Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 89 / 4

Use the polynomial graded ring $R[x_0,x_1,x_2,x_3]$, as in the PDF; the doubled brackets in the converted TeX are an OCR error. Let $I$ denote the homogeneous defining ideal. The four cubic coordinates in the proposed map never vanish simultaneously on ${\mathbb{P}}_F^1$, since $y_0^3$ and $y_1^3$ already have that property. Substitution makes each generator of $I$ zero, so they define a morphism into the generic fibre.

The two standard opens $D_{+}(x_0)$ and $D_{+}(x_3)$ cover $X$. Indeed, a homogeneous prime containing $x_0,x_3$ contains $x_2$ by $x_0{x}_3^2=x_2^3$, and then contains $x_1$ by $x_1^2={\pi }^2{x}_0{x}_2$. Such a prime contains the irrelevant ideal and gives no point of Proj.

On $D_{+}(x_0)$ put $a=x_1/x_0$, $b=x_2/x_0$, $c=x_3/x_0$. Over $F$, where $\pi$ is invertible, the equations giveThus the generic-fibre chart is exactly $\mathrm{Spec}F[t]$. The proposed map restricts to this isomorphism with $t=y_1/y_0$. On $D_{+}(x_3)$ put $s=x_2/x_3$. The equations giveso this chart is $\mathrm{Spec}F[s]$, and the map has $s=y_0/y_1$. On the overlap $s=t^{-1}$. These are the standard two charts of the projective line, proving the scheme-level isomorphismEquivalently, rescaling $x_1$ by ${\pi }^{-1}$ gives the usual twisted cubic. The chart proof also shows that the extra cubic relation introduces no additional generic-fibre scheme structure.

On $D_{+}(x_3)$, write $r=x_0/x_3$, $q=x_1/x_3$, and $s=x_2/x_3$. The equations include $q=\pi s^2$ and $r=s^3$; after making these substitutions, the remaining equations are identities. ThereforeThe other chart isThere is a homomorphism $B\to R[t]$ defined by $a\mapsto \pi t$, $b\mapsto t^2$, $c\mapsto t^3$. To prove injectivity rather than merely exhibit a parametrization, reduce monomials using the four relations. Powers of $a$ reduce to at most one $a$; any remaining product of $a$ with $b$ or $c$ eliminates $a$; powers of $c$ reduce to at most one $c$. Thus every element is an $R$-linear combination ofTheir images are $1,\pi t,t^{2n},t^{2n+3}$, with pairwise distinct degrees. Since $R$ is a domain and $\pi \ne 0$, these images are $R$-linearly independent. Hence they are a basis, the map is injective, andIn particular $B$ is torsion-free, indeed free, as an $R$-module.

The two charts cover $X$ by the prime-ideal argument in part i. Both coordinate rings are torsion-free over the discrete valuation ring $R$, hence flat: over a principal ideal domain a torsion-free module is a filtered union of finitely generated free submodules, and filtered colimits preserve flatness. Thus the morphism $X\to \mathrm{Spec}R$ is flat on an open cover and therefore flat everywhere. It is proper because $X$ is a closed subscheme of the proper projective space ${\mathbb{P}}_R^3$. Consequently $X$ is both proper and flat over $R$.

The overlap of the two charts is obtained by inverting $c=t^3$. Since $c^2=b^3$, this also inverts $b=t^2$; then $t=c/b$ and $t^{-1}=b/c$. Thus its coordinate ring is $R[t,t^{-1}]$. The other chart has coordinate $s=t^{-1}$. A global section is therefore an element in the intersectioninside $R[t,t^{-1}]$: a Laurent polynomial with both only nonnegative and only nonpositive exponents is constant.

Let $k=R/(\pi )$. Reducing the presentation of $B$ givesAs a vector space and as a ring extension, this iswhere $a^2=0$ and $a$ is annihilated by $b,c$. In particular $a\ne 0$: the free basis of $B$ in part ii remains a basis after tensoring with $k$. Although $a$ mapped to $\pi t$ before reduction, reducing the subring is not the same operation as taking its image inside $k[t]$; that distinction is exactly where the extra section arises.

On the overlap, $c$ is invertible and $ac=0$, so $a$ restricts to zero. Therefore the section $a$ on the first chart glues to zero on the second chart and defines a nonzero global nilpotent section $\eta$ on $X_s$. More precisely the reduced first-chart ring is $k[t^2,t^3]$, andin $k[t,t^{-1}]$. The only extra compatible sections are multiples of $a$. HenceThe map printed in the question reduces $R$ modulo $\pi$ and includes its image as constants. Its kernel is $\pi R$ and its cokernel is $k\eta$. More strongly, even the usual base-change map after tensoring,is not surjective. This is global-section base-change failure in a flat projective family, despite the properness and flatness proved above.