\), the tangent cone is \((y-x)^2\), so the singularity is not an ordinary node. The birational calculation above does not divide by \(2\) and remains valid in either characteristic.
Solution
1. Every conic in \(\PP^2\) is a rational curve.
By P-AGH31CONICS, every projective conic is isomorphic to \(\PP^1\). An isomorphism is in particular a birational equivalence. Hence every conic is rational, proving (a).
2. The cuspidal cubic\[ C=Z(y^2-x^3)\subseteq\AA^2 \]has function field \(K(C)\cong k(t)\).
As in P-AGH317NORMAL, its coordinate ring is \[ A(C)\cong k[t^2,t^3] \] under \[ x\longmapsto t^2, \qquad y\longmapsto t^3. \] Therefore \[ K(C)=\operatorname{Frac}k[t^2,t^3]. \] This fraction field contains \[ t=\frac{t^3}{t^2}, \] and conversely \(t^2,t^3\in k(t)\). Thus \[ K(C)=k(t), \] a purely transcendental extension of \(k\) of degree one. Hence \(C\) is rational, proving (b).
3. Projection from\[ P=[0:0:1] \]to the line \(z=0\) is the rational map\[ \varphi:Y\dashrightarrow\PP^1, \qquad [x:y:z]\longmapsto[x:y]. \]
This is the coordinate form of projection from a point established in P-AGH314PROJPOINT. On \(Y\), the simultaneous equations \(x=y=0\) give only the point \(P\). Thus the displayed formula is a morphism on \(Y\setminus\{P\}\) and represents the desired rational map.
4. On the dense open subset\[ U=Y\cap D_+(x), \]the map \(\varphi\) is an isomorphism onto\[ V=\{[1:t]\in\PP^1:t^2\ne1\}. \]
If \([x:y:z]\in U\), then \(z\ne0\): otherwise the cubic equation with \(z=0\) would give \(x^3=0\), contradicting \(x\ne0\). Normalize to \(z=1\) and put \[ t=\frac yx. \] The equation \[ y^2=x^2(x+1) \] then gives \[ t^2=x+1, \] so \[ x=t^2-1, \qquad y=t(t^2-1). \] Because \(x\ne0\), we have \(t^2\ne1\).
Conversely define \[ \psi:V\to Y, \qquad [1:t]\longmapsto[t^2-1:t(t^2-1):1]. \] Substitution gives \[ t^2(t^2-1)^2=(t^2-1)^2((t^2-1)+1), \] so the image lies in \(Y\); because \(t^2\ne1\), it lies in \(U\). Both coordinate formulas are regular on the indicated affine opens. Finally, \[ \varphi(\psi([1:t]))=[t^2-1:t(t^2-1)]=[1:t], \] while the formulas derived above show \(\psi(\varphi(Q))=Q\) for \(Q\in U\). Thus \(U\cong V\).
5. The projection \(\varphi\) is birational and \(Y\) is rational.
Step 4 gives an isomorphism between nonempty open subsets of \(Y\) and \(\PP^1\). Therefore \(\varphi\) is a birational map. Hence \(Y\) is a rational curve, proving (c).
6. Q.E.D.
Steps 1, 2, and <1>3–<1>5 prove (a), (b), and (c), respectively.
Source Collections
Dependencies
P-AGH31CONICSConics, and varieties that are not isomorphic to one anotherusesP-AGH317NORMALNormal varieties and the normalization of an affine varietyusesP-AGH314PROJPOINTProjection from a point, and the cuspidal cubic image of the twisted cubicuses