# Eisenstein and Quadratic Reciprocity as a consequence of Artin Reciprocity, and Composition of Reciprocity Laws

Question 1: I've heard that Eisenstein and Quadratic Reciprocity can be derived from the Artin Reciprocity by applying it to certain field extensions. But I haven't seen on any reference an explicit description of this, and I am here asking for one.

Question 2: I have seen quartic, octic, and sextic reciprocity laws. They look just like by applying some kind of power reciprocity in fields satisfying another power reciprocity law. So is there a general procedure to construct reciprocity of arbitrary powers based on composition of reciprocity of degree 2 and odd primes (Since their product generate the natural numbers)? If not, what is the obstruction?

• For a derivation of cubic reciprocity from Artin reciprocity, see Theorem 2.3.5 (p. 63) of Noah Snyder's senior thesis on Artin L-functions, which is currently available from his Columbia Univ. webpage.
– KCd
Dec 29, 2012 at 1:41

To derive quadratic reciprocity from Artin reciprocity, consider the field extensions $$\mathbb{Q} \subset F=\mathbb{Q}(\sqrt{p^*}) \subset K=\mathbb{Q}(\zeta)$$ where $p$ is a prime $\ne 2$, $\zeta$ is a primitive $p$th root of unity, and $p^* = (-1)^{(p-1)/2} p$.

(To see that $F$ is contained in $K$, look up Gauss sums.)

Let $q$ be another prime, $q \ne p, q \ne 2$. We know that $p^*$ is a square mod $q$ iff $q$ splits in $F$ iff the Artin symbol of $q$ in $F/\mathbb{Q}$ is trivial. (All Galois groups considered here are abelian, so the Artin map depends only on the base prime $q$ and not on any particular prime in the upper field.)

The Artin symbol of $K/\mathbb{Q}$ over the prime $q$ is the element $\sigma_q: \zeta \mapsto \zeta^q$ in the Galois group of $K/\mathbb{Q}$.

But Artin($F/\mathbb{Q}$) is the restriction of Artin($K/\mathbb{Q}$) to $F$, so Artin($F/\mathbb{Q}$) is trivial iff $\sigma_q$ is in the kernel of this restriction map. Both Galois groups are cyclic, and so this restriction map is the unique nontrivial homomorphism from $(\mathbb{Z}/p\mathbb{Z})^{\times}$ to $\{\pm 1\}$. It is easy to check that $\sigma_q$ will be in the kernel of the restriction map iff $q$ is a square mod $p$.

Conclusion: $p^*$ is a square mod $q$ iff $q$ is a square mod $p$. This is quadratic reciprocity.

For other reciprocity laws, you'll have to choose the field $F$ differently. For example, for cubic reciprocity, choose $F$ to be the unique degree 3 extension in $K/\mathbb{Q}$. (Of course, this only exists if $p \equiv 1$ (mod 3), but if that condition fails then cubic reciprocity is not very interesting: everything will be a cube mod $p$.) Things get more complicated because the analog of the step "$p^*$ is a square mod $q$ iff $q$ splits in $F$" is not as simple. But it's the same idea.

Edit: This last paragraph is not quite correct. The proper extensions to consider for cubic reciprocity is $L \subset L(p^{1/3})$ where $L = \mathbb{Q}(\omega)$, $\omega$ a primitive cube root of unity. The Artin conductor of this extension is a divisor of $3p$, and given a prime $q \equiv 1$ (mod 3) in $L$ (relatively prime to $3p$), the image of the ideal $(q)$ given by Artin reciprocity is essentially the unique nontrivial map from the cyclic group $(O_L / q)^{\times}$ to the cyclic group Gal($L(p^{1/3})/L$) of order 3. For details, see the senior thesis of Noah Snyder referenced by KCd in the comments above. For general $n$th power reciprocity, use $L \subset L(p^{1/n})$ where $L$ is the cyclotomic field generated by a primitive $n$th root of unity.

• How to determine the Artin symbol of $K/Q$ over the prime $q$? Jan 2, 2013 at 14:16
• By definition, the Artin symbol of an unramified prime $\mathfrak{q}$ (in $K$) over a prime $q$ (in $\mathbb{Q}$) is the unique automorphism $\sigma \in$ Gal($K/\mathbb{Q}$) such that $\sigma(x) \equiv x^q$ (mod $\mathfrak{q}$) for all $x \in K$. The map $\sigma_q$ above satisfies this property.
– Ted
Jan 2, 2013 at 18:19
• A small remark: if you are already going to use all this algebraic number theory for a proof of quadratic reciprocity, I think it is cleaner to argue that $F\subset K$ by observing that $\mathbb{Q}(\zeta_p)$ must contain a unique quadratic subfield that ramifies only over $p$, hence is of the form $\mathbb{Q}(\sqrt{d})$ where $d\equiv 1\bmod 4$ and has only the prime $p$ as prime divisor. This forces $d=p^{*}$. Sep 20, 2019 at 13:51

For a nice simple example use the extension $\mathbb{Q}(i)/\mathbb{Q}$.

Notice that for an odd prime $p$ (i.e. unramified) the Frobenius element of $p$ relative to this extension is the identity if and only if $p\equiv 1 \bmod 4$.

When showing this algebraically you find that:

Frob$_p(a+ib) \equiv a + (-1)^{\frac{p-1}{2}}bi \bmod \mathfrak{p}$

However, by Eulers criterion the RHS is congruent to $a + \left(\frac{-1}{p}\right)bi\bmod \mathfrak{p}$.

So really the Frobenius element of $p$ here encodes the value of the Legendre symbol $\left(\frac{-1}{p}\right)$.

Thus $\left(\frac{-1}{p}\right) = 1$ if and only if Frob$_p$ is the identity, which is equivalent to $p \equiv 1 \bmod 4$.

This gives one of the supplementary laws. The rest of quadratic reciprocity comes from changing the above extension to something similar (as shown by the other replies).

• $@fretty$ if you don't mind, could you explain how $Frob_{p}(a+ib)\equiv a+(-1)^{\frac{p-1}{2}}bi\pmod{\mathfrak{p}}$ , is it by definition? and does $\frak{p}$ lies above $p$. if so what happens for $Q(\sqrt{-3})/Q$ Oct 24, 2013 at 5:59
• Yes, $\mathfrak{p}$ lies above $p$ and so when working mod $\mathfrak{p}$ (i.e. in the residue field) you will be working in characteristic $p$. By definition Frob$_p (a+ib) \equiv (a+ib)^{p} \bmod \mathfrak{p}$. You get what I write from this... Oct 24, 2013 at 6:39
• If you do it for $\mathbb{Q}(\sqrt{-3})/\mathbb{Q}$ you will find the Legendre symbol $\left(\frac{-3}{p}\right)$ hidden in there for unramified $p$. I gather the workings will not be as nice. You should read D.Cox's book "Primes of the form $x^2 + ny^2$", there is a nice section on Frobenius elements, class field theory, Artin reciprocity and how the weak/strong reciprocity laws follow from it (it is these ones that explicitly give you quadratic, cubic, quartic, Eisenstein reciprocity laws plus many others). Oct 24, 2013 at 6:43
• Thanks Fretty, Thank you very much Oct 24, 2013 at 6:48
• You should try the calculations for this extension by yourself...it is not too hard. The hard bit is in understanding the general proof of quadratic reciprocity above. Oct 24, 2013 at 6:51