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
$$ (x + ysqrt{-5},)(x - ysqrt{-5},) = z^2 $$
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 $(x + ysqrt{-5}, x - ysqrt{-5},)$ is either $(1)$ or
${mathfrak p}$; thus
$$ (x + ysqrt{-5},) = {mathfrak a}^2, quad (x - ysqrt{-5},) =
{mathfrak b}^2 $$
in the first and
$$ (x + ysqrt{-5},) = {mathfrak p}{mathfrak a}^2, quad
(x - ysqrt{-5},) = {mathfrak p}{mathfrak b}^2 $$
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 ${mathfrak a} = (a + bsqrt{-5},)$
is principal. Since the only units in ${mathcal O}_K$ are $pm 1$,
this gives $x + y sqrt{-5} = pm(a+bsqrt{-5},)^2$ and hence
$$ x = pm (a^2 - 5b^2), quad y = pm 2ab. $$
If ${mathfrak a}$ is not principal, then
${mathfrak p}{mathfrak a} = (a+bsqrt{-5},)$ is,
and from $({mathfrak p}{mathfrak a})^2 = 2(x+ysqrt{-5},)$ 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
$$ (x + ysqrt{-5},)(x - ysqrt{-5},) = z^2 $$
and the fact that the factors on the left hand side are
coprime we deduce that $x + ysqrt{-5} = varepsilon alpha^2$ 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
$alpha = frac{a + bsqrt{-5}}{2^t}$, where we may assume that $a$
and $b$ are not both even, we get
$$ x + y sqrt{-5} = varepsilon frac{a^2 - 5b^2 + 2absqrt{-5}}{2^{2t}}. $$
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
$$ (x + ysqrt{-5},)(x - ysqrt{-5},) = z^2 $$
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
$varepsilon in {1, i, omega, iomega }$. 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 ${mathfrak a} = (a+bsqrt{-5},)$ is principal in $K$,
or ${mathfrak p} {mathfrak a} = (a+bsqrt{-5},)$ is; in the second
case we observe
that ${mathfrak p} = (1+i)$ becomes principal in ${mathcal O}_L$.
Thus either
$$ x + y sqrt{-5} = (a+bsqrt{-5},)^2 quad text{or} quad
x + y sqrt{-5} = i Big(frac{a+bsqrt{-5}}{1+i},Big)^2, $$
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
$$ X + Y sqrt{-5} = frac{a+bsqrt{-5}}{a-bsqrt{-5}}
= frac{a^2 - 5b^2 + 2absqrt{-5}}{a^2 + 5b^2}. $$
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
$$ x^2 + 5y^2 = frac14Big(A^2 + 5b^2Big)^2
= Big(frac{A^2 - 5b^2}2Big)^2 + 5(Ab)^2. $$
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