What would it mean to understand this Galois group? You could mean several things.
You could mean trying to give the group in terms of some smallish generators and relations. This would be nice, and help to answer questions like the inverse Galois problem that Greg Muller mentioned, and having a certain family of "generating" Galois automorphisms would allow you to study questions about e.g. the representation theory in quite explicit terms. However, the Galois group is an uncountable profinite group, and so to give any short description in terms of generators and relations leads you into subtle issues about which topology you want to impose.
You could also ask for a coherent system of names for all Galois automorphisms, so that you can distinguish them and talk about them on an individual basis. One system of names comes from the dessins d'enfant that Ilya mentioned: associated to a Galois automorphism we have some associated data.
- We have its image under the cyclotomic character, which tells us how it acts on roots of unity. By the Kronecker-Weber theorem this tells us about the abelianization of the Galois group.
- We also have an element in the free profinite group on two generators, which (roughly speaking) tells us something about how abysmally acting on the coefficients of a power series fails to commute with analytic continuation.
These two names satisfy some relations, called the $2$-, $3$-, and $5$-cycle relation, which are conjectured to generate all relations (at least the last time I checked), but it is difficult to know whether they actually do so. If they do, then the Galois group is the so-called Grothendieck-Teichmüller group.
The problem with this perspective is that the names aren't very explicit (and we don't expect them to be: we may need the axiom of choice to show they exist, and there are only two Galois automorphisms of $\mathbb{C}$ that are measurable functions!) and it seems to be a difficult problem to determine whether the Grothendieck-Teichmuller group really is the whole thing. (Or it was the last time I checked.)
However, the cyclotomic character is a nice, and fairly canonical, name associated for Galois automorphisms. We could try to generalize this: there are Kummer characters telling us what a Galois automorphism does to the system of real positive roots of a positive rational number number (these determine a compatible system of roots of unity, or equivalent an element of the Tate module of the roots of unity). This points out one of the main difficulties, though: we had to make choices of roots of unity to act on, and if Galois theory taught us nothing else it is that different choices of roots of an irreducible polynomial should be viewed as indistinguishable. Different choices differ by conjugation in the Galois group.
This brings us to the point JSE was making: if we take the "symmetry" point of view seriously, we should only be interested in conjugacy-invariant information about the Galois group. Assigning names to elements or giving a presentation doesn't really mesh with the core philosophy.
So this brings us to how many people here have mentioned understanding the Galois group: you understand it by how it manifests, in terms of its representations (as permutations, or on dessins, or by representations, or by its cohomology), because this is how it's most useful. Then you can study arithmetic problems by applying knowledge about this. If I have two genus $0$ curves over $\mathbb{Q}$, what information distinguishes them? If I have two lifts of the same complex elliptic curve to $\mathbb{Q}$, are they the same? How can I get information about a reduction of an abelian variety mod $p$ in terms of the Galois action on its torsion points? Et cetera.
EDIT: Hendrik Lenstra emailed me a proof of Conjecture 2. I'll append it below. So Jagy's question is now solved.
OK so I think that Jagy wants to make the following conjecture:
CONJECTURE 1: an integer $C$ is not representable by the form F(x,y,z)=2x^2+xy+3y^2+z^3-z if, and only if, $C$ is odd and $27C^2-4=23D^2$ with $D$ an integer.
[EDIT/clarification: Jagy only asks one direction of the iff in his question, and this answer below gives a complete answer to the question Jagy asks. I came back to this question recently though [I am writing this para a year after I wrote the original answer] and tried to fill in the details of the argument in the other direction (proving that if C was not an odd integer solution to $27C^2-4=23D^2$ then $C$ was represented by the form) and I failed. So the "hole" I flag in the answer below still really is a hole, and this post still remains an answer to Jagy's question, but not a complete proof of Conjecture 1, which should still be regarded as open.]
I have a proof strategy for this. I am too lazy to fill in some of the details though, so maybe a bit of it doesn't work, but it should be OK. However, I am also reliant on a much easier-looking conjecture (which I've tested numerically so should be fine, but I can't see why it's true):
CONJECTURE 2: if $C$ is odd and $27C^2-4=23D^2$, then there's no prime p
dividing D of the form $2x^2+xy+3y^2$.
So I am claiming Conj 2 implies the "only if" version of Conj 1. I don't know how to prove Conj 2
but it looks very accessible [edit: I do now; see below]. Note that the Pell equation is related to units
in $\mathbf{Q}(\sqrt{69})$ and the $2x^2+xy+3y^2$ is related to factorization
in $\mathbf{Q}(\sqrt{-23})$. I've seen other results relating the arithmetic
of $\mathbf{Q}(\sqrt{D})$ and $\mathbf{Q}(\sqrt{-3D})$.
Ok, so assuming Conjecture 2, let me sketch a proof of the "only if" part of Conjecture 1.
The Pell equation is intimately related to the recurrence relation
$$t_{n+2}=25t_{n+1}-t_n$$
with various initial conditions. For example the positive $C$s which
are solutions to $27C^2-4=23D^2$ are all generated by this recurrence
starting at $C_1=C_2=1$, and the $D$s are all generated by the same
recurrence with $D_1=-1$ and $D_2=1$. Note that $C_n$ is even iff $n$
is a multiple of 3, and (by solving the recurrence explicitly) one
checks easily that $C_{3n}=(3C_{n+1})^3-(3C_{n+1})$, so we've represented
the even solutions to the Pell equation as values of $F$ (with $x=y=0$).
Let's then consider the odd solutions to the Pell equation. Say $C$
is one of these. We want to prove that there is no solution in
integers $x,y,z$ to
$$2x^2+xy+3y^2=z^3-z+C.$$
Let's do it by contradiction. Consider the polynomial $Z^3-Z+C$. First
I claim it's irreducible. This is because it is monic, of degree 3,
and has no integer root, because $C$ is odd. Next I claim that
the splitting field contains $\mathbf{Q}(\sqrt{-23})$. This is
because of our Pell assumption and the fact that the discriminant
of $Z^3-Z+C$ is $4-27C^2$. Next I claim that the splitting
field of $Z^3-Z+C$ is in fact the Hilbert class field of
$\mathbf{Q}(\sqrt{-23})$. I only know an ugly way of seeing this:
if $\theta$ is a root of $Z^3-Z+1=0$ then I know recurrence relations
$e_n$, $f_n$ and $g_n$ (all defined using the relation above but with
different initial conditions) with $e_n\theta^2+f_n\theta+g_n$ a root of
$Z^3-Z+C_{3n+1}$, and other relations giving roots of $Z^3-Z+C_{3n+2}$
and $Z^3-Z-C_{3n+1}$ and $Z^3-Z-C_{3n+2}$. Most unenlightening but it
does the job because it embeds $\mathbf{Q}(\theta)$ into the splitting
field, and the Galois closure of $\mathbf{Q}(\theta)$ is the Hilbert
class field of $\mathbf{Q}(\sqrt{-23})$.
Right, now for the contradiction, assuming Conjecture 2. Let's assume
that $C$ is a solution to the Pell, and $z^3-z+C$ can be written $2x^2+xy+3y^2$.
Now $C$ is odd so $z^3-z+C$ isn't zero, and hence it's positive,
so it's the norm of a non-principal ideal~$I$ in the integers $R$ of
$\mathbf{Q}(\sqrt{-23})$. This ideal $I$ is a product of prime ideals,
and $I$ isn't principal, so one of the prime ideals had better also not
be principal. Say this prime ideal has norm $p$. We conclude that $p$
divides $z^3-z+C$ and $p$ is of the form $2x^2+xy+3y^2$. Note in
particular that this implies $p\not=23$. Also $p\not=3$, because $C$
is odd and (because of general Pell stuff) hence prime to 3.
CASE 1: $p$ is coprime to $D^2$ (with $27C^2-4=23D^2$). In this
case the polynomial $Z^3-Z+C$ has non-zero discriminant mod $p$
(because $p\not=23$) and furthermore has a root $Z=z$ mod $p$.
Hence mod $p$ the polynomial either splits as the product of a linear
and a quadratic, or the product of three linears. This tells us
something about the factorization of $p$ in the splitting field
of $Z^3-Z+C$: either $p$ remains inert in $\mathbf{Q}(\sqrt{-23})$,
or it splits into 6 primes in the splitting field and hence splits
into two principal primes in $\mathbf{Q}(\sqrt{-23})$ (because the
principal primes are the ones that split completely in the Hilbert
class field). In either case $p$ can't be of the form $2x^2+xy+3y^2$,
so this case is done.
CASE 2: This is simply Conjecture 2.
In both cases we have our contradiction, and
so we have proved, so far, assuming Conjecture 2, that a solution $C$ to $27C^2-4=23D^2$
is representable as $2x^2+xy+3y^2+z^3-z$ iff it's even.
Note that Conjecture 2 can be verified by computer for explicit values
of $C$, giving unconditional results---for example I checked in just
a few seconds that any odd $C$ with $|C|<10^{72}$ and satisfying the
Pell equation was not representable by the form, and that result
does not rely on anything. At least that's something concrete for Jagy.
OK so what about the other way: say $27C^2-4$ is not 23 times a square.
How to go about representing $C$ by our form? Well, here I am going to
be much vaguer because there are issues I am simply too tired to deal
with (and note that this is not the question that Jagy asked anyway).
Here's the idea. Look at the proof of Theorem 2 in Jagy's pdf Mordell.pdf.
Here Mordell gives a general algorithm to represent certain integers
by (quadratic in two variables) + (cubic in one variable). If you
apply it not to the form we're interested in, but to the following
equation:
$$x^2+xy+6y^2=z^3-z+C$$
then, I didn't check all the details, but I convinced myself that they
could easily be checked if I had another hour or two, but I think that
the techniques show that whatever the value of $C$ is, this equation
has a solution. The idea is to fix $C$, let $\theta$ be a root of
the cubic on the right (which we can assume is irreducible, as if it
were reducible then we get a solution with $x=y=0$), to rewrite the right
hand side as $N_{F/\mathbf{Q}}(z-\theta)$, with $F=\mathbf{Q}(\theta)$
and now to try and write $z-\theta$ as $G^2+GH+2H^2$ with
$G,H\in\mathbf{Z}[\theta]$. Mordell does this explicitly (in a slightly
different case) in the pdf. The arguments come out the same though,
and we end up having to check that a certain cubic in four variables
has a solution modulo~23 with a certain property. I'll skip the painful
details. The cubic depends on $C$ mod 23, and so a computer calculation
can deal with all 23 cases.
Once this is done properly we have a solution to $x^2+xy+6y^2=z^3-z+C$,
so we have written $z^3-z+C$ as the norm of a principal ideal in
the integers of $\mathbf{Q}(\sqrt{-23})$. What we need to do now is
to write it as the norm of a non-principal ideal, and of course we'll
be able to do this if we can find some prime $p$ dividing $z^3-z+C$
which splits in $\mathbf{Q}(\sqrt{-23})$ into two non-principal
primes, because then we replace one of the prime divisors above $p$
in our ideal by the other one. What we need then is to show that
if the discriminant of $z^3-z+C$ is not $-23$ times a square,
then there is some prime $p$ of the form $2x^2+xy+3y^2$ dividing
some number of the form $z^3-z+C$ which is the norm of a principal
ideal. This should follow from the Cebotarev density theorem, because
Mordell's methods construct a huge number of solutions to $x^2+xy+6y^2=z^3-z+C$
which are "only constrained modulo 23", and so one should presumably
be able to find a prime which splits in $\mathbf{Q}(\sqrt{-23})$,
splits completely in the splitting field of $z^3-z+C$ and doesn't
split completely in the splitting field of $z^3-z+1$. I have run out
of energy to deal with this point however, so again there is a hole here.
This issue seems analytic to me, and I am not much of an analytic guy.
[edit: I came back to this question a year later and couldn't do it,
so this should not be regarded as a proof of the "if" part of Conj 1]
EDIT: OK so here, verbatim, is an email from Lenstra in which he establishes
Conjecture 2.
(EDIT: dollar signs added - GM)
Fact. Let $\theta$ be a zero of $X^3-X+1$, let $\eta$ in ${\bf Z}[\theta]$ be
a zero of $X^3-X+C$ with $C$ in $\bf Z$ odd, and let $p$ be a prime
number that is inert in ${\bf Z}[\theta]$. Then $p$ does not divide
index$({\bf Z}[\theta]:{\bf Z}[\eta])$.
Proof. By hypothesis, ${\bf Z}[\theta]/p{\bf Z}[\theta]$ is a field of size $p^3$.
Let $e$ be the image of $\eta$ in that field. Since $X^3-X+C$ is
irreducible in ${\bf Z}[X]$ (even mod 2), it is the characteristic
polynomial of $\eta$ over $\bf Z$. Hence its reduction mod $p$ is the
characteristic polynomial of $e$ over ${\bf Z}/p{\bf Z}$. If now $e$ is in
${\bf Z}/p{\bf Z}$, then that characteristic polynomial also equals $(X-e)^3$,
so that in ${\bf Z}/p{\bf Z}$ we have $3e = 0$ and $3e^2 = -1$, a contradiction.
Hence $e$ is not in ${\bf Z}/p{\bf Z}$, so $({\bf Z}/p{\bf Z})[e] = {\bf Z}[\theta]/p{\bf Z}[\theta]$,
which is the same as saying ${\bf Z}[\theta] = {\bf Z}[\eta] + p{\bf Z}[\theta]$. Then
$p$ acts surjectively on the finite abelian group ${\bf Z}[\theta]/{\bf Z}[\eta]$,
so the order of that group is not divisible by $p$. End of proof.
Best Answer
Ok so it looks like I misjudged this and the community seem happy to have the question here, at least at present. So I figured I'd pass on the comments which Serre sent the LMS.
p.135, part b) of Lemma 4. Replace $H^q(H,M)$ by $H^q(K/H,M^H)$ and replace $\hat{H}^0(H,M)$ by $\hat{H}^0(K/H,M^H)$.
p.135, line 6 from bottom. Replace "to $G/H$" by "to $H$".
p.145. In Prop.6, replace "of Proposition 3" by "of Proposition 5".
p.225, line 11 (second line after Th.7). After "can be taken to be rational integers" add the following parenthesis : (provided one does not insist that the extensions $K/\Omega_{\ell}$ be cyclic).