“Der Wille zum Glück” von Thomas Mann

Am Freitag habe ich eine Erzählung von Thomas Mann gelesen; die heißt “Der Wille zum Glück”. Es gibt einen kranken Mann, dessen Herz medizinisch schwach ist. Die Doktoren sagen, es sei eine Wunder, er lebe noch. Er hat sich in einem Mädchen verliebt, und sie liebte ihn auch. Der Vater des Mädchens aber hatte die Hand seiner Tochter versagt, weil er krank war, und konnte sehr bald sterben. Der Junge ist danach abgereist, und kehrte niemals zurück, und das Mädchen hat versprochen, sie würde niemals einem anderen Mann die Hand reichen als ihm.

Die Zeit floss weiter, und es ist ihm selbst aufgefallen, dass er immer noch am Leben war.

“Ich lebe doch noch immer. Ich bin beinahe täglich am Ende…Das Herz klopft mir bis in den Hals…Ich glaube, dass ich in diesen Jahren tausendmal schon den Tod von Angesicht zu Angesicht gesehen habe. Ich bin nicht gestorben. Mich hält etwas.”

Nach fünf Jahren, aber, kriegte der Junge einen Brief von dem Vater; seine Tochter wurde sich verlobt. Der Junge ist kurz danach gestorben.

“…gestorben am Morgen nach der Hochzeitsnacht, – beinahe in der Hochzeitsnacht. Es mußte so sein. War es nicht der Wille, der Wille zum Glück allein, mit dem er so lange den Tod bezwungen hatte? Er mußte sterben, ohne Kampf und Widerstand sterben, als seinem Willen zum Glück Genüge geschehen war; er hatte keinen Vorwand mehr zu leben.”

Über diesen letzten Satz musste ich nachdenken. Wieso leben und streben die Menschen? Ist es nur für das Selbstglück? Was über das Glück der anderen? Kann man nicht für das Glück der anderen streben? Es gibt Zeiten und Momenten, wenn ich persönlich nicht glücklich bin, aber dann versuche ich, mich zu erinnern, es gibt andere glückliche Leute. Ich bin nur ein Mensch, und meine Rolle in der Welt, in der Gesellschaft, ist auch sehr wichtig. Ja, auf der einer Seite bin ich nur ein Mensch. Aber auf der anderer Seite bin ich ein kleines Teil der gesamten Schöpfung. Glücklichkeit ist irgendwie durch die Schöpfung ausgebreitet, und die Glücklichkeit der anderen könnte auch meine Glücklichkeit sein. Auf dieser Weise hat meiner Meinung nach jeder einen Grund zu leben.

Breaking up the Hard Step

This post continues the discussion of Mazur’s Theorem, started here a while ago.

Last time we proved the theorem, assuming a rather hard step, which I called the Hard Step.

In this post we will reduce the following proposition to another hard step, the formal immersion criterion (to be explained below).

Proposition. (Corollary 4.3 in [M]) Let K be a number field, and N a square-free number. Let \mathfrak{p} be a prime of K, of characteristic p (possibly dividing N) such that the ramification index at p satisfies the inequality

e_\mathfrak{p}(K/Q) < p- 1.

Let E/K be an elliptic curve possessing a K-rational cyclic subgroup C of order N. Let x=j(CN,E) \in X_0(N)(K). Suppose there exists an optimal quotient f : Jo(N)^{new} \to A such that f (x) is of finite order in A(K). (This is necessarily true if the Mordell-Weil group A(K) is finite.) Then E has potentially good reduction at \mathfrak{p}.

When K = \mathbb{Q}, the assumption in this Proposition is satisfied, by the following major theorem, to be found in Mazur’s 1977 paper.

Theorem. The jacobian J_0(N) of X_0(N) admits an abelian variety quotient J, called the Eisenstein quotient, which over \mathbb{Q} has rank 0 (for all prime N \geq 11).

Thus the Hard step from last time follows from this proposition and theorem.

So let’s get started on the Proposition.

Suppose for a contradiction that E has potentially multiplicative reduction at \mathfrak{p}. That is, over every extension of K, E will continue to have bad reduction. This means that our section x on X_0(N), viewed as a smooth scheme over Spec O_K, is a cusp when specialised at \mathfrak{p}. WLOG we may suppose that this cusp is \infty_\mathfrak{p}; this is because the Atkin-Lehner operators act transitively on the cusps.

Embedding X_0(N) into J_0(N) using the Abel-Jacobi map with \infty as the base point, then passing to the new part of the jacobian, and finally composing with the f as in the proposition, we get a map f_\mathbb{Z} : X_0(N)_\mathbb{Z} \longrightarrow A_\mathbb{Z} on Néron Models.

Let me just quickly mention that the sections x and \infty, although they cross at \mathfrak{p}, are not the same; for x lives in the affine part Y_0(N) of X_0(N).

Now clearly \infty maps under f_\mathbb{Z} to the zero section on A, because we used \infty as our base point for the Abel-Jacobi map.

The same thing can’t be said about x a priori. But in fact it’s true; it also maps under f_\mathbb{Z} to the zero section.

Why is that, then?

For this, you need the Specialisation Lemma of Raynaud.

Specialisation Lemma. Let e(K/\mathbb{Q}_p) be < p - 1, and let O be the valuation ring of K. Let G_O be a finite fiat group scheme, and x \in G(O) a section. Then the order of x equals the order of the specialization of x to k (denoted X/k) in G(k).

Now f(x) is of finite order, and by this lemma, it must be the order of the specialisation at \mathfrak{p}, which is 1. That is, f(x) is the zero section.

So now we have two distinct sections, \infty and x, which both map under f to the zero section.

The contradiction arises from the fact, the formal immersion criterion, that f is a formal immersion at \infty_\mathfrak{p}.

What does this mean? There is a notion of formally completing a smooth scheme along a section, which gives the completion of the local ring at the section. If you have a map f : X \to Y of Noetherian schemes, and a point x \in X, then you get an induced map on said completions \hat{O}_{Y,f(x)} \rightarrow \hat{O}_{X,x}. You then say that f is a formal immersion at x if this induced map is surjective.

If you just accept that our f is indeed a formal immersion at \infty_\mathfrak{p}, then the contradiction arises thus. By some analogue of Tate uniformisation, the formal completion of X_0(N) at \infty_\mathfrak{p} is the power series ring \hat{O}_\mathfrak{p}[[q]]. Under this identification, the section \infty is the homomorphism q \mapsto 0, and the section x is the homomorphism q \mapsto something else q_0 say.

Letting \mathcal{A} denote the formal completion of A at 0_\mathfrak{p}, we get a map \mathcal{A} \to \hat{O}_{\mathfrak{p}}[[ q ]]. Since this map is surjective, our two homomorphisms q \mapsto 0,q_0 lift to maps from \mathcal{A}. But now we must conclude that the two maps are the same, because on A the two sections are the same. This is the contradiction.

Next time we’ll look into why the f is a formal immersion.

I am confused about Integers in a Compositum. Please help me.

Let K and L be two number fields with rings of integers O_K and O_L. We define the compositum KL of K and L to be the field over \mathbb{Q} generated by K \cup L, all taking place in \overline{\mathbb{Q}}. We similarly define the compositum O_K \cdot O_L of O_K and O_L as the ring over \mathbb{Z} generated by O_K \cup O_L, all taking place in \overline{\mathbb{Z}}.

One can ask about the difference between O_{KL} and O_K \cdot O_L. Clearly, O_K \cdot O_L is contained inside O_{KL}, and is actually of finite index. In other words, O_K \cdot O_L is a full rank sublattice of KL.

But when are they the same?

Well, the obvious keywords thrown into Google brings up a paper in fact from 1980 by Kataoka, where in the first paragraph it is stated that they are the same if K and L are linearly disjoint over \mathbb{Q}.

What does this mean? There are several definitions, but the one that confuses me is this one:

Definition. Two number fields K and L are linearly disjoint over \mathbb{Q} if K \otimes_{\mathbb{Q}} L is a field (and hence the same as KL).

Why does this confuse me?

Well, let K = \mathbb{Q}(\sqrt{2}) and L = \mathbb{Q}(\sqrt{3}). Clearly [1], K \otimes_{\mathbb{Q}} L = \mathbb{Q}(\sqrt{2},\sqrt{3}) and hence they are linearly disjoint, so we get

O_{KL} = O_K \cdot O_L = \mathbb{Z}[\sqrt{2},\sqrt{3}].

Here comes the confusion: Shouldn’t this also work if I replace 3 by 7?

That is, if K = \mathbb{Q}(\sqrt{2}) and L = \mathbb{Q}(\sqrt{7}), then shouldn’t I have

O_{KL} = O_K \cdot O_L = \mathbb{Z}[\sqrt{2},\sqrt{7}]?

The problem is that I shouldn’t; the first equality in this equation is not valid; indeed, \frac{1}{2}(\sqrt{2} + \sqrt{7}) is in O_{KL}, and not in the compositum O_K \cdot O_L. Based on what Kataoka said, we must conclude that K and L are not linearly disjoint.

So, here’s what the question boils down to:

Why is \mathbb{Q}(\sqrt{2}) \otimes_\mathbb{Q} \mathbb{Q}(\sqrt{3}) = \mathbb{Q}(\sqrt{2},\sqrt{3}), but \mathbb{Q}(\sqrt{2}) \otimes_\mathbb{Q} \mathbb{Q}(\sqrt{7}) \neq \mathbb{Q}(\sqrt{2},\sqrt{7})?

[1]: (As ever, “Clearly” is synonymous with “I should probably check that”)

Gestern Abend

Gestern Abend haben meine Mitbewohnerin und ich einige Freunde zum Abendessen herzlich eingeladen. Seit drei Wochen haben wir alle zusammen am Samstag gegessen, und danach einen gewissen Brettspiel, der einfach großartig ist und für bis zu 8 Personen geeignet ist, gespielt (Übrigens heißt der Spiel “Dominion”). Letzte Woche sind wir zu Ihnen gegangen, und gestern waren wir dran.

Du musst begreifen, dass meine Mitbewohnerin eine abenteuerliche Köchin ist, und ist voller Stolz darauf, eine gnädige Gastgeberin zu sein. Alles was Du kannst, das kann ich viel besser. Und deswegen hatte sie beschlossen, eine Fondue-Party zu geben.

Es war mein erstes Mal, Fondue zu haben. Wir hatten Paprika, Würstchen, Salami, Chorizo, verschiedene Steaksorte, Brot und viel mehr. Als Substanzen hatten wir Fett, Käse und Schokoladenfondue.

Während des Abends hatte ich vielleicht ein bisschen zu viel zu trinken, und möglicherweise machte mich zum Narren. Aber nur die anderen Anwesenden können das bestätigen.

Mazur’s Theorem on Rational Isogenies of Prime Degree


(Firstly let me apologise for the poor presentation of \LaTeX in this post. This is my first ever blog post, and unfortunately I can’t blog using \LaTeX, but rather I must use HTML, which I’ve never before used. Advice on how I can make the formulae bigger would be most welcome.)


In 1978, Mazur proved the following theorem, which in this blog post will be referred to as the `Main Theorem’.

Theorem. Let E/\mathbb{Q} be an elliptic curve which possesses a \mathbb{Q}-rational isogeny of prime degree N. Then

N \in \left\{\mbox{primes }\leq 19\right\} \cup \left\{37,43,67,163\right\}

The proof contains several key ideas that were generalised and used in the proof of the uniform boundedness theorem.

Theorem (Uniform Boundedness Theorem).Let K be a number field of degree d. Then there is a constant B(d) depending only on d such that, if E/K is an elliptic curve with a K-rational torsion point of order N, then N < B(d).

It is fair to say that the strategy of proof of the uniform boundedness theorem closely follows Mazur’s proof; there are of course further subtleties and ideas needed in the Uniform Boundedness Theorem. In particular, making B(d) effective was made possible by work of Oesterlé and Parent, and is a great story in its own right. Here, and in the subsequent blog posts, however, I’d like to focus on Mazur’s 1978 theorem.

I will not explain all of the details; think of these posts more as a roadmap, a sketch of the landscape, a museum exhibit. They’re not a substitute for truly understanding all of the mechanics. References will always be to Mazur’s paper “Rational Isogenies of Prime Degree”, Inventiones math. 44, 129 – 162, 1978, hereafter referred to as [M].

The Hard Step

By far the hardest step of the proof is establishing the following innocent looking result.

Theorem. (Corollary 4.4 in [M])
Let N = 11 or a prime \geq 17, and suppose we have an elliptic curve E/\mathbb{Q} with a \mathbb{Q}-rational N-isogeny. Then E has potentially good reduction at all odd primes.

Future blog posts will reveal the depths of this theorem. For this post, we will see how the main theorem follows from this one. Hence, for the remainder of the post, assume this theorem is true.

Fix a number field K (which will ultimately be taken to be \mathbb{Q}), and suppose given an elliptic curve E/K with a K-rational N-isogeny, for N a rational prime. We wish to bound N in terms of only K (and hopefully, only on its degree).

The First Step – Study the Isogeny Character

The first step is to study the isogeny character associated to E. The kernel of the isogeny is a K-rational subgroup C_N of order N. Galois acts on this subgroup, giving the isogeny character

r : Gal(\bar{K}/K)^{ab} \longrightarrow Aut(C_N) \cong (\mathbb{Z}/N\mathbb{Z})^\ast.

Observe that it is the top-left entry of the image of the mod-N representation attached to E:

Im(\bar{\rho}_{E,N}) \cong \left( \begin{array}{cc} r & \ast \\ 0 & \chi/r \end{array} \right);

Here \chi is the mod-N cyclotomic character.

Mazur proves that, when N is inert in K and E has potentially good reduction at N (which will be the case when K=\mathbb{Q}), the isogeny character is `rigid’ in a strong sense. To make this precise, make the following definitions.

1.m = (N - 1)/2

2.n = numerator of \left(\frac{N-1}{12}\right)

3.t=m/n

For a reason to do with “exceptional automorphisms of the pair (C_N,E)” (see Lemma 5.1), we must restrict to the primes N that are inert in K. We then have the following statement.

Proposition. (Proposition 5.1 in [M])
If E has potentially good reduction in characteristic N, the isogeny character r may be written in the form

r = \alpha \cdot \chi^k,

where \alpha^{2t} is unramified everywhere, \chi : Gal(\bar{K}/K)^{ab} \longrightarrow Gal(\bar{\mathbb{Q}}/\mathbb{Q})^{ab} \longrightarrow \mathbb{F}_N^{\ast} is the cyclotomic character, and where the integer k takes on only these values modulo m:

a.k \equiv 0 \ or \ 1 \mod m

b.k \equiv 1/2 \mod m (only possible if m \neq 0 \mod 2)

c.k \equiv 1/3 \ or \ 2/3 \mod m (only possible if m \neq 0 \mod 3)

To recap the situation we’re in: we’ve fixed a number field K, an elliptic curve E/K with K-rational N-isogeny; we’re assuming that N is inert in K, that E has potentially good reduction at N; and we’ve remarked that these assumptions hold when K = Q. We’ve also written the isogeny character r as \alpha \cdot \chi^k, as in the above proposition.

We now bring in another prime \mathfrak{p} different to N, and again assume that E has potentially good reduction at \mathfrak{p}.

It is not too hard to show (see bottom of page 152, top of page 153 in [M]) that, in fact, E has good reduction at \mathfrak{p} (not merely potentially good reduction), and further, that the isogeny character of the reduced curve (which also has an N-isogeny) is of the form

b_{\mathfrak{p}} \cdot \chi^k

where b_{\mathfrak{p}} is an unramified character.

We now have an elliptic curve \tilde{E} over a finite field \kappa(\mathfrak{p}) with a \kappa(\mathfrak{p})-rational N-isogeny. We also have a tight control on what its isogeny character must look like. This finiteness is crucial, as you’ll see in the next two sections.

The Second Step – Congruences mod N

Continuing from the last section, let q_0 denote the cardinality of \kappa(\mathfrak{p}), and let \sigma_{\mathfrak{p}} \in Gal(\overline{\kappa(\mathfrak{p})}/\kappa(\mathfrak{p})) be the Frobenius automorphism. Let’s compute the trace of \sigma_\mathfrak{p} acting on \tilde{E}[N]. We know the image of the mod-N Galois representation is

Im(\bar{\rho}_{\tilde{E},N}) \cong \left( \begin{array}{cc} b_\mathfrak{p} \chi^k & \ast \\ 0 & b_{\mathfrak{p}}^{-1} \chi^{1-k} \end{array} \right) \in GL_2(\mathbb{F}_N);

since \chi(\sigma_\mathfrak{p}) = q_0, we get that the trace of Frobenius \sigma_\mathfrak{p} is congruent mod N to

b_\mathfrak{p}(\sigma_\mathfrak{p})q_0^k + b_{\mathfrak{p}}^{-1}(\sigma_\mathfrak{p})q_0^{1-k}.

But there are only finitely many possible values for the trace of \sigma_\mathfrak{p} of an elliptic curve over a finite field \kappa(\mathfrak{p})! In fact, if E/k is an elliptic curve defined over a finite field k, and we regard E as lying over a bigger finite field K, and we then compute the trace of Frobenius over this bigger field K, then there are still only finitely many values that the trace of Frobenius can take. What’s more, it is really easy to compute these possible values. If k has q_0 elements, and K has q_0^v elements, then the following SAGE code will give you the possible values. 



def foo(q0,v):


    b = floor(2*sqrt(q0))

    w = []


    for a in range(-b,b+1):

        t = expand(((a + sqrt(a^2 – 4*q0))/2)^v + ((a – sqrt(a^2 – 4*q0))/2)^v)

        w.append(t)


    X = Set(w)

    Y = X.list()


    return Y

Let a(\mathbb{F}_{q_0^v}/\mathbb{F}_{q_0}) denote these possible values. Here are some examples.

a.a(\mathbb{F}_{9^6}/\mathbb{F}_9) \in \left\{1458, 478, 658, -1358, -782, -1458\right\}

b.a(\mathbb{F}_{3^{12}}/\mathbb{F}_3) \in \left\{1458, 658, -1458\right\}

c.a(\mathbb{F}_{17^4}/\mathbb{F}_{17}) \in \left\{322, -574, 47, 578, -254, -497, -353, 511 \right\}

So, using this new a notation, we proved above that

b_\mathfrak{p}(\sigma_\mathfrak{p})q_0^k + b_\mathfrak{p}^{-1}(\sigma_\mathfrak{p})q_0^{1-k}

is congruent mod N to one of the a(\mathbb{F}_{q_0}/\mathbb{F}_{q_0}).

We can get rid of the b_\mathfrak{p} if we know its order v say. Viewing \tilde{E} as a curve over \mathbb{F}_{q_0^v} instead of over \mathbb{F}_{q_0}, and computing the trace of Frobenius over the bigger field, we get that

q_0^{vk} + q_0^{v-vk}

is congruent mod N to one of the a(\mathbb{F}_{q_0^v}/\mathbb{F}_{q_0}).

Finally, we set K = Q; so \mathfrak{p} = q_0 = p say. In this case, one can check (Lemma 5.3 in [M]) that the order of \alpha in Proposition 3.1 above is a divisor of 12; it follows that the order of b_p, which I called v, also divides 12, and hence we get the following statement; I repeat the assumptions for convenience.

Key Proposition. (Corollary 6.1 in [M])
Let E/\mathbb{Q} be an elliptic curve with a \mathbb{Q}-rational N-isogeny, for N = 11 or \geq 17 prime. Let p \neq N be an odd prime, and write the isogeny character of E as \alpha \cdot \chi^k according to proposition 3.1. Then p^{12k} + p^{12-12k} is congruent to one of the a(\mathbb{F}_{p^{12}}/\mathbb{F}_p) mod N.

This is the Key Proposition that will give us the Main Theorem.

The Third and Final Step – Analysing the Cases of k

We know that k has to be congruent to one of 0,1/3,1/2,2/3,1. But since the canonical involution w on X_0(N) interchanges k and 1-k, we only need to consider three cases: k \equiv 0,1/3,1/2 \mod m.

In the k \equiv 0 case, we apply Key Proposition with p=3. We get

1 + 3^{12} \equiv 1458, 658, -1458 \mod N

which implies

N = 2,3,5,7,13,19,37,97.

For our range of N, this gives three possibilities, namely, N = 19,37,97. There is in fact an elliptic curve over Q with a 37-isogeny; we can rule out 19 and 97 by running key proposition with p = 5 and getting a similar list of N, which doesn’t contain either 19 or 97.

In the k \equiv 1/3 case, we apply Key Proposition with p=3 again to get N = 11 and 17, both of which occur.

In the k \equiv 1/2 case, we run into trouble, because one of a(\mathbb{F}_{p^{12}}/\mathbb{F}_p) is always 2p^6. So we can’t use our wonderful key proposition.

Nevertheless, Mazur works around this. We can suppose that 2k \equiv 1 + m \mod N-1, t = 1 or 3, and N \equiv -1 \mod 4. Under these assumptions, he proves this claim:

Claim. For all odd primes p < N/4, we have \left(\frac{p}{N}\right) = -1.

He then proves that this claim implies that \mathbb{Q}(\sqrt{-N}) has class number 1 (see page 155); the theorem of Heegner-Baker-Stark then tells us that N must be 11,19,43,67 or 163.

Next Time

Next Time we’ll make a start on the Hard Step.


Acknowledgements/Shout-Outs. I thank Lloyd Yu-West for giving me the idea for writing a series of posts on this theme. I hope, if I have time, to talk about the Uniform Boundedness Theorem, and the work of Kamienny, Merel, Oesterlé and Parent. I also thank Martin Orr, from whose blog I shamelessly stole the HTML coding. Oh, and Hi Liya!