Takagi's goal is the following:
- show the existence of sufficiently many cyclic extensions defined
by division values of sn $u$ (the lemniscatic sine);
- prove that each abelian extension of ${\mathbb Q}(i)$ is contained in
the compositum of these fields.
Step 1 is analogous to the construction of the fields of $p^n$-th roots
of unity (in particular, proving the irreducibility of the corresponding
cyclotomic polynomials and determining their dirscriminants), whereas
Step 2 is the analogue of the theorem of Kronecker-Weber.
Step 1
Takagi (thesis, Sect. 6) constructs the following cyclic extensions
of ${\mathbb Q}(i)$:
- For each Gaussian prime $\mu$ such that $p^h$ is the largest power
of the prime $p$ that divides $(N\mu-1)$, there exists a cyclic
extension unramified outside $\mu$ with degree $p^h$.
- For each Gaussian prime $\pi$ with odd prime norm $p \equiv 1 \bmod 4$
and each integer $\lambda \ge 1$ there exists a cyclic extension of
prime power degree $p^\lambda$ unramified outside $\pi$.
- For each prime number $q \equiv 3 \bmod 4$ and each integer
$\lambda \ge 1$ there exist $q^\lambda +1$ cyclic extensions with
degree $q^\lambda$ unramified outside $q$.
- For each Gaussian prime $\mu$ with $N\mu-1 = 2^{h+2} u$, where
$u$ is an odd integer, there is a cyclic extension of degree $2^{h+2}$
unramified outside $(1+i)\mu$ for which the subfield of degree $2^h$
is unramified outside $\mu$.
- For each integer $\lambda > 1$ there are $2^\lambda +1$ cyclic
extensions of degree $2^\lambda$ unramified outside $1+i$.
Takagi also proves the decomposition law in these extensions.
Step 2
In Sect. 9, Takagi begins proving the analogue of the
theorem of Kronecker-Weber. His approach is the one used by Hilbert
in his proof of the theorem of Kronecker-Weber: Given an abelian
extension $L/{\mathbb Q}(i)$, we write $L$ as a compositum of cyclic
extensions of prime power degree. If the odd Gaussian prime $\mu$ is
ramified in $L$, form the compositum of $L$ and the cyclic extension
$K/{\mathbb Q}(i)$ unramified outside $\mu$ constructed in Step 1 and
show that the compositum $KL$ contains a cyclic extension
$M/{\mathbb Q}(i)$ unramified at $\mu$ such that $L$ is contained
in $KM$; the case of wild ramification requires a careful investiagtion
of the ramification subgroups.
By induction, this process reduces the proof to cyclic extensions that
are unramified everywhere. There are various ways of proving that
such extensions do not exist:
We can use Takagi's Lemma, according to which every cyclic extension
of ${\mathbb Q}(i)$ that is normal over ${\mathbb Q}$ is a cyclotomic
field. This is how Takagi proves this claim.
By Hilbert's Theorem 94, unramified cyclic extensions of prime
degree $\ell$ of a number field $K$ exist only if $K$ has class
number divisible by $\ell$.
Minkowski's bounds show that ${\mathbb Q}(i)$ does not
admit any nontrivial unramified extension.
P.S. Takagi's thesis is contained in his collected papers. I also would like to refer you to a beautiful article by Cox and Hyde on The Galois theory of the lemniscate.
I was just thinking about this the other day! Here is one solution.
As in the original post, let $K = \mathbb{Q}(\sqrt{-3})$ and let $\theta$ and $\pi$ be distinct primary primes. Also, set $L = K(\sqrt[3]{\theta \pi})$ and $M = K(\sqrt[3]{\theta}, \sqrt[3]{\pi})$.
We note that $M/L$ is unramified. The only primes which might ramify are those lying over $(\sqrt{-3})$, $(\pi)$ and $(\theta)$ in $K$. We can check that there is no ramification at these places using, respectively, a $3$-adic computation and the fact that $\theta$ and $\pi$ are primary, the presentation $M = L[T]/(T^3-\theta)$ and the presentation $M = L[U]/(U^3-\pi)$.
Let $\pi'$ be the prime of $L$ lying over $\pi$, and let $\theta'$ be the prime of $L$ lying over $\theta$. Now, $\theta' \pi'$ is the principal ideal generated by $\sqrt[3]{\theta \pi}$. So Artin reciprocity gives the equation
$$\mathrm{Frob}_{M/L}(\theta') \mathrm{Frob}_{M/L}(\pi') = \mathrm{Id} \qquad (\ast)$$
in $\mathrm{Gal}(M/L)$.
Now, $(\sqrt[3]{\pi})^{N(\pi')} = (\sqrt[3]{\pi})^{N(\pi)} \equiv \left[ \frac{\pi}{\theta} \right]_3 \sqrt[3]{\pi} \bmod \theta$. So
$$\mathrm{Frob}_{M/L}(\theta')(\sqrt[3]{\pi}) = \left[ \frac{\pi}{\theta} \right]_3 \sqrt[3]{\pi}.$$
Since $\sqrt[3]{\pi} \sqrt[3]{\theta} \in L$, it is fixed by $\mathrm{Frob}_{M/L}(\theta')$, so we compute
$$\mathrm{Frob}_{M/L}(\theta')(\sqrt[3]{\theta}) = \left[ \frac{\pi}{\theta} \right]_3^{-1} \sqrt[3]{\theta}.$$
Similarly,
$$\mathrm{Frob}_{M/L}(\pi')(\sqrt[3]{\theta}) = \left[ \frac{\theta}{\pi} \right]_3 \sqrt[3]{\theta} \quad
\mathrm{Frob}_{M/L}(\pi')(\sqrt[3]{\pi}) = \left[ \frac{\theta}{\pi} \right]_3^{-1} \sqrt[3]{\pi} .$$
Plugging into $(\ast)$ and acting on $\sqrt[3]{\pi}$ we get
$$ \left[ \frac{\pi}{\theta} \right]_3 \left[ \frac{\theta}{\pi} \right]_3^{-1} = 1$$
as desired.
On a linguistic note, I recently realized that "reciprocal" is a word, like "moot" or "cleave", which can denote two opposite things:
Life cannot subsist in society but by reciprocal concessions. (Samuel Johnson)
involves the exact opposite meaning from
...the pressures and expansions [volumes] to be in reciprocal relation. (Boyle, p. 60)
The higher reciprocity laws use reciprocal in the former sense.
Best Answer
As GH has already remarked, the same thing happened a lot later after Taniyama and Shimura asked whether elliptic curves defined over the rationals are modular. To begin with your last question, there were no other candidates for the Artin isomorphism; reciprocity laws at the time were intimately connected to power residue symbols. Actually Euler had formulated the quadratic reciprocity law in the correct way (namely that the symbol $(\Delta/p)$ only depends on the residue class of $p$ modulo $\Delta$), but Legendre's formulation prevailed. I'm pretty sure that Hasse would not have laughed had Artin shown him right away that the general reciprocity law is equivalent to the known power reciprocity laws. Hasse did not have to wait for Chebotarev's density law to be convinced that Artin was right - by the time Chebotarev's ideas appeared it was clear that Artin must have been right. And, by the way, Chebotarev's article did a lot more than prove the importance of the Frobenius element - it provided Artin with the key idea for the proof of his reciprocity law, namely Hilbert's technique of abelian crossings.
Let me also add that in 1904, Felix Bernstein conjectured a reciprocity law in 1904 that is more or less equivalent to Artin's law in the special case of unramified abelian extensions. Its technical nature shows that it was not at all easy to guess a simple law such as Artin's from the known power reciprocity laws.