Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 21 (Algebraic Geometry)

Use the usual algebraically closed field convention for classical algebraic sets, or equivalently interpret their points geometrically. The proof also gives the scheme-theoretic statement over any field; the difference from an arbitrary-field rational-point image is explained below.

Choose an affine open cover $X={\bigcup }_i{U}_i$. On each $U_i$, the subset $Y_i=Y\cap (U_i\times {\mathbb{P}}^n)$ is closed relative to $U_i\times {\mathbb{P}}^n$. If the affine-base assertion holds, its projected image is closed in $U_i$. ButThus the complement of $\pi (Y)$ is open on each member of the open cover and hence open in $X$. Therefore it suffices to prove closedness for affine $X$. No finite-cover assumption is needed for this localization argument.

Put $I=(f_1,\dots ,f_r)$ and $\mathfrak{m}=(y_0,\dots ,y_n)$. If ${\mathfrak{m}}^N\subseteq I$, a projective common zero would have some coordinate $y_j\ne 0$. For $N>0$, the polynomial $y_j^N\in I$ could not vanish there; for $N=0$, the inclusion puts $1$ in $I$, also impossible. Thus the projective zero set is empty.

Conversely, over an algebraically closed field, emptiness of the projective zero set means the affine common zero set is contained in the origin. Every $y_j$ vanishes there, so the Hilbert Nullstellensatz puts $y_j$ in $\sqrt{I}$. Choose integers $e_j\ge 1$ with $y_j^{e_j}\in I$, and setEvery degree-$N$ monomial has some exponent at least the corresponding $e_j$, by the pigeonhole principle, and is consequently in $I$. Those monomials generate ${\mathfrak{m}}^N$, provingThe argument includes the unit-ideal case. This proves the Projective Nullstellensatz criterion directly from its affine counterpart.

Over a general field, this same criterion concerns geometric zeros: apply the proof after extension to an algebraic closure. Inclusion of each degree-$N$ monomial descends back to $k$, since its class in $k[y]/I$ is zero if and only if it becomes zero after a field extension; tensoring a vector space with an extension field is faithful. If instead one counts only $k$-rational projective points, the assertion is false. For example $y_0^2+y_1^2$ has no real projective zero but has the complex zero $[1:i]$, so its ideal cannot contain any power of $\mathfrak{m}$.

For an affine base, use the homogeneous equations furnished by the question, with fixed degrees $d_i$ in the projective coordinates. At a point $x$, the fiber $Y_x$ is the projective zero set of their specializations; a specialization may be zero, but is still assigned its original degree. LetThen $x\in \pi (Y)$ exactly when $Y_x$ is nonempty. Part (b) says this is equivalent to $I_x$ containing no power of the irrelevant ideal of projective space. In logical form, noncontainment must hold for every $N$, soFor schemes over an arbitrary field, replace $k$ in this fiber calculation by the residue field $\kappa (x)$ and use geometric fiber points. A point belongs to the scheme-theoretic topological image exactly when its fiber is nonempty, which is equivalent to nonemptiness after algebraic closure of that residue field.

By part (a), work first with an affine base and its homogeneous equations. Put $S=k[y_0,\dots ,y_n]$, write $S_N$ for its finite-dimensional degree-$N$ vector space, and use $S_j=0$ for $j<0$. At each base point define the graded multiplication map for a projective fiberIts image is exactly $(I_x{)}_N$, since taking the degree-$N$ part of any expression in the homogeneous ideal discards all other degrees. Moreoverbecause degree-$N$ monomials generate ${\mathfrak{m}}^N$.

Choose monomial bases for the domain summands and codomain. Every entry of the resulting matrix is a coefficient of one of the $f_i$, hence a regular function of $x$. If $h_N=\dim S_N$, failure of surjectivity is the vanishing of all $h_N$-by-$h_N$ matrix minors. It is therefore a closed rank-defect locus $C_N$. If the matrix has fewer than $h_N$ columns, the failure is automatic and $C_N=X$, also closed. Constants and $N=0$ are covered by the same construction.

Part (c) gives $\pi (Y)={\bigcap }_{N\ge 0}{C}_N$, an intersection of closed subsets, hence closed. Apply part (a) to pass back to arbitrary $X$. Thus projection from $X\times {\mathbb{P}}^n$ sends every closed subset to a closed subset. This proves the closedness of projection from projective space without assuming the properness result being proved.

The field convention matters: a rational-point projection need not be Zariski closed. Over $\mathbb{R}$, the closed equation $y_0^2=t{y}_1^2$ in ${\mathbb{A}}^1\times {\mathbb{P}}^1$ has real-point image $[0,\infty )$. Indeed $y_1=0$ would force $y_0=0$, so a solution exists precisely when $t$ is a square in $\mathbb{R}$. This infinite proper subset of the affine line is not Zariski closed. The valid arbitrary-field theorem uses scheme or geometric images, for which the matrix argument remains valid at every residue field.

Here the degree-zero Picard group of a curve uses tensor product as its group operation. The map is the Abel map of a pointed smooth projective curve. If $\alpha (p)=\alpha (q)$, then $\mathcal{O}(p-q)$ is trivial. By the divisor description of line bundles, there is a nonzero rational function $f$ with principal divisor $(f)=p-q$.

If $p\ne q$, this function has exactly one pole, of order one. It defines a nonconstant morphism of algebraic varieties $X\to {\mathbb{P}}^1$: locally at a pole one uses $1/f$ as the target coordinate. The degree of a morphism of curves is the total pole degree, so it is one. The map is finite, since a nonconstant map of projective integral curves has finite fibers and is proper, and a degree-one finite map to the normal projective line is an isomorphism. Algebraically, on a target affine chart the finite coordinate algebra lies inside the common function field and is integral over the integrally closed coordinate ring, so it equals that ring. Thus $X\cong {\mathbb{P}}^1$ and its genus is zero, contradicting $g\ge 1$. ConsequentlyThis spells out why a principal divisor with one simple zero and one simple pole cannot occur on a curve of positive genus.

If $g=0$, the chosen point $p_0$ is enough to identify $X$ with ${\mathbb{P}}^1$, even over a nonclosed field. By the Riemann-Roch theorem, $h^0(\mathcal{O}(p_0))=2$, since the complementary canonical twist has degree $-3$ and no nonzero section. Besides constants there is a rational function with its sole pole a simple pole at $p_0$, giving the same degree-one isomorphism. On ${\mathbb{P}}^1$, every degree-zero divisor is principal: over an algebraically closed field write a product of powers of linear forms for its finite point coefficients, with the degree-zero condition giving the coefficient at infinity. Over a general field the same construction uses irreducible polynomial factors for closed points, weighted by their degrees. Therefore

Let $L$ be any degree-zero line bundle on the genus-one smooth projective curve. The twist $M=L\otimes \mathcal{O}(p_0)$ has degree one. The Riemann-Roch theorem givesThe second line bundle has degree $2g-2-1=-1$, so it has no nonzero section: the zero divisor of such a section would be effective of negative degree. Hence $h^0(M)=1$. A nonzero section has an effective zero divisor of degree one, which is a single point $p$. It is a $k$-rational point even over a general field, since a degree-one effective divisor has residue degree one. Thus $M\cong \mathcal{O}(p)$ and $L\cong \mathcal{O}(p-p_0)=\alpha (p)$.

This proves surjectivity; part (a) gives injectivity. Hence $\alpha$ is a bijection, as in the Abel-Jacobi map of a genus-one curve. DefineTensor product of line bundles is associative and commutative, so this is an abelian group law on $X(k)$. Its identity is $p_0$, since $\alpha (p_0)={\mathcal{O}}_X$. Equivalently, $p\oplus q$ is the unique point $r$ with $\mathcal{O}(r-p_0)\cong \mathcal{O}(p+q-2p_0)$.

First use the algebraically closed field convention. Every line bundle on a smooth projective curve has a nonzero rational section of a line bundle: trivialize it on a nonempty open set and take its generic generator. The orders of its zeros and poles give a finite divisor $D$, and the local trivializations identify the line bundle with $\mathcal{O}(D)$. For a degree-zero line bundle writesince every closed point is a $k$-point. ThereforeOnly finitely many factors occur, and negative tensor powers mean dual line bundles. This provesFor genus zero the group is trivial, so the conclusion remains valid.

For a general field, a closed point need not be rational, and its contribution to degree is weighted by its residue degree. Consequently rational points need not generate the degree-zero Picard group of a curve. Here is an explicit counterexample to that stronger reading. Let $C$ be the smooth projective curve over ${\mathbb{F}}_2$ obtained by completing the affine equationThe polynomial is irreducible over ${\overline{\mathbb{F}}}_2(x)$: a function $r^2+r$ has even pole order at infinity when it has a pole there, while the right side has pole order five. Its affine model is smooth because the partial derivative with respect to $y$ is one. The normalized projective completion is therefore a geometrically integral smooth projective curve.

There is exactly one point $p_{\infty }$ above infinity, and it is rational. Indeed the map $x:C\to {\mathbb{P}}^1$ has degree two. At a point above infinity put $v(x)=-e$, where the ramification index $e$ is at most two. The equation forces $v(y)<0$ and $2v(y)=5v(x)=-5e$. Hence $e=2$ and $v(y)=-5$. The degree formula then leaves exactly one point of residue degree one. Thus $x$ and $y$ have pole orders two and five there. There are no affine rational points: for either $x\in {\mathbb{F}}_2$, the right side is one, while $y^2+y=0$ for every $y\in {\mathbb{F}}_2$. Hence $C({\mathbb{F}}_2)=\{p_{\infty }\}$ and the image of its Abel map of a pointed smooth projective curve is trivial.

The equations $y=0$, $x^2+x+1=0$ define a degree-two closed point $D$, since $x^5+x+1$ vanishes modulo $x^2+x+1$. The divisor $D-2p_{\infty }$ has degree zero but is not principal. To prove this, a function with poles bounded by $2p_{\infty }$ is regular on the affine curve and hence has a unique expression $A(x)+B(x)y$. The possible pole orders of its two terms are $2\mathrm{deg}A$ and $2\mathrm{deg}B+5$, of opposite parity, so they cannot cancel. The pole bound forces $B=0$ and $\mathrm{deg}A\le 1$. But no nonzero $a+bx$ over ${\mathbb{F}}_2$ vanishes at $D$, whose residue class of $x$ has irreducible quadratic minimal polynomial. Thus no function has principal divisor $D-2p_{\infty }$, proving its class is nonzero. This establishes the rational points need not generate the degree-zero Picard group qualification with a complete obstruction, while the generation proof above supplies the intended algebraically closed-field answer.

Over ${\overline{\mathbb{F}}}_3$, take the projective cubicIn characteristic three its partial derivatives are $F_X=Z^2$, $F_Y=2YZ$ and $F_Z=Y^2+2XZ$. If they vanished simultaneously, $Z=0$ and then $Y=0$. The equation of $C$ would force $X=0$, impossible for a projective point. Thus the Jacobian criterion proves smoothness everywhere, including infinity.

It is an irreducible cubic: on the affine chart $Z=1$, the polynomial is $y^2-(x^3-x)$, and $x^3-x$ is not a square in ${\overline{\mathbb{F}}}_3(x)$ because its pole at infinity has odd order three. A quadratic in $y$ would factor over that field only if its right side were a square. Gauss lemma for polynomials then gives affine irreducibility, and homogenization introduces no factor at infinity because $Z$ does not divide $F$. This supplies the smooth plane cubic in characteristic three. In particular, choosing a Fermat cubic here would fail, because its cubic partial derivatives vanish in characteristic three.

Take the projective hypersurfaceAs a polynomial in $X_0$ over $\mathbb{C}[X_1,X_2,X_3]$, it is primitive because $X_1$ and $X_2^2$ are coprime. It is linear and irreducible over the fraction field, so Gauss lemma for polynomials proves irreducibility. Thus it is an irreducible surface.

Its gradient is $(X_1,X_0,-2X_2,0)$. The simultaneous zero set on $S$ is exactlyThe Jacobian criterion proves every other point is smooth. At that vertex the affine equation in the chart $X_3=1$ has no linear term, so the tangent space has dimension three whereas the surface has dimension two; the vertex is genuinely singular. This is a projective quadric cone with one singular vertex.

Take the plane quartic with exactly two nodesPut $d=X^2-Z^2$. Its partial derivatives areOn $Y=0$, the equation gives $X^2=Z^2$, producing exactly $P_{+}=[1:0:1]$ and $P_{-}=[-1:0:1]$, and their gradients vanish.

There are no singular points with $Y\ne 0$. Normalize $Y=1$. The derivative equations give $X=4dZ$ and $(16d^2+1)Z=0$. If $Z=0$, then $X=0$ and $F_Y=4\ne 0$, so $Z\ne 0$ and $d^2=-1/16$. The remaining equation gives $XZ=-2$, hence $Z^2=-1/(2d)$. But $d=X^2-Z^2=-2Z^2=1/d$, forcing $d^2=1$, a contradiction. Thus the singular set is exactly $\{P_{+},P_{-}\}$.

In the chart $Z=1$ near $P_{\epsilon }$, put $X=\epsilon +u$, $Y=v$ with $\epsilon =\pm 1$. The lowest-degree part is $4u^2+\epsilon v^2$, which splits into two distinct complex linear factors. Each singularity is therefore an ordinary double point, with two smooth transverse branches.

It remains to prove irreducibility. A repeated polynomial factor would make every point on that component singular, contradicting the finite singular set, so $F$ is reduced. If it were reducible, partition its irreducible factors into two coprime factors of degrees $r$ and $4-r$, $1\le r\le 3$. Every intersection of the two components would be singular on $C$, hence one of the two ordinary double points. At an ordinary double point shared by them, the two branches are transverse, so its intersection multiplicity is one. Their total intersection multiplicity would therefore be at most two. The Bézout theorem instead makes it $r(4-r)\ge 3$. This contradiction proves the displayed quartic is irreducible and has exactly two singular points.

Being constant on the two-element orbits is equivalent to $f(-x,-y)=f(x,y)$. In characteristic zero, comparing monomial coefficients shows that precisely the terms of even total degree survive. Thus a vector space basis isSet $u=x^2$, $v=xy$, $w=y^2$. The two basis types are $u^a{w}^b$ and $v{u}^a{w}^b$, so they prove $R=k[u,v,w]$. Hence $R$ is a finitely generated algebra. It is an integral domain as a subring of the integral domain $k[x,y]$.

The generators satisfy $uw-v^2=0$. To prove this is the whole relation ideal, reduce any polynomial in $U,V,W$ modulo the monic relation $V^2-UW$ toIts image in $k[x,y]$ is $A(x^2,y^2)+xyB(x^2,y^2)$. The first summand has both exponents even and the second both odd; their distinct monomials are linearly independent. Thus the image vanishes only when both $A,B$ are zero. ConsequentlyThis gives the requested closed embedding and the quadratic cone invariant ring presentation.

The associated morphism of algebraic varieties is explicitlyIt is surjective. For $(u,v,w)\in Y$, if $u\ne 0$, choose $x$ with $x^2=u$ using algebraic closedness and put $y=v/x$. Then $y^2=v^2/u=w$. If $u=0$, the relation forces $v=0$; take $x=0$ and choose $y^2=w$. This covers every point of $Y$. Except over the origin, the two preimages differ by the simultaneous sign change, consistent with the orbit description.

Finally the gradient of $UW-V^2$ is $(W,-2V,U)$, so the Jacobian criterion makes every nonzero point smooth. At the origin, the maximal ideal modulo its square has the three independent classes of $U,V,W$, since the defining relation is quadratic. The tangent space there has dimension three, while $Y$ has dimension two. Therefore

Use the standard convention that a projective variety is nonempty and irreducible, over an algebraically closed field. Let $f:X\to {\mathbb{A}}^1$ be a regular function, and compose with the inclusion into ${\mathbb{P}}^1$. Its graph is closed in $X\times {\mathbb{P}}^1$: on an affine open of $X$, with homogeneous target coordinates $[T_0:T_1]$, it is given by $T_1-f(x)T_0=0$. This equation has no solution at $T_0=0$, and the local closedness statements glue.

Embed $X$ as a closed subvariety of ${\mathbb{P}}^m$. The graph is then closed in ${\mathbb{P}}^m\times {\mathbb{P}}^1$. Part 1, after swapping factors, makes its projected image $f(X)$ closed in ${\mathbb{P}}^1$. Since it avoids infinity, it is a proper closed subset of the projective line and hence finite. A continuous image of an irreducible space is irreducible, and a nonempty finite closed subset over an algebraically closed field is irreducible only if it is a singleton. HenceThis proves global regular functions on an irreducible projective variety are constant, using the closed-projection result rather than assuming it. If one allows disconnected projective algebraic sets instead of irreducible varieties, different components can carry different constants; irreducibility is the relevant convention here.

Let $\varphi :C\setminus S\to {\mathbb{P}}^n$ be the given algebraic morphism of algebraic varieties. Its value at the generic point supplies homogeneous coordinates $[h_0:\cdots :h_n]$ with $h_i\in k(C)$, not all zero. Thus it defines a rational map of projective varieties on $C$.

Fix a missing point $p\in S$. The local ring of a smooth algebraic curve at $p$ is a discrete valuation ring $A={\mathcal{O}}_{C,p}$, with uniformizer $t$ and valuation $v_p$. PutZero coordinates remain zero. Every $a_i$ is in $A$, and some $a_j$ is a unit because its valuation is zero. Choose a neighborhood $U$ of $p$ on which all $a_i$ are regular and $a_j$ remains nonvanishing; this is possible because membership in the local ring means regularity on a neighborhood. The affine target-chart coordinatesare regular on $U$, so they define a morphism of algebraic varieties $U\to {\mathbb{P}}^n$. At $p$ its homogeneous value is $[a_0(p):\cdots :a_n(p)]$, well-defined because $a_j(p)\ne 0$.

Multiplication of all homogeneous coordinates by the same nonzero rational function does not change their projective point. Thus this local morphism agrees with the original one on the common dense open where their coordinates were chosen. They agree throughout the overlap: the target's diagonal is closed, being defined in homogeneous coordinate pairs by $T_i{U}_j-T_j{U}_i=0$ for all $i,j$. Thus the equalizer is closed and contains that dense open. Apply this construction at each point of the finite set $S$ and glue with the original morphism. The same closed-diagonal argument proves uniqueness. ThereforeThis gives the extension of a rational map from a smooth projective curve by explicit local coordinates. The mechanism is the one-dimensional smooth local ring: it permits removal of the smallest common pole order while leaving at least one unit coordinate.