Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 126 (Abelian Varieties)

For $A\to B\to C=B/I$, defineandThe first map is well-defined because $d(ij)=idj+jdi$ becomes zero after tensoring with $C$ when $i,j\in I$. The second is induced by the universal derivation $B\to C\stackrel{d}{\to }{\Omega }_{C/A}$ and is surjective because the elements $d\bar{b}$ generate ${\Omega }_{C/A}$.

The composite is zero since $\bar{i}=0$. Conversely, quotienting ${\Omega }_{B/A}{\otimes }_{B}C$ by the $di$ for $i\in I$ imposes exactly the relations needed for the derivation of $B$ to descend to $C$. The universal property of the Module of Kähler differentials therefore identifies that quotient with ${\Omega }_{C/A}$, proving the Conormal exact sequence for Kähler differentials

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Every $k$-derivation $D:K[t]\to M$ is uniquely determined by its restriction to $K$ and by $D(t)$. Conversely, a $k$-derivation $K\to M$ and an arbitrary element $m\in M$ extend uniquely byRepresenting this natural decomposition of derivations gives

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Present $L=K[t]/(f)$ by sending $t$ to $b$. Applying the Conormal exact sequence for Kähler differentials and part ii givesWriting $f(t)={\sum }_i{a}_i{t}^i$, the image of $1$, represented by $f$, isThis has the required form $(\ast ,f^{\prime }(b)dt)$.

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If $L/K$ is finite separable, the primitive element theorem writes it as $K(b)$ with $f^{\prime }(b)\ne 0$. The relation in part a has a nonzero $dt$ component, so projection along that relation gives an isomorphism

For an arbitrary simple finite extension, part a presents ${\Omega }_{L/k}$ as a quotient of a vector space of dimension ${\dim }_K{\Omega }_{K/k}+1$ by the image of a space of dimension at most one. Its dimension is therefore at least ${\dim }_K{\Omega }_{K/k}$. Applying this one generator at a time through a finite tower proves the same inequality for every finite extension.

Strict inequality occurs in characteristic $p$. Take $k=K={\mathbb{F}}_p(u)$ and $L=K(u^{1/p})$. Then ${\Omega }_{K/k}=0$, while the relation $b^p-u=0$ has zero differential relative to $k$, sohas dimension one.

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The relevant theorem supplies, locally on $S=\mathrm{Spec}A$, a bounded complex $K^{\bullet }$ of finite free $A$-modules such thatnaturally for every $A$-module $M$. In particular the fiber cohomology at $s$ is computed by $K^{\bullet }{\otimes }_A\kappa (s)$.

The Euler characteristic of a finite-dimensional complex equals the alternating sum of the dimensions of its terms. Hencewhich is constant wherever one complex works. It is therefore locally constant on $S$, and is constant when $S$ is connected.

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Use the same finite complex computing cohomology in a proper flat family. In fixed bases its differentials are matrices over $A$. The condition that such a matrix have rank at most $r$ is closed, being defined by its $(r+1)\times (r+1)$ minors. Sincethe condition that this dimension be at least $n$ is a finite union of intersections of closed rank loci. It is therefore closed. This is the Semicontinuity theorem for coherent cohomology.

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For every integer $t$,For $t\ge 0$ this is the dimension of the homogeneous polynomials of degree $t$, since the higher cohomology vanishes; polynomiality then identifies the Hilbert polynomial.

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A smooth plane curve of degree $m$ has genusThe line bundle ${\mathcal{O}}_X(D)(t)$ has degree $d+mt$. The Riemann-Roch theorem therefore gives

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A useful form of the Mumford rigidity lemma says that if $X$ is complete and connected and $h:X\times Y\to Z$ is constant on one fiber $X\times \{y_0\}$, then under the pointed separated hypotheses it factors through $Y$.

For $f:X\to G$ with $f(e_X)=e_G$, defineAt $x=e_X$ this is constantly $e_G$. Applying rigidity with the complete factor $X$ in the $y$ variable shows that $h(x,y)$ is independent of $y$. At $y=e_X$ its value is $f(x)$, soThus $f$ is a homomorphism of group varieties.

Completeness is essential. On the additive group variety ${\mathbb{G}}_a$ over a field of characteristic different from two, the morphism $f(t)=t^2$ fixes zero but is not additive.

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For $x:S\to G$, defineThen $T_{x/S}=(T_x,{\mathrm{pr}}_2)$ is an $S$-morphism, and translation by $x^{-1}$ is its inverse.

Because an isomorphism preserves relative differentials and $T_{x/S}$ lies over $S$,Using base change, ${\Omega }_{G\times S/S}\cong {\mathrm{pr}}_1^{\ast }{\Omega }_{G/k}$, while the left side is $T_x^{\ast }{\Omega }_{G/k}$. Pulling this isomorphism back along the identity section $e\times 1_S$ givesFinally take $S=G$ and $x=1_G$. This yields the invariant differential on a group scheme trivializationso ${\Omega }_{G/k}$ is a free ${\mathcal{O}}_G$-module.

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The Theorem of the square states that for a line bundle $L$ on an abelian variety $X$ and $x,y\in X(k)$,

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Define the homomorphism associated to a line bundle on an abelian varietyThe Theorem of the square gives ${\varphi }_L(x+y)={\varphi }_L(x)\otimes {\varphi }_L(y)$, so this is a homomorphism.

Tensor products satisfy ${\varphi }_{L\otimes M}={\varphi }_L\otimes {\varphi }_M$ and ${\varphi }_{L^{-1}}={\varphi }_L^{-1}$. Thereforeis a subgroup of the Picard group. If $M={\varphi }_L(x)$, translations commute and the theorem of the square givesfor every $y$. Hence $M\in {\mathrm{Pic}}^{0}X$, proving $\mathrm{im}{\varphi }_L\subseteq {\mathrm{Pic}}^{0}X$.

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The mapis a homomorphism. Pullback along $i_1(x)=(x,e_2)$ recovers $L_1$, up to tensoring with a fixed one-dimensional vector space, which is a trivial line bundle; similarly $i_2^{\ast }$ recovers $L_2$. Thus $\tau$ is injective.

It need not be surjective. For an elliptic curve $E$, the line bundle ${\mathcal{O}}_{E\times E}(\Delta )$ of the diagonal restricts to $E\times \{y\}$ as ${\mathcal{O}}_E(y)$, whose class varies with $y$. A line bundle pulled back separately from the two factors has constant class on these fibers. Hence $\mathcal{O}(\Delta )$ is not in the image of $\tau$.

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Part iii already proves injectivity. Pullback along $i_j$ sends translation-invariant line bundles to translation-invariant line bundles, so it definesClearly $\rho \tau =1$.

For $L\in {\mathrm{Pic}}^0(X_1\times X_2)$, put $L_j=i_j^{\ast }L$ andThen $M$ is trivial on both coordinate axes. Since $L$ and the two correcting factors lie in ${\mathrm{Pic}}^0$, translation by $(x_1,e_2)$ leaves $M$ invariant. Hence every restriction $M{|}_{\{x_1\}\times X_2}$ is trivial. The Seesaw theorem and triviality on $X_1\times \{e_2\}$ imply $M\cong {\mathcal{O}}_X$. Thus $L=\tau (L_1,L_2)$, so

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