Algebraic Symmetries
Symmetries IV
This means that, from an abstract point of view,
is simply a number
that satisfies the equation
and no other. But 
, 
and 
have exactly the same
property. Thus for
strictly algebraic purposes we could take
to be rather than 
. Thus is
replaced by . Since all algebraic relations are to be preserved, this entails
replacing 
by
and 
by
. Once again,
because all algebraic relations are to be respected, the number
is to be replaced by
In general a number
is replaced by
which is a number of the same kind. For example
Forms of algebraic symmetries
The symmetry just examined can be viewed in two ways:
1) It takes the sequence 
of all roots of
to a sequence formed from the same numbers but in a different order
2) It takes any number
to a number
of the same kind.
This is the kind of symmetry that was later investigated in general by
Galois. We have to spend some time growing accustomed to it. Suppose we
apply the symmetry twice. Then
Thus applying the basic symmetry twice leads to the first
symmetry considered,
complex conjugation. We apply it again.
Yet again!
So the symmetry when repeated four times comes back where
it began. It is
a four-fold symmetry.
Anticipating Galois and his successors
No matter which of the numbers , , , we take
to be, the
collection of numbers
,a, b, c, d all
fractions
is the same. Modern mathematicians usually call the
collection a field. the
sum and the product of two numbers of this sort are again numbers of the
same sort. This we have seen already. I give another example.
Any symmetry of this collection that respect the algebraic
operations will
take 0 to 0 and 1 to 1. Then adding and dividing it takes any fraction a/b to
a/b. Moreover any root of
will be taken to another root. Thus will be taken to
, , or .
In other words, the symmetry will be one of the four (including the trivial
symmetry!) we already have. Denote the one taking to by the letter
.
Then the repeating to obtain
we obtain the symmetry taking to
. Repeating again, we obtain
which takes to . Repeating
again, we find that 
is the trivial symmetry.
Inside the collection of numbers there is a smaller collection of numbers
that have a special symmetry. We met them before. They are those that
are not affected by
, thus by complex conjugation. They are the numbers
a+ bw, w = +. We were able to construct by succesive square roots, by
first singling out this special collection of numbers, finding that any number
in it satisfied a quadratic equation with fractions as coefficients, in particular
that w2+w−1 = 0, so that 
, and then solving
.
We now apply these ideas, which I hope are clear, to the heptadecagon!
The Heptadecagon
From the Disquisitiones Arithmeticae
There is a famous remark from the introduction to the seventh
and last chapter of the Disquisitiones that I quote here. What it
anticipates is the study of the division points on elliptic curves, in
the remark a special elliptic curve, a study that led over the course
of the nineteenth and twentieth century to many things, especially
complex multiplication and l-adic representations, that are relevant
to the Shimura-Taniyama-Weil conjecture,
Ceterum principia theoriae, quam exponere aggredimur, multo
latius patent, quam hic extenduntur. Namque non solum ad functiones
circulares, sed pari successu ad multas alias functiones transscendentes
applicari possunt, e. g. ad eas, quae ab integrali
pendent, praetereaque etiam ad variam congruentarium
genera: sed
quoniam de illis functionibus transscendentibus amplum opus peculiare
paramus, de congruentibus autem in continuatione disquisitionum
arithmeticarum copiose tractabitur, hoc loco solas functiones
circulares considerare visum est.
Lecture 7
A proof by Gauss (beginning)
Recall that we want to show that is a root of the equation
but of no equation of the forms
in which a, b c and d are fractions.
The impossibility of the last equation is clear because is not a fraction.
If it were a root of the first, then using long division to divide (I) by
, we would find
Substitute to find that is also a root of
Then, as we just observed e is not 0, unless e = f = g =
0. If e is not 0,
divide by it. Thus either
or satisfies an equation of type (II). If it satisfies
(II), then perform a long
division to obtain
Since e + d cannot be 0 unless e = d = 0, we conclude
that
We now show that the factorizations of (IV) and (V) are
impossible. We
first observe a very important fact.
An equation of degree n
cannot have more than n roots!
Suppose (VI) has a root e. Using long division, divide by
Z − e. The
result is
Substitute e to see that f = 0. Now any other root e' not
equal to e of (VI)
must be a root of
so that if (VI) had more than n roots, then (VII) would
have more than n−1.
All we have to do now is continue, working our way down to lower and lower
degree until we arrive at an equation of degree one
that clearly has only one root.