Discussion and solution to Exa1, Q10.
The 3rd hint is misleading. The right way to think about it is as
an extension of Dedekind domain, so locally an extension of DVRs.
Theorem
(I) Let A be a DVR with parameter z. Then the extension ring
B = A[x]/(x^3-z) is a DVR with parameter x
(and the element z has valuation 3 in B).
(II) Let u in A^x be a unit, and assume that 3 is invertible in A.
Write k = A/(z) for the residue class ring, and ubar for the image
of u.
Then X^3-ubar in k[X] is a separable polynomial. It may
(1) be irreducible, or
(2) split as linear*quadratic, or
(3) split as three distinct linear factors (X-al1)*(X-al2)*(X-al3).
Here separable and distinct roots depend on the assumption 1/3 in A.
Then the extension ring B = A[x]/(x^3-u) is semilocal, with 1 or 2
or 3 maximal ideals depending on the 3 cases for (cube root of
ubar). The localisation of B at each maximal ideal is a DVR (with
the same parameter z as A).
===
Corollary
Let A be a Dedekind domain with 1/3 invertible (for example
ZZ[1/3] or k[z] for any field k of char <> 3).
Let a,b in A[x] be square-free and coprime, as in Exa1, Q10.
Then the ring
B = A[x,y]/(x^2-a*y, x*y-a*b, y^2-b*x)
is again a Dedekind domain. In particular, it is integrally
closed.
===
Although somewhat embarrassing for me, this illustrates the
point that normalisation (integral closure) is a complicated
process. And it also makes the point that commutative algebra
gets its strength from localisation.
===
More geometrically: If a,b in CC[z] have distinct simple
roots, the equations
x^2 = a*y, x*y = a*b, y^2 = b*x in AA^3
define a nonsingular curve. The localisation at each maximal ideal is a
DVR, therefore integrally closed. In fact if a*b <> 0, the equations
x^3 = a^2*b and y^3 = a*b^2 already give x and y as unramified functions
of z. If we are at a root of a then b <> 0 and a has a simple root so
b*x = .. and b*a = ..
are implicit equations for x and z, and y is a local parameter.
===
I don't really know what happens without 1/3. Computer algebra
experiments over ZZ seem to say that B = ZZ[x,y]/(above ideal) is
normal, even if a or b have 3 as a factor.
What are the possibility for cubic extensions of DVRs either in
pure characteristic 3, or in mixed characteristic at 3? A Galois
ZZ/3 action might remain separable (Artin-Schreier), or might have
a wild degeneration to an inseparable action of mu_3 or al_3. But
even then, it might still happen that the ring B over A is a DVR,
so that the extension remains a Dedekind domain.
This must be solved somewhere in the literature, but I don't have
time to search it out.
// Magma seems to say that the ring of integers of the field
// QQ(cuberoot (a*b^2)) has basis (1, x, y) over ZZ
// with x^3 = a*b^2 and y^3 = a^2*b
a := 51; b := 385; IsCoprime(a,b); a*b^2; a^2*b; a*b; //
R := PolynomialRing(Rationals());
L := NumberField(x^3 - a*b^2);
A := RingOfIntegers(L);
A eq MaximalOrder(L);
MinimalPolynomial(A.1); [A.1*m eq m : m in [a,b,A.1,A.2,A.3]]; // A.1 is 1
MinimalPolynomial(A.2); a*b^2; // A.2 is CubeRoot of a*b^2
MinimalPolynomial(A.3); a^2*b; // A.3 is CubeRoot of a^2*b
[MinimalPolynomial(k/3*A.1 + (i/3)*A.2 + (j/3)*A.3)
: i,j,k in [0..2] | [i,j,k] ne [0,0,0]];
// combinations with 3 in denominator are not integral over ZZ.