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Recall the following theorem, due to Hilbert. Let ${k}$ be a field, and ${S=k[x_0,...,x_n]}$ considered as a graded ring, and take ${M}$ any finitely generated graded ${S}$-module. Then there is a free resolution of ${M}$ of length at most ${n+2}$. This theorem is important because applying it in the case ${M=S/I}$ for ${I}$ a homogeneous ideal (i.e. to a projective variety), it tells us that the Hilbert polynomial exists and gives an explicit bound on its degree.

I had always assumed this theorem was in some sense fairly deep or hard, but in Chan’s scheme theory course today we saw that it’s actually a small amount of some fairly basic homological algebra.

Step 1: Do the case $M=k$, where each ${x_i}$ acts as the zero homomorphism. I would rather not do the details of this in a blog post, but it isn’t very hard, and you if you try to write down a resolution that’s as efficient as possible, you should end up with a resolution of length ${n+2}$, which usually seems to be given the intimidating name the Koszul complex’.

Step 2: Let ${M}$ now be arbitrary, and consider a minimal free resolution ${... F_n \rightarrow F_{n-1} \rightarrow ... \rightarrow F_0 \rightarrow M \rightarrow 0}$ (in the sense that the first map has the smallest possible number of generators, then the second one does, and so on). By minimality, for all ${k\geq 1}$ a basis element of ${F_k}$ must be mapped to an element with no constant term in any coefficient. Indeed, if ${e \in F_{k+1}}$ is mapped to ${(p_1,...,p_l) \in F_k}$, each ${p_i}$ is homogeneous (since the maps are maps of graded modules) and if some ${p_i}$ is constant, since ${k}$ is a field and ${(p_1,...,p_l)}$ is in the kernel of ${F_k \rightarrow F_{k-1}}$, we see that the basis vector ${e_i}$ maps to an element that is expressible as a linear combination of the image of the other ${e_j}$s, contradicting minimality. What this observation buys us is that if we take such a complex and tensor it with the ${S}$-module ${k}$, we get a complex

$\displaystyle ... \rightarrow F_n \otimes_S k \rightarrow ... \rightarrow F_1 \otimes_S k \rightarrow F_0 \otimes_S k \rightarrow M \otimes_S k \rightarrow 0,$
and all of the maps save the first one are the zero map.

Step 3: Use the fact that ${A \otimes B = B \otimes A}$, or more generally that ${Tor_i(A,B) = Tor_i(B,A)}$. This really is the key ingredient. Knowing that the above maps are zero, we see that the rank of ${F_j}$ is just the rank of ${Tor_j^S(M,k)}$. And this is the rank of ${Tor_j^S(k,M)}$, which can be computed by as the homology of the complex obtained by tensoring the Koszul complex with ${M}$. In particular, we see that it vanishes for ${i>n+1}$, so Hilbert’s theorem is proved.

In this post, in advance of a talk I am giving tomorrow as part of the Harvard Mazur Torsion seminar we shall give an overview of Mazur and Tate’s proof that there are no elliptic curves /${{\mathbb Q}}$ with a rational 13-torsion point. [WARNING: I apologise that this post is probably currently littered with mistakes, and certainly with abuses of notation, having been written fairly hastily. I will try to come back and make it better later.]

The theory of modular curves had been well-developed at the time of the paper, so there is a smooth affine curve over ${{\mathbb Z}[1/13]}$ out there called ${Y_1(13)}$ whose rational points are precisely isomorphism classes of elliptic curves over ${{\mathbb Q}}$ together with a rational point of order 13. The problem is therefore reduced to proving some curve has no rational points. It turns out to be nicer to work with the compactification ${X_1(13)}$, which has 12 extra points (cusps’), six of which are rational.

For quite a few numbers that aren’t 13, one is rather lucky and finds that either ${X_1(N)}$ is an elliptic curve (so write down its equation, do 2-descent to show it has rank 0, and work out that all the rational points are cusps) or more likely one finds that some other closely related curve like ${X_0(N)}$ or a quotient of this by an Atkin-Lehner involution is an elliptic curve of rank 0, and thinking about fibres of maps like ${X_1(N) \rightarrow X_0(N)}$ can again find all the rational points.

However, 13 is unlucky for some, and indeed it is unlucky for us. The curve ${X=X_1(13)}$ we are interested in has genus 2, but the only logical thing to map it to, ${X_0(13)}$, has genus 0 and a rational point, hence loads of rational points. We must therefore attack the curve ${X_1(13)}$ directly in some way.

One natural thing to do is to consider its Jacobian ${J=Jac(X_1(13))}$, a 2-dimensional abelian variety whose geometric points correspond to elements of the 0-Picard group of ${X(k^{sep})}$. In particular it is equipped with, for any choice ${P_0}$ of rational cusp, an embedding defined over ${{\mathbb Q}}$ ${X \hookrightarrow J}$ given by ${P \mapsto (P)-(P_0) \in Div(X)}$. In particular, if we can show that the Jacobian has Mordell-Weil rank zero and analyse its rational torsion we’ll be done.

The first progress here was made by Ogg, who took one of the other rational cusps and discovered that it was a torsion point of order 19, and the subgroup it generates intersects the curve ${X}$ precisely at the rational cusps. Given our initial bad luck, this is nothing short of a miracle! Since ${X}$ and hence ${J}$ has good reduction at 2, the Riemann hypothesis tells us that ${|J({\mathbb Q})_{tor}| \leq |J(\mathbb{F}_2)| \leq (1+\sqrt{2})^4 < 38}$, so after Ogg’s work, we know that the only thing left to rule out is a point of infinite order in ${J({\mathbb Q})}$.

If you have an elliptic curve, the standard trick for computing its Mordell-Weil rank is to take your favourite isogeny ${\pi}$ (usually one whose kernel is a well-understood Galois module) and analyse the cokernel of the map ${\pi: E(K) \rightarrow E(K)}$ by interpreting it as a subgroup of the first Galois cohomology group ${H^1(G_K, (Ker \pi)(\bar{K}))}$. There is no reason this shouldn’t work for abelian varieties, but finding an isogeny whose kernel we can understand is somewhat harder: for elliptic curves the multiplication by 2 map will usually do nicely, because the 2-torsion can be read off from an equation in Weierstrass form, but we have no such trick here.

What we do have here however are a bewildering family of interesting automorphisms of ${J}$. In particular, we have the diamond operators ${\gamma_m: (E,P) \mapsto (E,mP)}$ (for ${m \in ({\mathbb Z}/13{\mathbb Z})^*/\{\pm\}}$), and the Atkin-Lehner involutions ${\tau_{\zeta}}$ which map ${(E,P)}$ to the pair ${(E/

, [Q])}$

where ${Q}$ is some 13-torsion point of ${E}$ such that it pairs with ${P}$ under the Weil pairing to give ${\zeta}$. The Galois group ${G_{\mathbb Q}}$ commutes with the former and acts on the latter in the obvious way (${\sigma(\tau_{\zeta}) = \tau_{\sigma(\zeta)}}$). These operators generate a subgroup ${\Delta \subset End(J)}$ isomorphic to the dihedral group, and with the noted Galois action they satisfy ${\sigma(g.x) = \sigma(g)\sigma(x)}$.

Of great interest to us is the subring ${{\mathbb Z}[\gamma_2]}$. One can easily verify that ${\gamma_2}$ acts with order 6 on the cusps of ${J}$, and since ${J}$ is simple (there aren’t many elliptic curves with good reduction at 2 and a 19-torsion point) one deduces that ${{\mathbb Z}[\gamma_2] \cong {\mathbb Z}[X]/(X^2-X+1)}$, i.e. it’s the Eisenstein integers, a rather nice quadratic PID. We also have a rational subgroup ${V(1) \subset J[19]}$, so it’s natural to consider trying to find a 19-isogeny, and we now have one staring us in the face: ${19}$ factors into two distinct primes ${19=\pi \bar{\pi}}$ in ${{\mathbb Z}[\gamma_2]}$, giving an idempotent decomposition

$\displaystyle J[19] = J[\pi] \oplus J[\bar{\pi}].$

Relabelling as necessary we may assume ${V(1) \subset J[\bar{\pi}]}$. One can then show that any Atkin-Lehner involution swaps these two eigenspaces and maps ${V(1)}$ onto a subset ${V(\gamma) \subset J[\pi]}$ with a faithful action of ${Gal({\mathbb Q}(\zeta_{13})^+/{\mathbb Q})}$. Moreover, by considering the ${\gamma_2}$ action on the Weil pairing one shows that the self-duality of ${J[19]}$ (all Jacobians admit a principal polarisation) induces a Cartier duality between ${J[\pi]}$ and ${J[\bar{\pi}]}$, and so one ends up with a short exact sequence of ${G_{\mathbb Q}}$-modules:

$\displaystyle 0 \rightarrow V(\gamma) \rightarrow J[\pi] \rightarrow \mu_{19} \rightarrow 0.$

We are now in good shape to attempt a ${\pi}$-descent. We already know the Mordell-Weil theorem for abelian varieties, so ${J({\mathbb Q})}$ is a finitely generated ${{\mathbb Z}}$-module, in particular a finitely generated ${{\mathbb Z}[\gamma_2]}$-module. Since this is a PID, the structure theorem tells us that if we can show ${\pi: J({\mathbb Q}) \rightarrow J({\mathbb Q})}$ is surjective, then it has rank 0. We will therefore want to consider some kind of Galois cohomology which will give us a small ${H^1(J[\pi])}$ (our dream would be if this could vanish, but that’s maybe rather unrealistic).

Recall that Galois cohomology is just the étale cohomology of abelian sheaves on the spectrum of a field. If instead of allowing our sections to only be defined on a field (an `arithmetic point’) we spread them out across something like ${Spec({\mathbb Z}[1/13])}$, it figures that maybe the cohomology groups will get smaller. However, we must be careful: we are being led to study the cohomology of 19-primary finite flat group schemes over a base where 19 is not invertible. In particular, we will later need a Kummer exact sequence to analyse ${\mu_{19}}$, and this fails to be exact as a sequence of étale sheaves. The solution is instead to consider the fppf topology which is finer than the étale topology but still coarser than the fpqc topology (so all schemes are fppf sheaves).

Let’s do that. Recall that ${J}$ was defined over ${{\mathbb Z}[1/13]}$ so we get a naturally defined exact sequence

$\displaystyle 0 \rightarrow J[\pi] \rightarrow J \rightarrow J \rightarrow 0,$

which can be viewed as a sequence of fppf sheaves on ${{\mathbb Z}[1/13]}$. Taking global sections, we get a long exact sequence containing a chunk:

$\displaystyle J({\mathbb Q}) \rightarrow J({\mathbb Q}) \rightarrow H^1({\mathbb Z}[1/13], J[\pi]),$

where the first map is the one we wish to show is surjective.

We secretly know, as with 2-descent, that all the juicy stuff (say, the data in the image of the second map) probably happens at the bad primes, which in this case is just the prime 13. It’s therefore maybe natural to base change to ${{\mathbb Q}_{13}}$ and analyse the situation there.

In fact, the situation there is rather nice: it’s not too hard to prove that ${\pi: J({\mathbb Q}_{13}) \rightarrow J({\mathbb Q}_{13})}$ is in fact an isomorphism. By taking a Neron model (best possible smooth – but not necessarily proper – group scheme whose generic fibre returns ${J}$) ${A/{\mathbb Z}_{13}}$ for ${J}$, in particular with ${A({\mathbb Z}_{13}) = J({\mathbb Q}_{13})}$. We can take the reduction, and notice that ${\pi}$ acts invertibly on the (pro-13) kernel of reduction, so by the snake lemma it suffices to check that ${\pi}$ is injective: in other words, that there are no nontrivial ${G_{{\mathbb Q}_{13}}}$-invariants of ${J[\pi]}$. Given our explicit description of this Galois module, this boils down to the fact that ${13}$ doesn’t split completely in ${{\mathbb Q}(\zeta_{13})^+/{\mathbb Q}}$ (it’s totally ramified) or ${{\mathbb Q}(\zeta_{19})/{\mathbb Q}}$ (${13 \not\equiv 1 \mod 19}$).

How do we transfer this information back up to ${{\mathbb Z}[1/13]}$? Well, if we can prove that ${H^1({\mathbb Z}[1/13], J[\pi]) \rightarrow H^1({\mathbb Q}_{13}, J[\pi])}$ is injective then that will be enough (draw a diagram, and note that by what we just showed ${J({\mathbb Q}_{13}) \rightarrow H^1({\mathbb Q}_{13}, J[\pi])}$ is the zero map). At this point it seems worthwhile to want to reconstruct our explicit description of ${J[\pi]}$ and verify that this map is injective on each piece. I.e. recall we proved earlier that the generic fibre of ${J[\pi]}$ is an extension of a module which becomes the trivial ${{\mathbb Z}/19{\mathbb Z}}$ over ${{\mathbb Q}(\zeta_{13})^+}$ by the Galois module ${\mu_{19}}$. But by a general theorem in a paper by Oort and Tate concerning finite flat group schemes, since 19 isn’t very ramified in ${{\mathbb Q}}$, we can identify 19-primary finite flat group schemes by their generic fibre. So taking the Zariski closure, we get a short exact sequence

$\displaystyle 0 \rightarrow E \rightarrow J[\pi] \rightarrow \mu_{19} \rightarrow 0,$

where ${E}$ is some finite flat group scheme whose base change to ${{\mathbb Z}[1/13, \zeta_{13}]}$ is isomorphic to the constant group scheme ${{\mathbb Z}/19{\mathbb Z}}$.

So we now are reduced to showing that the induced maps on the first cohomology of these pieces when we base change to ${{\mathbb Q}_{13}}$ are injective. Firstly, in fact we can show that ${H^1({\mathbb Z}[1/13], E) = 0}$. A nonzero element of this group is a nontrivial ${E}$-torsor on ${{\mathbb Z}[1/13]}$, which under base change corresponds to a nontrivial ${{\mathbb Z}/19{\mathbb Z}}$-torsor on ${{\mathbb Z}[1/13, \zeta_{13}]}$. But the class number of ${{\mathbb Q}(\zeta_{13})}$ has been computed and is in fact 1, so in particular this field doesn’t admit any nontrivial degree 19 covers unramified away from 13.

Finally, we must prove that ${H^1({\mathbb Z}[1/13], \mu_{19}) \rightarrow H^1({\mathbb Q}_{13}, \mu_{19})}$ is injective. Since we had the foresight to use fppf cohomology, we get the Kummer exact sequence

$\displaystyle 0 \rightarrow \mu_{19} \rightarrow \mathcal{O}^* \rightarrow \mathcal{O}^* \rightarrow 0.$

Taking the long exact sequence of cohomology, and noting that both of these rings are PIDs, so have trivial picard group (${=H^1(\mathcal{O}^*)}$ by a generalisation of Hilbert 90), we get isomorphisms of these cohomology groups with ${\mathcal{O}^*/\mathcal{O}^{*19}}$, which one can check explicitly have an induced injection.

And that’s it, we read off that our original ${\pi}$ was surjective, and so we’re done: there are no ${{\mathbb Q}}$-torsion points of order 13 on elliptic curves over ${{\mathbb Q}}$.