Let \(Y\) be the curve \(y^2 = x^3 - x\) in \(\AA^2\), and assume \(\operatorname{char} k \neq 2\). This exercise shows that \(Y\) is not a rational curve, and hence that \(K(Y)\) is not a purely transcendental extension of \(k\).
Show that \(Y\) is nonsingular, and deduce that \(A = A(Y) \cong k[x,y]/(y^2 - x^3 + x)\) is an integrally closed domain.
Let \(k[x]\) be the subring of \(K = K(Y)\) generated by the image of \(x\) in \(A\). Show that \(k[x]\) is a polynomial ring, and that \(A\) is the integral closure of \(k[x]\) in \(K\).
Show that there is an automorphism \(\sigma: A \to A\) sending \(y\) to \(-y\) and fixing \(x\). For \(a \in A\), define the norm of \(a\) to be \(N(a) = a \cdot \sigma(a)\). Show that \(N(a) \in k[x]\), that \(N(1) = 1\), and that \(N(ab) = N(a) N(b)\) for all \(a, b \in A\).
Using the norm, show that the units in \(A\) are precisely the nonzero elements of \(k\). Show that \(x\) and \(y\) are irreducible elements of \(A\), and that \(A\) is not a unique factorization domain.
Prove that \(Y\) is not a rational curve.
Solution
Put \[ F=y^2-x^3+x,qquad A=k[x,y]/(F),qquad K=\operatorname{Frac}(A). \] We write \(x,y\) also for their images in \(A\).
1. The affine curve \(Y=V(F)\) is nonsingular and irreducible; hence \(A\) is an integrally closed domain.
The partial derivatives are \[ F_x=1-3x^2,qquad F_y=2y. \] At a singular point we would have \(y=0\), because \(2\ne0\) in \(k\), and the equation of the curve would give \[ x(x-1)(x+1)=0. \] At \(x=0\) we have \(F_x=1\). At \(x=\pm1\) we have \(F_x=-2\), which is also nonzero because \(\operatorname{char}k\ne2\). Thus \(Y\) has no singular point.
The polynomial \(F\), viewed as a monic quadratic in \(y\) over \(k[x]\), is irreducible. Indeed, reducibility would make \(x^3-x=x(x-1)(x+1)\) a square in \(k[x]\); its three roots are distinct when \(\operatorname{char}k\ne2\), so this is impossible. Therefore \(A\) is a domain.
Every local ring of the nonsingular curve \(Y\) is a regular local ring of dimension one, hence a discrete valuation ring and therefore integrally closed. Normality is local, so the affine coordinate ring \(A\) is integrally closed in its fraction field. This proves part (a).
2. The subring generated by \(x\) is a polynomial ring \(k[x]\), and \(A\) is finite integral over it.
If a polynomial \(p(T)\in k[T]\) satisfied \(p(x)=0\) in \(A\), then \(p(x)\) would lie in the principal ideal \((F)\subseteq k[x,y]\). Division by the monic polynomial \(F\) in the variable \(y\) leaves \(p(x)\) as its own remainder, so \(p=0\). Thus \(x\) is transcendental over \(k\) and the indicated subring is a copy of \(k[x]\).
The relation \[ y^2=x^3-x \] is monic in \(y\). Hence \(y\) is integral over \(k[x]\) and \[ A=k[x]\oplus k[x]y \] as a \(k[x]\)-module. In particular, \(A\) is finite and integral over \(k[x]\).
3. The ring \(A\) is the integral closure of \(k[x]\) in \(K\).
2.Conversely, let \(u\in K\) be integral over \(k[x]\).The same monic equation also has its coefficients in the larger ring \(A\), so \(u\) is integral over \(A\).Step 1 says that \(A\) is integrally closed in \(K\), hence \(u\in A\).Thus \(A\) is exactly the integral closure, proving part (b).
4. The involution \(\sigma(x)=x\), \(\sigma(y)=-y\) defines an automorphism of \(A\), and for \(a=p(x)+yq(x)\),\[ \boxed{N(a)=p(x)^2-(x^3-x)q(x)^2}. \]
The substitution \(y\mapsto-y\) fixes the relation \(F\), so it induces a \(k\)-algebra automorphism \(\sigma\) of \(A\) with \(\sigma^2=1\). The decomposition in step 2 is unique, so every \(a\in A\) is \(p+yq\) with \(p,q\in k[x]\). Then \[ N(a)=a\sigma(a)=(p+yq)(p-yq)=p^2-y^2q^2=p^2-(x^3-x)q^2\in k[x]. \] Also \(N(1)=1\), and because \(A\) is commutative and \(\sigma\) is an automorphism, \[ N(ab)=ab\,\sigma(a)\sigma(b)=N(a)N(b). \] This proves part (c).
5. The units of \(A\) are precisely \(k^\times\).
4.If \(q\ne0\), the two terms of\[ N(a)=p^2-(x^3-x)q^2 \]have degrees \(2\deg p\) and \(3+2\deg q\) when nonzero.One degree is even and the other odd, so their leading terms cannot cancel; the norm then has positive degree.Thus \(q=0\).Now \(N(a)=p^2\) is constant, so \(p\in k^\times\).Conversely every nonzero constant is a unit.
6. The elements \(x\), \(x-1\), \(x+1\), and \(y\) are irreducible in \(A\).
5.Moreover, the norm formula in step 4 shows that a nonconstant norm cannot have degree one: if \(q=0\) its degree is the positive even integer \(2\deg p\), while if \(q\ne0\) its degree is the maximum of an even integer and the odd integer \(3+2\deg q\), hence at least two.
For \(c\in\{0,1,-1\}\), \[ N(x-c)=(x-c)^2. \] If \(x-c=ab\) with both \(a,b\) nonunits, multiplicativity would give a factorization of this degree-two polynomial into the two nonconstant polynomials \(N(a),N(b)\), each of degree at least two, which is impossible. Thus each \(x-c\) is irreducible.
Similarly, \[ N(y)=-(x^3-x)=-x(x-1)(x+1) \] has degree three. If \(y=ab\) with both factors nonunits, then \(N(a)\) and \(N(b)\) would again both have degree at least two, so their product would have degree at least four, contradiction. Hence \(y\) is irreducible.
7. The domain \(A\) is not a unique factorization domain.
6.By step 5 the only units are nonzero constants, so \(y\) is not associate to any of \(x,x-1,x+1\), and those three are pairwise nonassociate.Thus the two sides are genuinely different factorizations into irreducibles.Therefore \(A\) is not a unique factorization domain, completing part (d).
8. The curve \(Y\) is not rational.
The curve is nonsingular by step 1 and affine by definition. It is not isomorphic to \(\PP^1\): the nonconstant element \(x\in A=\Gamma(Y,\OO_Y)\) is a global regular function, whereas every global regular function on \(\PP^1\) is constant.
7.Hence \(Y\) is not rational.Since a rational curve has function field \(k(t)\), it follows equivalently that \(K(Y)\) is not purely transcendental over \(k\).This proves part (e).
9. Q.E.D.
Step 1 proves part (a), steps <1>2–<1>3 prove part (b), step 4 proves part (c), steps <1>5–<1>7 prove part (d), and step 8 proves part (e).