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?

  • 2
    $\begingroup$ 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. $\endgroup$
    – KCd
    Dec 29, 2012 at 1:41

2 Answers 2


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.

  • $\begingroup$ How to determine the Artin symbol of $K/Q$ over the prime $q$? $\endgroup$
    – Ash GX
    Jan 2, 2013 at 14:16
  • $\begingroup$ 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. $\endgroup$
    – Ted
    Jan 2, 2013 at 18:19
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    $\begingroup$ 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 $d=p^{*}$. $\endgroup$
    – Tim.ev
    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).

  • $\begingroup$ $@fretty$ 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 $Q(\sqrt{-3})/Q$ $\endgroup$
    – Math123
    Oct 24, 2013 at 5:59
  • $\begingroup$ 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... $\endgroup$
    – fretty
    Oct 24, 2013 at 6:39
  • $\begingroup$ 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). $\endgroup$
    – fretty
    Oct 24, 2013 at 6:43
  • $\begingroup$ Thanks Fretty, Thank you very much $\endgroup$
    – Math123
    Oct 24, 2013 at 6:48
  • $\begingroup$ 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. $\endgroup$
    – fretty
    Oct 24, 2013 at 6:51

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