Suppose you have a diophantine problem whose solution is connected with the structure of the p-class group of a number field K. Then you have the following options:
- Use ideal arithmetic in the maximal order OK
- Replace OK by a suitable ring of S-integers with trivial p-class group
- Replace K by the Hilbert class field, which (perhaps) has trivial p-class group.
Experience with descent on elliptic curves has shown me that ultimately, the equations you have to solve in methods 1 and 2 are the same; moreover, the approach using ideals is a lot less technical than using factorial domains in S-integers (the class group relations come back in through the larger rank of the group of S-units). I am certain that the route via the Hilbert class field is even more technical: again, the unit group in the class field will produce more difficulties than a trivial class group will eliminate.
Edit. As an example illustrating my point in a very simple example, let me solve the diophantine equation $x^2 + 5y^2 = z^2$ in several different ways. I will always assume that $\gcd(x,y) = 1$.
1. Elementary Number Theory
The basic idea is factoring: from $5y^2 = (z+x)(z-x)$. Since $d = \gcd(z-x,z+x) = \gcd(2z,z+x) \mid 2$ we have $d = 1$ or $d = 2$; moreover we clearly have $z-x > 0$.
This gives $z+x = da^2$, $z-x = 5db^2$ or $z+x = 5da^2$, $z-x = db^2$. Solving for $x$ and $z$ yields
$$ x = \pm \frac d2 (a^2 - 5b^2), \quad y = dab. $$
2. Parametrization
Set $X = \frac xz$ and $Y = \frac yz$; then $X^2 + 5Y^2 = 1$. Take the line $Y = t(X+1)$ through the obvious point $(-1,0)$; the second point of intersection is given by
$$ X = \frac{1-5t^2}{1+5t^2}, \quad Y = \frac{2t}{1+5t^2}. $$
Dehomogenizing using $t = \frac ba$ and $X = \frac xz$ etc. gives
the projective parametrization
$$ (x:y:z) = (a^2-5b^2:2ab:a^2+5b^2). $$
If $ab$ is odd, all coordinates are even, and we find
$$ x = \frac12(a^2 - 5b^2), \quad y = ab; $$
if $a$ or $b$ is even we get
$$ x = a^2 - 5b^2, \quad y = 2ab $$
as above.
3. Algebraic Number Theory
Consider the factorization
$$ (x + y\sqrt{-5}\,)(x - y\sqrt{-5}\,) = z^2 $$
in the ring of integers of the number field $K = {\mathbb Q}(\sqrt{-5}\,)$.
The class number of $K$ is $2$, and the ideal class is generated by
the prime ideal ${\mathfrak p} = (2,1+\sqrt{-5}\,)$.
The ideal $(x + y\sqrt{-5}, x - y\sqrt{-5}\,)$ is either $(1)$ or
${\mathfrak p}$; thus
$$ (x + y\sqrt{-5}\,) = {\mathfrak a}^2, \quad (x - y\sqrt{-5}\,) =
{\mathfrak b}^2 $$
in the first and
$$ (x + y\sqrt{-5}\,) = {\mathfrak p}{\mathfrak a}^2, \quad
(x - y\sqrt{-5}\,) = {\mathfrak p}{\mathfrak b}^2 $$
in the second case.
The second case is impossible since the left hand side as well as
${\mathfrak a}^2$ are principal, but ${\mathfrak p}$ is not. We
could have seen this immediately since $x$ and $y$ cannot both be odd.
In the first case, assume first that ${\mathfrak a} = (a + b\sqrt{-5}\,)$
is principal. Since the only units in ${\mathcal O}_K$ are $\pm 1$,
this gives $x + y \sqrt{-5} = \pm(a+b\sqrt{-5}\,)^2$ and hence
$$ x = \pm (a^2 - 5b^2), \quad y = \pm 2ab. $$
If ${\mathfrak a}$ is not principal, then
${\mathfrak p}{\mathfrak a} = (a+b\sqrt{-5}\,)$ is,
and from $({\mathfrak p}{\mathfrak a})^2 = 2(x+y\sqrt{-5}\,)$ we
similarly get
$$ x = \pm \frac12(a^2 - 5b^2), \quad y = \pm ab. $$
4. S-Integers
The ring $R = {\mathbb Z}[\sqrt{-5}\,]$ is not a UFD, but $S = R[\frac12]$ is;
in fact, $S$ is even norm-Euclidean for the usual norm in $S$
(the norm is the same as in $R$ except that powers of $2$ are dropped).
It is also easily seen that $S^\times = \langle -1, 2 \rangle$. From
$$ (x + y\sqrt{-5}\,)(x - y\sqrt{-5}\,) = z^2 $$
and the fact that the factors on the left hand side are
coprime we deduce that $x + y\sqrt{-5} = \varepsilon \alpha^2$ for some unit
$\varepsilon \in S^\times$ and some $\alpha \in S$. Subsuming squares into
$\alpha$ we may assume that $\varepsilon \in \{\pm 1, \pm 2\}$. Setting
$\alpha = \frac{a + b\sqrt{-5}}{2^t}$, where we may assume that $a$
and $b$ are not both even, we get
$$ x + y \sqrt{-5} = \varepsilon \frac{a^2 - 5b^2 + 2ab\sqrt{-5}}{2^{2t}}. $$
It is easily seen that we must have $t = 0$ and $\varepsilon = \pm 1$ or
$t = 1$ and $\varepsilon = \pm 2$; a simple calculation then yields the
same formulas as above.
5. Hilbert Class Fields
The Hilbert class field of $K$ is given by $L = K(i)$. It is not
too difficult to show that $L$ has class number $1$ (actually it is
norm-Euclidean), and that its unit group is generated by $i = \sqrt{-1}$
and $\omega = \frac{1+\sqrt{5}}2$ (we only need to know that these
units and their product are not squares). From
$$ (x + y\sqrt{-5}\,)(x - y\sqrt{-5}\,) = z^2 $$
and the fact
that the factors on the left hand side are coprime in ${\mathcal O}_K$
we deduce that $x + y \sqrt{-5} = \varepsilon \alpha^2$. Subsuming
squares into $\alpha^2$ we may assume that
$\varepsilon \in \{1, i, \omega, i\omega \}$. Applying the nontrivial
automorphism of $L/K$ to $x + y \sqrt{-5} = \varepsilon \alpha^2$ we find
$\varepsilon \alpha^2 = \varepsilon' {\alpha'}^2$. Since the ideal
${\mathfrak a} = (\alpha)$ is fixed and since $L/K$ is unramified,
the ideal ${\mathfrak a}$ must be an ideal in ${\mathcal O}_K$.
Thus either ${\mathfrak a} = (a+b\sqrt{-5}\,)$ is principal in $K$,
or ${\mathfrak p} {\mathfrak a} = (a+b\sqrt{-5}\,)$ is; in the second
case we observe
that ${\mathfrak p} = (1+i)$ becomes principal in ${\mathcal O}_L$.
Thus either
$$ x + y \sqrt{-5} = (a+b\sqrt{-5}\,)^2 \quad \text{or} \quad
x + y \sqrt{-5} = i \Big(\frac{a+b\sqrt{-5}}{1+i}\,\Big)^2, $$
giving us the same formulas as above.
Avoiding ideal arithmetic in $K$ and only using the fact that
${\mathcal O}_L$ is a UFD seems to complicate the proof even more.
Edit 2 For good measure . . .
6. Hilbert 90
Consider, as above, the equation $X^2 + 5Y^2 = 1$.
It shows that the element $X + Y \sqrt{-5}$ has norm $1$;
by Hilbert 90, we must have
$$ X + Y \sqrt{-5} = \frac{a+b\sqrt{-5}}{a-b\sqrt{-5}}
= \frac{a^2 - 5b^2 + 2ab\sqrt{-5}}{a^2 + 5b^2}. $$
Dehomogenizing via $X = \frac xz$ and $Y = \frac yz$ yields the same
projective parametrization as above, and we end up with the
familiar formulas.
7. Binary Quadratic Forms
The equation $x^2 + 5y^2 = z^2$ tells us that the form $Q_0(X,Y) = X^2 + 5Y^2$
with fundamental discriminant $\Delta = -20$ represents a square;
this implies that $Q_0$ lies in the principal genus (which is trivial
since $Q_0$ is the principal form), and that the representations of
$z^2$ by $Q_0$ come from composing representations of $z$ by forms
$Q_1$ with $Q_1^2 \sim Q_0$ with themselves.
There are only two forms with discriminant $\Delta$ whose square is
equivalent to $Q_0$: the principal form $Q_0$ itself and the form
$Q_1(X,Y) = 2X^2 + 2XY + 3Y^2$. Thus either
$$ z = Q_0(a,b) = a^2 + 5b^2 \quad \text{or} \quad
z = Q_1(a,b) = 2a^2 + 2ab + 3b^2. $$
The formulas for Gauss composition of forms immediately provide us with
expressions for $x$ and $y$ in terms of $a$ and $b$, but they can also
be checked easily by hand. In the first case, we get
$$ x^2 + 5y^2 = (a^2 + 5b^2)^2 = (a^2 - 5b^2)^2 + 5(2ab)^2, $$
and in the second case we can reduce the equations to this one
by observing that $2Q_1(a,b) = A^2 + 5b^2$ with $A = 2a+b$, which gives
$$ x^2 + 5y^2 = \frac14\Big(A^2 + 5b^2\Big)^2
= \Big(\frac{A^2 - 5b^2}2\Big)^2 + 5(Ab)^2. $$
A trick I have seen several times: If you want to show that some rational number is an integer (i. e., a divisibility), show that it is an algebraic integer. Technically, it is then an application of commutative algebra (the integral closedness of $\mathbb Z$, together with the properties of integral closure such as: the sum of two algebraic integers is an algebraic integer again), but since you define algebraic number theory as the theory of algebraic numbers, you may be interested in this kind of applications.
Example: Let $p$ be a prime such that $p\neq 2$. Prove that the $p$-th Fibonacci number $F_p$ satisfies $F_p\equiv 5^{\left(p-1\right)/2}\mod p$.
Proof: We can do the $p=5$ case by hand, so let us assume that $p\neq 5$ for now. Then, $p$ is coprime to $5$ in $\mathbb Z$. Let $a=\frac{1+\sqrt5}{2}$ and $b=\frac{1-\sqrt5}{2}$. The Binet formula yields $F_p=\displaystyle\frac{a^p-b^p}{\sqrt5}$. Now, $a^p-b^p\equiv\left(a-b\right)^p\mod p\mathbb Z\left[a,b\right]$ (by the idiot's binomial formula, since $p$ is an odd prime). Note that $p$ is coprime to $5$ in the ring $p\mathbb Z\left[a,b\right]$ (since $p$ is coprime to $5$ in the ring $\mathbb Z$, and thus there exist integers $a$ and $b$ such that $pa+5b=1$). Now,
$\displaystyle F_p=\frac{a^p-b^p}{\sqrt5}\equiv\frac{\left(a-b\right)^p}{\sqrt5}$ (since $a^p-b^p\equiv\left(a-b\right)^p\mod p\mathbb Z\left[a,b\right]$ and since we can divide congruences modulo $p\mathbb Z\left[a,b\right]$ by $\sqrt5$, because $p$ is coprime to $5$ in $p\mathbb Z\left[a,b\right]$)
$\displaystyle =\frac{\left(\sqrt5\right)^p}{\sqrt5}$ (since $a-b=\sqrt5$)
$=5^{\left(p-1\right)/2}\mod p\mathbb Z\left[a,b\right]$.
In other words, the number $F_p-5^{\left(p-1\right)/2}$ is divisible by $p$ in the ring $\mathbb Z\left[a,b\right]$. Hence, $\frac{F_p-5^{\left(p-1\right)/2}}{p}$ is an algebraic integer. But it is also a rational number. Thus, it is an integer, so that $p\mid F_p-5^{\left(p-1\right)/2}$ and thus $F_p\equiv 5^{\left(p-1\right)/2}\mod p$, qed.
Best Answer
This answer is a bit late; sorry for that.
Kummer's proof of the nonsolvability of $x^p + y^p = z^p$ for regular primes $p$ used “ideal numbers” (in present-day language: ideals) and was intact, at least basically. Hilbert in his Zahlbericht gave a modified proof. Both proofs cover not only rational integers but also numbers in $\mathbb{Z}[\zeta_p]$. On the other hand, Kummer’s second result concerning irregular primes that satisfy certain additional conditions covers just the rational integers (although Hilbert, in the very last section of Zahlbericht, erroneously says that Kummer had proven this result for $\mathbb{Z}[\zeta_p]$ as well). Thus one cannot exclude the possibility that there is indeed a solution $(x,y,z)$ for $p=37$. And because of "Kolyvagin's criterion" about $(2^{37}-2)/37$, this solution must belong to the second case, that is, at least one of these three numbers $x,y,z$ in $\mathbb{Z}[\zeta_{37}]$ must have a common factor with $37$ (as mentioned by George Lowther).
By the way, this criterion was also proven by Taro Morishima in 1935 (Japan. J. Math. 11, 241-252, Satz 1; but warning: Satz 2 or at least its proof is incorrect since it is based on some incorrect result of Vandiver).
I don’t know how to find such a solution $(x,y,z)$.