If $L\in {\mathrm{Pic}}^0(X)$, part (iii) makes $\Lambda (L)$ trivial. Pulling it back along $(f,g):Y\to X\times X$ givesTaking $Y=X$, $f={\mathrm{id}}_X$, and $g=i$, where $i$ is inversion, givesInduction with $f=[n]$ and $g={\mathrm{id}}_X$ proves $[n{]}^{\ast }L\simeq L^{\otimes n}$ for $n\ge 0$; combining this with inversion proves

Conversely, suppose $i^{\ast }L\simeq L^{\vee }$. For $M={\varphi }_L(x)$, part (ii) gives $M\in {\mathrm{Pic}}^0(X)$, so the result just proved yields $i^{\ast }M\simeq M^{\vee }$. On the other hand,whereas $i^{\ast }L\simeq L^{\vee }$ gives ${\varphi }_{i^{\ast }L}(x)={\varphi }_{L^{\vee }}(x)=M^{\vee }$. Hence $M^{\otimes 2}$ is trivial for every $x$, so ${\varphi }_{L^{\otimes 2}}$ is trivial and $L^{\otimes 2}\in {\mathrm{Pic}}^0(X)$. The torsion-freeness proved in part (ii) now implies

LetIf $L\in {\mathrm{Pic}}^0(X)$, then its restriction to $X\times \{y\}$ is $T_y^{\ast }L\otimes L^{\vee }={\varphi }_L(y)$, up to a constant one-dimensional factor, and is therefore trivial. Its restriction to $\{e\}\times X$ is also trivial. The Seesaw theorem now implies that $\Lambda (L)$ is trivial on $X\times X$.

Conversely, if $\Lambda (L)$ is trivial, restricting it to $X\times \{y\}$ shows that ${\varphi }_L(y)$ is trivial for every $y\in X(k)$. Thus

For a line bundle $L$ on $X$, define the homomorphism associated to a line bundle on an abelian varietyThe Theorem of the square givesso ${\varphi }_L$ is a homomorphism. Pullback distributes over the tensor product of sheaves, and thereforeIterating the homomorphism law in $x$ gives

Suppose $L^{\otimes n}\in {\mathrm{Pic}}^0(X)$. Then ${\varphi }_L(nx)$ is trivial for every $x$. The multiplication-by-n morphism on an abelian variety is surjective, so ${\varphi }_L$ is trivial and $L\in {\mathrm{Pic}}^0(X)$. Thus the Néron-Severi groupis torsion-free.

Finally, put $M={\varphi }_L(x)$. For every $y$,which is trivial by the Theorem of the square. Hence

The Theorem of the square says that for an abelian variety $X$, a line bundle $L$, and $x,y\in X(k)$,where $T_x$ denotes translation by $x$.

One form of the Mumford rigidity lemma says that if $X$ is a complete variety, $S$ is connected, and a morphism $F:X\times S\to Z$ maps $X\times \{s_0\}$ to one point, then $F$ factors through the projection to $S$. In particular, if $F$ also maps $\{x_0\}\times S$ to that point, then $F$ is constant.

Choose $s_0\in S(k)$ and put $g=f{|}_{X\times \{s_0\}}$. To see that the pointed morphism $g:X\to Y$ is a homomorphism, apply rigidity toIt vanishes on $X\times \{e\}$, so it factors through the second projection; it also vanishes on $\{e\}\times X$, so it is identically zero. Now defineThen $F(x,s_0)=e$ for every $x$ and $F(e,s)=e$ for every $s$. Rigidity forces $F$ to be identically $e$, so

A group scheme over $k$ is a $k$-scheme $G$ with multiplication $m_G:G{\times }_{k}G\to G$, identity $e_G:\mathrm{Spec}k\to G$, and inversion $i_G:G\to G$ satisfying the group axioms as identities of morphisms. A homomorphism of group schemes $f:H\to G$ is a $k$-morphism satisfyingand it then preserves the identity and inversion.

Assume $G$ and $H$ are commutative. The group $G(H)={\mathrm{Mor}}_k(H,G)$ has pointwise additionFor every $k$-algebra $R$ and $x,y\in H(R)$,where commutativity permits the middle terms to be reordered. Hence $f+g$ is a homomorphism. The zero morphism and pointwise inverse are also homomorphisms, so ${\mathrm{Hom}}_k(H,G)$ is a subgroup of $G(H)$. The definition immediately gives

Repeated pointwise addition gives $(nf{)}_R(x)=n{f}_R(x)$. Since $f_R$ is a group homomorphism,The Yoneda lemma turns equality on all $R$-valued points into equality of morphisms, proving

Take an affine open subscheme $U=\mathrm{Spec}A\subseteq Y$. Properness and flatness survive base change, and $A$ is reduced because $Y$ is reduced. A bounded complex of finite locally free modules computes the cohomology of $\mathcal{F}$ on $X_U$.

If some $H^p(X_U,\mathcal{F})$ were nonzero, choose the largest such $p=n$. All groups above degree $n$ would vanish, while every fiber group in degree $n$ vanishes by hypothesis. Part (iii) would force $H^n(X_U,\mathcal{F})=0$, a contradiction. Hencefor every affine $U\subseteq Y$ and every $p\ge 0$. These groups compute the sections of the higher direct images over affine opens, so $R^p{f}_{\ast }\mathcal{F}=0$ for all $p$, including $f_{\ast }\mathcal{F}=0$ when $p=0$. The Leray spectral sequence now gives

The finite complex computing cohomology in a proper flat family gives a bounded complex $K^{\bullet }$ of finite locally free $A$-modules such that, for every $A$-module $M$,In particular, $M=A$ computes $H^p(X,\mathcal{F})$ and $M=\kappa (s)$ computes $H^p(X_s,\mathcal{F}(s))$.

Since $H^p(K^{\bullet })=0$ for $p>n$, the finite exact tail above degree $n$ can be split successively: its last differential is surjective onto a projective module, hence splits, and induction moves left. Removing the resulting contractible summands leaves a finite locally free complex ending in degree $n$. Thereforeand after tensoring with $\kappa (s)$ the same formula computes the fiber cohomology.

If $H^n(X,\mathcal{F})=0$, the last differential is surjective, and remains so after every base change; hence every $H^n(X_s,\mathcal{F}(s))$ vanishes. Conversely, if all fiber groups vanish, the finitely generated cokernel $C$ satisfies $C{\otimes }_A\kappa (s)=0$ for every $s\in \mathrm{Spec}A$. Localizing and applying Nakayama lemma gives $C_s=0$ for every $s$, so $C=0$. Thus

For $X=U\cup V$, the Mayer-Vietoris sequence for sheaf cohomology is the long exact sequence

We prove the required vanishing by induction on the number $m$ of open sets. The case $m=1$ is an assumption. Put $U=U_1\cup \cdots \cup U_{m-1}$ and $V=U_m$. The induction hypothesis gives $H^p(U,\mathcal{F})=0$ for every $p$. The intersections $U_i\cap U_m$ cover $U\cap V$, and every nonempty finite intersection among them is one of the intersections in the hypothesis, so the same induction gives $H^p(U\cap V,\mathcal{F})=0$. We also have $H^p(V,\mathcal{F})=0$. Exactness of the Mayer-Vietoris sequence now yields

For an affine morphism $f:X\to Y$ and a quasi-coherent sheaf $\mathcal{F}$, every inverse image of an affine open $V\subseteq Y$ is affine. Higher cohomology of a quasi-coherent sheaf on an affine scheme vanishes, soThe Leray spectral sequence therefore has only its zeroth row, and its edge maps give

A morphism of schemes $f:X\to Y$ is a flat morphism when every local-ring map ${\mathcal{O}}_{Y,f(x)}\to {\mathcal{O}}_{X,x}$ makes ${\mathcal{O}}_{X,x}$ a flat module.

In (i), the coordinate map is $k[t]\to k[y]$, $t\mapsto y^2$, andIt is therefore a free module of rank two and the morphism is flat, including in characteristic two.

In (ii), $k[x]$ is finite over the cusp ring $R=k[x^2,x^3]$ and has generic rank one. Were it flat, finite flatness over the local ring at the cusp would make it free of rank one. Its fiber there is insteadwhich has dimension two, so this morphism is not flat.

In (iii), the base coordinate $t$ acts as $x$, and the nonzero element $y$ satisfies $ty=xy=0$. Thus the coordinate ring has torsion as a $k[t]$-module. Since $k[t]$ is a principal ideal domain and a module over it is flat exactly when it is torsion-free, this morphism is not flat. Consequently

For a ring homomorphism $A\to B$, the Module of Kähler differentials ${\Omega }_{B/A}$ is the $B$-module generated by symbols $db$, subject toEquivalently, it represents $A$-linear derivations:For $A\to B\to C$, the Transitivity exact sequence for Kähler differentials isIf $B\to C=B/I$ is surjective, the Conormal exact sequence for Kähler differentials is

Let $L/K$ be a finite field extension. By the primitive element theorem, its maximal separable field extension $L_s/K$ is simple, and transitivity reduces the calculation to a simple algebraic extension. If $L=K(\alpha )$ with minimal polynomial $m_{\alpha }$, thenThus a separable simple extension has zero differentials. Conversely, if $L/K$ is not separable, the purely inseparable part has a generator whose minimal polynomial has zero formal derivative in positive characteristic, producing a nonzero differential. Hence

Now let $\mathrm{char}K=p>0$. If $L=K(a)$, $a^p\notin K$, and $a^{p^2}\in K$, then the minimal polynomial is $T^{p^2}-a^{p^2}$ and has zero derivative, soIf $L=K(a,b)$, where $a\notin K$, $b\notin K(a)$, and $a^p,b^p\in K$, then $[K(a):K]=[L:K(a)]=p$. Both defining equations have zero derivative, and

For the three morphisms, compute the Sheaf of relative Kähler differentials on coordinate rings.

If $\mathrm{char}k\ne 2$, this is $k[y]/(y)dy$ and its support is the origin $V(y)=V(x,y)$. If $\mathrm{char}k=2$, it is free of rank one and its support is all of $X$.

Its support is the origin $V(x)$ in every characteristic; the module is $k[x]/(x)$ unless $\mathrm{char}k=2$, when it is $k[x]/(x^2)$.

Its support is the entire component $V(x)\subset X$.

The three-isogeny descent connecting map, identified through the Weil pairing with $H^1(\mathbb{Q},{\mu }_3)={\mathbb{Q}}^{\times }/({\mathbb{Q}}^{\times }{)}^3$, isThe long exact sequence attached to $0\to E[\varphi ]\to E\stackrel{\varphi }{\to }{E}^{\prime }\to 0$ makes it a group homomorphism withThe function $f=y$ from part (b) gives the explicit formulaAt $T$, the value $d^{-1}$ is the leading coefficient of $y$ relative to the local parameter $x$, since $y(y+d)=x^3$ gives $y/x^3\to d^{-1}$. At $-T=(0,-d)$ the ordinary formula gives $-d$, whose class is the inverse of $d^{-1}$ because $-1$ is a cube.

Let $S$ be the primes dividing $d$. For a prime $\ell \notin S$, useIf $v_{\ell }(y)>0$, then $y+d$ is an $\ell$-adic unit, so $v_{\ell }(y)=3v_{\ell }(x)$. If $v_{\ell }(y)<0$, then $v_{\ell }(y+d)=v_{\ell }(y)$, so $2v_{\ell }(y)=3v_{\ell }(x)$. In either case $v_{\ell }(y)$ is divisible by three; the special values at $O$ and $T$ have the same property. ThereforeWhen $d=1$, this power-class group is trivial: a rational number whose valuation at every prime is divisible by three is a cube up to sign, and $-1=(-1{)}^3$. Thus $\alpha$ is trivial, its kernel is all of $E^{\prime }(\mathbb{Q})$, and

The principal divisor criterion on an elliptic curve says that $D=\sum n_P(P)$ is principal exactly when $\mathrm{deg}D=0$ and $\sum [n_P]P=O$.

On $E^{\prime }:y^2+dy=x^3$, takeThe line $y=0$ meets the cubic three times at $T$, while $y$ has a triple pole at the point at infinity. HenceWith $g=u/v\in \mathbb{Q}(E)$, part (a) givesTaking divisors and cancelling the factor three yieldsPullback on degree-zero divisor classes is the dual isogeny, so the pulled-back class is represented by $\hat{\varphi }(T)$. It is principal, and therefore

An isogeny of elliptic curves is a nonconstant morphism preserving identity points; it is automatically a finite surjective group homomorphism. On the affine chart $v\ne 0$, putThe equation of $E$ is $t^3+d=s^3$, and the proposed map isIt lands on $E^{\prime }$ becauseThe rational formulas extend across $v=0$ to a morphism sending $O_E$ to $O_{E^{\prime }}$. It is nonconstant, hence an isogeny. On function fields, $t$ satisfies $t^3=y$, so the degree is at most three; generically the three cube roots give three distinct preimages. Equivalently, the points with $v=0$ form its three-element geometric kernel. Therefore

For a reduced rational number $x=a/b$ with $b>0$, the height of a rational number isCondition (i) holds because only finitely many coprime integer pairs have bounded maximum.

Condition (iii) also holds. The standard height inequalitygivesCondition (ii) fails: for positive integers $m$,whose absolute value is unbounded. Thus precisely conditions $\boxed{\text{(i) and (iii)}}$ hold. This is consistent with $\mathbb{Q}/n\mathbb{Q}=0$ although the additive group $\mathbb{Q}$ is not finitely generated.

If $A$ is finitely generated, the structure theorem for finitely generated modules over a principal ideal domain immediately makes $A/nA$ finite.

Conversely, first replace the given height by a quadratic one. SetCondition (ii) makes this limit converge and givesThus $\hat{h}$ still has finite bounded subsets. Condition (i) also gives a global lower bound for $h$, so $\hat{h}\ge 0$. Applying condition (iii) to $2^{r}P,2^{r}Q$, dividing by $4^r$ and passing to the limit gives one direction of the parallelogram identity. Applying the same inequality to $P+Q$ and $P-Q$, and using $\hat{h}(2P)=4\hat{h}(P)$, gives the reverse direction. Henceand induction yields $\hat{h}(mP)=m^2\hat{h}(P)$ for every integer $m$.

Now suppose $A/nA$ is finite and choose representatives $R_1,\dots ,R_s$. Put $R={\max }_i\hat{h}(R_i)$. For any $P$, write $P=nQ+R_i$. Nonnegativity and the parallelogram identity giveso, since $n\ge 2$,Repeated division modulo $nA$ therefore reaches the finite set $B=\{P:\hat{h}(P)\le R+1\}$. Reversing the recursion expresses every element of $A$ using $B$ and the finitely many $R_i$. This is the height descent lemma, and proves

The natural maphas finite image by hypothesis. It remains to bound its kernel. If $P\in E(K)$ becomes $nQ$ for $Q\in E(L)$, thenis a one-cocycle for $G=\mathrm{Gal}(L/K)$. Changing $Q$ by an $n$-torsion point changes this cocycle by a coboundary, producing a well-defined map from the kernel toIf its cohomology class is zero, subtracting the corresponding torsion point from $Q$ makes $Q$ Galois fixed, so $P\in nE(K)$. The map is therefore injective. Both $G$ and $E(L)[n]$ are finite, so this group cohomology set is finite. A finite kernel and finite image give

An integral Weierstrass equation has good reduction outside the finitely many primes dividing its nonzero discriminant. This proves finiteness of the set of bad primes. To prove finiteness of rational torsion, choose two distinct good primes. The reduction of torsion points on an elliptic curve injects each primary component at a good prime of different residue characteristic, so the two finite reduced point groups bound every primary component of $E(\mathbb{Q}{)}_{\mathrm{tors}}$.

Forthe displayed equation is minimal andIts bad primes are therefore exactlyThe good reductions at $11$ and $13$ haveTheir coprime orders exclude every rational torsion primary component, including the residue-characteristic components by using the other prime. Hence

For a minimal integral Weierstrass equation, let $\tilde{E}$ be the reduced cubic and ${\tilde{E}}_{\mathrm{ns}}(k)$ its nonsingular points, with their induced group law. Define the filtration of elliptic-curve points over a local field byandThe parameter $z=-x/y$ identifies $E_1(K)$ with the formal group of an elliptic curve on $\pi {\mathcal{O}}_K$. Part (a) therefore gives $E_1(K)[n]=0$. Reduction restricts to the exact sequenceIts restriction to $n$-torsion has trivial kernel, yielding the injection