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I know it seems excessive, but I have been trying to understand the relationship between two concepts:

  • Galois cohomology
  • Fermat Descent

The first one is very abstract and I know very little about it. Many sources indicate there should be some kind of relationship e.g. Wikipedia

In this way, abstractly-defined cohomology groups in the theory become identified with descents in the tradition of Fermat.

and I can go find real math textbooks that indicate the same, but none of them really indicate how cohomology was (implicitly used). The same Wikipedia article on descent cites two examples:

  • $\sqrt{2} \notin \mathbb{Q}$
  • $p = a^2 + b^2$ iff $p = 4k+1$

There's lots of discussion and staring at the result I can see "cohomological" things about them, but none of the discussions really link the abstract and concrete discussions.

I have seen Skorobogatov's book still doesn't show me how to approach this specific case.

Attaching names to all the concepts, we are looking for points in $X = \{ x^2 - 2y^2 = 0 \} \subseteq \mathbb{Q}^2 $. I could think of several approaches:

#1 $\sqrt{2} \in \mathbb{R}$, I forget if $\sqrt{2} \in \mathbb{Q}_2$, but certainly $\sqrt{2} \notin \mathbb{Q}_3, \mathbb{Q}_5$. Therefore $\sqrt{2} \notin \mathbb{Q}$ by Hasse principle.

In other words, $X(\mathbb{A}) \neq \varnothing$ but $X(\mathbb{Q}_5) = \varnothing$ so that $X(\mathbb{A}) = \varnothing$. This is cheating, but we can see that an $H^1$ was involved.

#2 There is a map $X(\mathbb{Z}) \stackrel{\div2}{\to} X(\mathbb{Z})$ by $(x,y) \to (y/2, x)$. Once can produce an $\mathbb{R}$-valued invariant (such as a height) that decreases indefinitely.

In all these examples I don't know what $H^1$ is being used? Perhaps there is a torsor $T$ and a field $\mathbb{Q}$ and we are computing $H^1( \mathbb{Q}, T)$ ?

I have been reading about strong approximation in it's various formulations.

$$ X(\mathbb{Q}) \to X(\mathbb{A}) \text{ is dense} $$

For my soft-ball examples, the use of cohomology should be very minimal, but I am failing to see the connection at all.

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  • $\begingroup$ There is an excellent book on Galois cohomology by Serre. $\endgroup$
    – Jason Starr
    Commented Jun 10, 2017 at 13:58
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    $\begingroup$ Certainly $\sqrt2 \not\in \mathbb Q_2$, because $\mathrm{ord}_2(2) = 1$ and $\mathrm{ord}$ is $\mathbb Z$-valued on $\mathbb Q_2$. I'm not sure if it's what you have in mind, but there is an identification $\mathbb Q^\times/(\mathbb Q^\times)^2 \cong \mathrm H^1(\mathbb Q, \mu_2)$ sending $k \mapsto (\sigma \mapsto (\sigma - 1)\sqrt k)$, that is, sending a square class to the coboundary in $\mathrm GL_1$ corresponding to either square root, which is necessarily $\mu_2$-valued. $\endgroup$
    – LSpice
    Commented Jun 10, 2017 at 14:00
  • $\begingroup$ supplement: 28 proofs of the irrationality of $\sqrt{2}$ $\endgroup$
    – john mangual
    Commented Jun 10, 2017 at 14:32

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