## An Obvious Claim

It’s been a while since I saw Serre’s “how to write mathematics badly” lecture, but I’m pretty sure there would have been something about the dangers of using the word “obvious.” After all, if something really is obvious, then it shouldn’t be too difficult to explain why. It is especially embarrassing when someone asks you to clarify a remark/claim in one of your papers which you claim is “obvious” and you find yourself having no idea what the implicit argument was supposed to be. Such a thing happened recently to me, when Toby asked me to explain why the following was true:

Claim: Let $N \equiv 3 \mod 4$ be prime, and let $\epsilon$ be the fundamental unit of $K = \mathbf{Q}(\sqrt{N})$. Then $\epsilon = a + b \sqrt{N}$ where $a$ is even and $b$ is odd.

Proof of Claim: Between Toby, Kevin, and myself, we managed to come up with the argument below, following a suggestion of Toby Rebecca Bellovin: It’s easy enough to see (obvious) that $a$ and $b$ are integers and $N(\epsilon) = 1$. Hence, it suffices to rule out the case that $b$ even and $a$ odd. Write $a^2 - N b^2 = 1$. It follows that $a^2 \equiv 1 \mod N$, and since $N$ is prime, that $a \equiv \pm 1 \mod N$. Assuming that $a$ is odd, write $a = 2NA \pm 1$, and $b = 2B$. Then the equation above becomes

$A(NA \pm 1) = B^2.$

Without loss of generality, assume that $A$ is positive. Then this equation implies that $A$ and $NA \pm 1$ are squares, say $A = d^2$ and $NA \pm 1 = c^2$. But then

$c^2 - N d^2 = (NA \pm 1) - N A = \pm 1,$

and hence $\eta = c + N \sqrt{d}$ is a (smaller) unit (in fact, $\eta^2 = \pm \epsilon$), contradicting the assumption that $\epsilon$ was a fundamental unit. $\quad \square$

This argument is really a 2-descent on the unit group. As Kevin remarked: “So this is a descent argument in a completely elementary situation which I don’t think I’d ever seen before and which proves something that I don’t think I knew … What’s ridiculous is that if the equation had been a cubic and we were after rational solutions then I would have instantly leapt on descent as one of my main tools for attacking it We live and learn!”

So what was I thinking when I wrote the paper? The actual claim in the paper is this: “If $H'$ is the (2 part of the) strict ray class group of $K$ of conductor $(2)$, then $H = H'$, where $H$ is the (2 part of the) class group. The “argument” is as follows:

The proof of [the above] is even more straightforward: it follows immediately from a consideration of the units in $\mathcal{O}^{\times}_K$ and the exact sequence

$\mathcal{O}^{\times}_K \rightarrow (\mathcal{O}_K/2 \mathcal{O}_K)^{\times} \rightarrow H' \rightarrow H \rightarrow 0.$

Well, at least the word obvious was only implicit here. I could try to place the blame on my co-author Matt here, but honestly the phrasing of the claim does sound a little like something I would write.

Next up: a report from Luminy!

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### 4 Responses to An Obvious Claim

1. TG says:

I think the suggestion of using descent was due to Rebecca Bellovin, in fact.

• Noted, thanks!

2. Florent says:

Using the descent argument in your “Proof of Claim” one can also show e.g. that the negative Pell equation $x^2-py^2=-1$ is soluble in integers $x,y$ if $p$ is a prime congruent to $1$ mod $4$. In a beautiful series of paper “Higher descent on Pell conics I, II and III” (available on the arXiv) Lemmermeyer gives historical background on these questions and seems to claim that the descent argument you use goes back to (at least) Legendre.
Also (if you don’t mind me self advertising a bit) together with E. Fouvry we recently made crucial use of the descent argument to study the size of the solutions to Pell’s equation and the (related question of the) size of regulators of real quadratic fields (e.g. here http://msp.org/pjm/2013/262-1/p05.xhtml).

• The claim about the existence of a norm $-1$ unit, on the other hand, is more directly obvious on the Galois side, since otherwise $\mathbf{Q}(\sqrt{p})$ would admit a quadratic extension unramified at all finite primes, which it does not. I’m sure there will be a similar argument for the claim above, except now one has to rule out the existence of a degree eight extension $E/\mathbf{Q}$ containing $\mathbf{Q}(\sqrt{N},\sqrt{-1})$ with limited ramification properties at two. I’m sure this is not so hard, but, perhaps, not “obvious.”