# 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 w^{2}+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.