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Is it possible to write the Legendre symbol as an element of the cohomology of some kind? We certainly have that it is multiplicative in both numerator and denominator:

$$ \left( \frac{a}{p} \right)\left( \frac{b}{p} \right) = \left( \frac{ab}{p} \right) $$

Technically I should say Jacobi symbol, since I don't mind if the denominator is composite. We have $(\frac{m}{n})$ for any two relatively prime ideals or numbers with $(m,n) = \mathbb{Z}$.

$$ \left( \frac{a}{p} \right)\left( \frac{a}{q} \right) = \left( \frac{a}{pq} \right) $$

and then we have the quadratic reciprocity relation:

$$ \left( \frac{p}{q} \right)\left( \frac{q}{p} \right) = (-1)^{\frac{p-1}{2}\frac{q-1}{2}}$$

In a way, can we just say that $\alpha = \frac{p}{q} \in \mathbb{Q}^\times$ an say that $(\cdot): \mathbb{Q}^\times \to \mu_2 = \{ +1, -1\}^\times$. Then reciprocity could say that $(\alpha)(\alpha^{-1}) = (-1)^{\dots}$ under the inversion relations as if some kind of cocycle.

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  • $\begingroup$ Browsing these notes of Lemmermeyer on Class Field Theory for a less clumsy formulation, also these notes of Milne. $\endgroup$
    – cactus314
    Commented Oct 26, 2018 at 22:14
  • $\begingroup$ Let $L/K$ be an abelian extension. GCFT says there is a reciprocity map $\mathrm{rec}: \mathbb{A}_K^{\times} \rightarrow \mathrm{Gal}(L/K)$, and $\mathrm{rec}(K^{\times}) \equiv 1$. If $p\equiv 1 \pmod{4}$ and you take $L = \mathbb{Q} (\sqrt{p})$ and $K=\mathbb{Q}$, then the $\mathrm{rec}$ maps reduces to $\mathbb{Q}^{\times} \rightarrow \{ \pm 1\}$ (which is what you have said) and if you work out how $\mathrm{rec}$ is defined, you can check that $\mathrm{rec}(q)=1$ precisely says the quadratic reciprocity. Maybe this is partly what you are looking for? $\endgroup$
    – dyf
    Commented Oct 27, 2018 at 1:13
  • $\begingroup$ @dalbouvet is the Jacobi symbol a first or second cohomology class of $\mathbb{Q}^\times$? perhaps $H^1(\mathbb{Q}^\times, \mu_2)$ $\endgroup$
    – cactus314
    Commented Oct 27, 2018 at 1:49
  • $\begingroup$ How would you define a $\mathbb{Q}^{\times}$-action on $\{\pm 1\}$? If you define it using the Jacobi symbol (regardless of whether this defines a well-defined action), then I don't think it satisfies the $1$-cocycle condition. It can never be an element in $H^2$ because a priori it's not a function from $\mathbb{Q}^{\times} \times \mathbb{Q}^{\times} \rightarrow \{ \pm 1\}$. $\endgroup$
    – dyf
    Commented Oct 27, 2018 at 2:13

1 Answer 1

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For a fixed prime $p$, the Legendre symbol $(\frac.p)$ originates in a particular case of the so-called Hilbert symbols of local CFT. Here is the general setting (see e.g. Serre's "Local Fields", chap. XIV):

Let $K$ be a finite extension of $\mathbf Q_p$, containing the group $\mu_n$ of $n$-th roots of unity. Throughout, we'll write $H^i(K,A)$ for the $i$-th cohomology group of the absolute Galois group of $K$ acting on $A$. Consider the cup-product $\cup: H^1(K,\mu_n)\otimes H^1(K,\mu_n) \to H^2(K,(\mu_n)^{\otimes 2})=H^2(K,\mu_n)$, where the last equality is due to the hypothesis that $K$ contains $\mu_n$. The image of $a\otimes b$ is the Hilbert symbol, denoted $(a,b)_v$, where $v$ is the valuation of $K$ (just a reminder). But $H^1(K,\mu_n)\cong K^*/K^*{^n}$ by Kummer theory, and $H^2(K_v,\mu_n)\cong \mu_n$ by local CFT (local theory of Brauer groups), so the Hilbert symbol can be viewed as a pairing $(a,b)\in K^*/K^*{^n}\otimes K^*/K^*{^n} \to (a,b)_v \in \mu_n$ . In our Kummerian case, the "explicit reciprocity laws" of local CFT amount to the explicit calculation of the Hilbert symbols. The tame case ($p\nmid n)$ is well known (loc. cit., XIV, §3). The wild case ($p \mid n)$ has been the subject of many studies (Artin, Hasse, etc.) which culminted, I think, in the reciprocity law of Bloch-Kato (circa 1990, but unfortunately, technically very difficult to prove, and even to state).

One recovers the Legendre symbol when taking $K=\mathbf Q_p$ and $n=2$. In the tame case ($p$ odd), if $a=a'p^\alpha$ and $b=b'p^\beta$, then $(a,b)_p=(-1)^{\frac {p-1}2\alpha \beta}(\frac {b'}p)^\alpha (\frac {a'}p)^\beta$ (loc. cit.). In particular, $(p,p)_p =(-1)^\frac {p-1}2$, and $(p,b)_p=(\frac {b}p)$ if $b$ is a unit of $\mathbf Q_p$. In the wild case ($p=2)$, a direct calculation shows the following: if $u$ is a unit of $\mathbf Q_2$, let $\omega(u)$ be the class mod $2$ of $\frac {u^2-1}8$, $\epsilon (u)$ be the class mod $2$ of $\frac {u-1}2$; then $(2,u)_2=(-1)^{\omega(u)}$ if $u$ is a unit, and $(u,v)_2=(-1)^{\omega(u)\epsilon (u)}$ if $u,v$ are units. This covers all the possible cases (loc. cit.). Of course the quadratic reciprocity law can be recovered from the previous explicit symbols.

As for the Jacobi symbol, it can be derived directly from the power residue symbol of global CFT (see Cassels-Fröhlich, Exercise 1-9), but ultimately, according to the "local-global principle" of CFT, this consists in putting together the local symbols ./.

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