Writting Legendre Symbol as an element of group cohomology of $\mathbb{Q}$

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.

• Browsing these notes of Lemmermeyer on Class Field Theory for a less clumsy formulation, also these notes of Milne. Commented Oct 26, 2018 at 22:14
• 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?
– dyf
Commented Oct 27, 2018 at 1:13
• @dalbouvet is the Jacobi symbol a first or second cohomology class of $\mathbb{Q}^\times$? perhaps $H^1(\mathbb{Q}^\times, \mu_2)$ Commented Oct 27, 2018 at 1:49
• 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 $\mathbb{Q}^{\times} \times \mathbb{Q}^{\times} \rightarrow \{ \pm 1\}$.
– dyf
Commented Oct 27, 2018 at 2:13

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.