# Please check my 6-line proof of Fermat's Last Theorem.

Kidding, kidding. But I do have a question about an $n$-line outline of a proof of the first case of FLT, with $n$ relatively small.

Here's a result of Eichler (remark after Theorem 6.23 in Washington's Cyclotomic Fields): If $p$ is prime and the $p$-rank of the class group of $\mathbb{Q}(\zeta_p)$ satisfies $d_p<\sqrt{p}-2$, then the first case of FLT has no non-trivial solutions. Once you know Herbrand-Ribet and related stuff, the proof of this result is even rather elementary.

The condition that $d_p<\sqrt{p}-2$ seems reminiscent of rank bounds used with Golod-Shafarevich to prove class field towers infinite. More specifically, a possibly slightly off (and definitely improvable) napkin calculation gives me that for $d_p>2+2\sqrt{(p-1)/2}$, the $p$-th cyclotomic field $\mathbb{Q}(\zeta_p)$ has an infinite $p$-class field tower. It's probably worth emphasizing at this point that by the recent calculation of Buhler and Harvey, the largest index of irregularity for primes less than 163 million is a paltry 7.

So it seems natural to me to conjecture, or at least wonder about, a relationship between the unsolvability of the first case of FLT and the finiteness of the $p$-class field tower over $\mathbb{Q}(\zeta_p)$. Particularly compelling for me is the observation that regular primes (i.e., primes for which $d_p=0$) are precisely the primes for which this tower has length 0, and have obvious historical significance in the solution of this problem. In fact, the mechanics of the proof would probably/hopefully be to lift the arithmetic to the top of the (assumed finite) p-Hilbert class field tower, and then use that its class number is prime to $p$ to make arguments completely analogously to the regular prime case.

I haven't seen this approach anywhere. Does anyone know if it's been tried and what the major obstacles are, or demonstrated why it's likely to fail? Or maybe it works, and I just don't know about it?

Edit: Franz's answer indicates that even for a relatively simple Diophantine equation (and relatively simple class field tower), moving to the top of the tower introduces as many problems as it rectifies. This seems pretty compelling. But if anyone has any more information, I'd still like to know if anyone knows or can come up with an example of a Diophantine equation which does benefit from this approach.

-
+1, even though your title made my blood pressure go up by about +50 ;) –  Yemon Choi Jan 29 '10 at 22:08
-1. Change the title –  Boris Bukh Jan 29 '10 at 23:00
+1: I thought it was funny –  Harry Gindi Jan 29 '10 at 23:04
+1: I thought it was very funny –  Julio César Salazar García Jan 30 '10 at 1:54
+i: Write the title backwards. –  Qiaochu Yuan Jan 30 '10 at 5:46

Suppose you have a diophantine problem whose solution is connected with the structure of the p-class group of a number field K. Then you have the following options:

1. Use ideal arithmetic in the maximal order OK
2. Replace OK by a suitable ring of S-integers with trivial p-class group
3. Replace K by the Hilbert class field, which (perhaps) has trivial p-class group.

Experience with descent on elliptic curves has shown me that ultimately, the equations you have to solve in methods 1 and 2 are the same; moreover, the approach using ideals is a lot less technical than using factorial domains in S-integers (the class group relations come back in through the larger rank of the group of S-units). I am certain that the route via the Hilbert class field is even more technical: again, the unit group in the class field will produce more difficulties than a trivial class group will eliminate.

Edit. As an example illustrating my point in a very simple example, let me solve the diophantine equation $x^2 + 5y^2 = z^2$ in several different ways. I will always assume that $\gcd(x,y) = 1$.

1. Elementary Number Theory

The basic idea is factoring: from $5y^2 = (z+x)(z-x)$. Since $d = \gcd(z-x,z+x) = \gcd(2z,z+x) \mid 2$ we have $d = 1$ or $d = 2$; moreover we clearly have $z-x > 0$. This gives $z+x = da^2$, $z-x = 5db^2$ or $z+x = 5da^2$, $z-x = db^2$. Solving for $x$ and $z$ yields $$x = \pm \frac d2 (a^2 - 5b^2), \quad y = dab.$$

2. Parametrization

Set $X = \frac xz$ and $Y = \frac yz$; then $X^2 + 5Y^2 = 1$. Take the line $Y = t(X+1)$ through the obvious point $(-1,0)$; the second point of intersection is given by $$X = \frac{1-5t^2}{1+5t^2}, \quad Y = \frac{2t}{1+5t^2}.$$ Dehomogenizing using $t = \frac ba$ and $X = \frac xz$ etc. gives the projective parametrization $$(x:y:z) = (a^2-5b^2:2ab:a^2+5b^2).$$ If $ab$ is odd, all coordinates are even, and we find $$x = \frac12(a^2 - 5b^2), \quad y = ab;$$ if $a$ or $b$ is even we get $$x = a^2 - 5b^2, \quad y = 2ab$$ as above.

3. Algebraic Number Theory

Consider the factorization $$(x + y\sqrt{-5}\,)(x - y\sqrt{-5}\,) = z^2$$ in the ring of integers of the number field $K = {\mathbb Q}(\sqrt{-5}\,)$. The class number of $K$ is $2$, and the ideal class is generated by the prime ideal ${\mathfrak p} = (2,1+\sqrt{-5}\,)$.

The ideal $(x + y\sqrt{-5}, x - y\sqrt{-5}\,)$ is either $(1)$ or ${\mathfrak p}$; thus $$(x + y\sqrt{-5}\,) = {\mathfrak a}^2, \quad (x - y\sqrt{-5}\,) = {\mathfrak b}^2$$ in the first and $$(x + y\sqrt{-5}\,) = {\mathfrak p}{\mathfrak a}^2, \quad (x - y\sqrt{-5}\,) = {\mathfrak p}{\mathfrak b}^2$$ in the second case.

The second case is impossible since the left hand side as well as ${\mathfrak a}^2$ are principal, but ${\mathfrak p}$ is not. We could have seen this immediately since $x$ and $y$ cannot both be odd.

In the first case, assume first that ${\mathfrak a} = (a + b\sqrt{-5}\,)$ is principal. Since the only units in ${\mathcal O}_K$ are $\pm 1$, this gives $x + y \sqrt{-5} = \pm(a+b\sqrt{-5}\,)^2$ and hence $$x = \pm (a^2 - 5b^2), \quad y = \pm 2ab.$$ If ${\mathfrak a}$ is not principal, then ${\mathfrak p}{\mathfrak a} = (a+b\sqrt{-5}\,)$ is, and from $({\mathfrak p}{\mathfrak a})^2 = 2(x+y\sqrt{-5}\,)$ we similarly get $$x = \pm \frac12(a^2 - 5b^2), \quad y = \pm ab.$$

4. S-Integers

The ring $R = {\mathbb Z}[\sqrt{-5}\,]$ is not a UFD, but $S = R[\frac12]$ is; in fact, $S$ is even norm-Euclidean for the usual norm in $S$ (the norm is the same as in $R$ except that powers of $2$ are dropped). It is also easily seen that $S^\times = \langle -1, 2 \rangle$. From $$(x + y\sqrt{-5}\,)(x - y\sqrt{-5}\,) = z^2$$ and the fact that the factors on the left hand side are coprime we deduce that $x + y\sqrt{-5} = \varepsilon \alpha^2$ for some unit $\varepsilon \in S^\times$ and some $\alpha \in S$. Subsuming squares into $\alpha$ we may assume that $\varepsilon \in \{\pm 1, \pm 2\}$. Setting $\alpha = \frac{a + b\sqrt{-5}}{2^t}$, where we may assume that $a$ and $b$ are not both even, we get $$x + y \sqrt{-5} = \varepsilon \frac{a^2 - 5b^2 + 2ab\sqrt{-5}}{2^{2t}}.$$ It is easily seen that we must have $t = 0$ and $\varepsilon = \pm 1$ or $t = 1$ and $\varepsilon = \pm 2$; a simple calculation then yields the same formulas as above.

5. Hilbert Class Fields

The Hilbert class field of $K$ is given by $L = K(i)$. It is not too difficult to show that $L$ has class number $1$ (actually it is norm-Euclidean), and that its unit group is generated by $i = \sqrt{-1}$ and $\omega = \frac{1+\sqrt{5}}2$ (we only need to know that these units and their product are not squares). From $$(x + y\sqrt{-5}\,)(x - y\sqrt{-5}\,) = z^2$$ and the fact that the factors on the left hand side are coprime in ${\mathcal O}_K$ we deduce that $x + y \sqrt{-5} = \varepsilon \alpha^2$. Subsuming squares into $\alpha^2$ we may assume that $\varepsilon \in \{1, i, \omega, i\omega \}$. Applying the nontrivial automorphism of $L/K$ to $x + y \sqrt{-5} = \varepsilon \alpha^2$ we find $\varepsilon \alpha^2 = \varepsilon' {\alpha'}^2$. Since the ideal ${\mathfrak a} = (\alpha)$ is fixed and since $L/K$ is unramified, the ideal ${\mathfrak a}$ must be an ideal in ${\mathcal O}_K$. Thus either ${\mathfrak a} = (a+b\sqrt{-5}\,)$ is principal in $K$, or ${\mathfrak p} {\mathfrak a} = (a+b\sqrt{-5}\,)$ is; in the second case we observe that ${\mathfrak p} = (1+i)$ becomes principal in ${\mathcal O}_L$.

Thus either $$x + y \sqrt{-5} = (a+b\sqrt{-5}\,)^2 \quad \text{or} \quad x + y \sqrt{-5} = i \Big(\frac{a+b\sqrt{-5}}{1+i}\,\Big)^2,$$ giving us the same formulas as above.

Avoiding ideal arithmetic in $K$ and only using the fact that ${\mathcal O}_L$ is a UFD seems to complicate the proof even more.

Edit 2 For good measure . . .

6. Hilbert 90

Consider, as above, the equation $X^2 + 5Y^2 = 1$. It shows that the element $X + Y \sqrt{-5}$ has norm $1$; by Hilbert 90, we must have $$X + Y \sqrt{-5} = \frac{a+b\sqrt{-5}}{a-b\sqrt{-5}} = \frac{a^2 - 5b^2 + 2ab\sqrt{-5}}{a^2 + 5b^2}.$$ Dehomogenizing via $X = \frac xz$ and $Y = \frac yz$ yields the same projective parametrization as above, and we end up with the familiar formulas.

7. Binary Quadratic Forms The equation $x^2 + 5y^2 = z^2$ tells us that the form $Q_0(X,Y) = X^2 + 5Y^2$ with fundamental discriminant $\Delta = -20$ represents a square; this implies that $Q_0$ lies in the principal genus (which is trivial since $Q_0$ is the principal form), and that the representations of $z^2$ by $Q_0$ come from composing representations of $z$ by forms $Q_1$ with $Q_1^2 \sim Q_0$ with themselves.

There are only two forms with discriminant $\Delta$ whose square is equivalent to $Q_0$: the principal form $Q_0$ itself and the form $Q_1(X,Y) = 2X^2 + 2XY + 3Y^2$. Thus either $$z = Q_0(a,b) = a^2 + 5b^2 \quad \text{or} \quad z = Q_1(a,b) = 2a^2 + 2ab + 3b^2.$$ The formulas for Gauss composition of forms immediately provide us with expressions for $x$ and $y$ in terms of $a$ and $b$, but they can also be checked easily by hand. In the first case, we get $$x^2 + 5y^2 = (a^2 + 5b^2)^2 = (a^2 - 5b^2)^2 + 5(2ab)^2,$$ and in the second case we can reduce the equations to this one by observing that $2Q_1(a,b) = A^2 + 5b^2$ with $A = 2a+b$, which gives $$x^2 + 5y^2 = \frac14\Big(A^2 + 5b^2\Big)^2 = \Big(\frac{A^2 - 5b^2}2\Big)^2 + 5(Ab)^2.$$

-
What is "(\ref{E5})"? (Very nice summary of methods!) –  darij grinberg Aug 30 '10 at 18:57
This is absolutely fantastic. –  Cam McLeman Aug 30 '10 at 19:27
@darji: corrected. thanks –  Franz Lemmermeyer Aug 31 '10 at 5:48
So certainly a lot of these techniques become hard/impossible as you move away from quadratic Diohpantine equations, right? I'm thinking methods 1, 2, and 7 in particular -- and 3 gets harder and harder as well. Maybe this should just be an exercise for myself, but it is clear to you that it would be rather difficult to find examples where the Hilbert Class Field approach would the best/only one? –  Cam McLeman Aug 31 '10 at 12:34
As you can see, I have used Galois theory to descend from the Hilbert class field back to K; without this descent, the approach via the class field seems to be too technical to be of any use. I haven't seen an approach for solving a specific diophantine equation directly with the Hilbert class field (as opposed to indirect methods: class fields are used in constructions of solutions using Heegner points), and I don't expect to see one. But I'd love to be surprised -) –  Franz Lemmermeyer Aug 31 '10 at 14:50

W. McCallum wrote a couple of papers in the early 90's connecting the index of irregularity and the rank of the Mordell-Weil of the Jacobian of the Fermat curve over the rationals. Then he used the (Skolem-)Chabauty-Coleman method to show that if $d_p<(p+5)/8$ then the Fermat curve had at most $2p-3$ rational points. No explicit connection with class field towers though.

-