54 CHAPTER 6

Varieties

Suppose we have ﬁxed our attention on a particular Z-equation.

We write S(Z) for the set of all integral solutions of that equation,

S(Q) for the set of all rational solutions of it, and so on. We call

S an “algebraic variety,” because for the various choices of number

systems, we get various sets of solutions. We say that the given

equation deﬁnes the variety S.

DEFINITION: The variety S deﬁned by a Z-equation (or a

system of Z-equations) is the function that assigns to any

number system A the set of solutions S(A) of the equation (or

system of equations).

For example, take the equation x

2

+ y

2

= 1. We know that the

set S(Z) cannot be very big, because x

2

≥ 0andy

2

≥ 0. Trial-and-

error with small numbers tells us the whole story: S(Z) contains

four elements, and it is {(1, 0), (−1, 0), (0, 1), (0, −1)}, where the pair

(a, b) stands for the solution x = a and y = b.

What about S(Q)? The answer can be found in Euclid, though not

written quite like this. It can be shown that

S(Q) =

1 − t

2

1 + t

2

,

2t

1 + t

2

: t any rational number

{(−1, 0)}.

The symbol

denotes the union of the two sets, that is, the set that

contains all elements of both sets.

How about S(R)? This is easier: If x is any number between −1

and 1, we can let y =

√

1 − x

2

or y =−

√

1 − x

2

. (By convention, if

a is positive, then

√

a always stands for the positive square root

of a.) We will in the future abbreviate the preceding two solutions

by writing y =±

√

1 − x

2

. Incidentally, there is a fancier way to write

this same solution set using trigonometry:

S(R) ={(cos θ ,sinθ ):0≤ θ<2π}.

Another example is S(C). Here, we can let x be any complex

number at all, and let y =±

√

1 − x

2

. The square roots will alw ays

exist, because we are allowing complex numbers.

EQUATIONS AND VARIETIES 55

What about modular arithmetic? If we want to compute S(F

2

),

for example, we can let x and y run through all of the elements of

F

2

(which does not take very long, because there are only the two

elements 0 and 1 in F

2

), and compute x

2

+ y

2

,andseeifweget1.We

ﬁnd out that S(F

2

) ={(1, 0), (0, 1)}. Similarly, we can compute that

S(F

3

) ={(1, 0), (2, 0), (0, 1), (0, 2)}.

EXERCISE: List all of the elements of S(F

5

).

SOLUTION: This is just trial-and-error. After running

through all of the possibilities, we get S(F

5

) ={(1, 0), (4, 0),

(0, 1), (0, 4)}.

EXERCISE: List all of the elements of S(F

7

).

SOLUTION: This is a bit more interesting, and again trial

and error gives all the solutions. We get S(F

7

) ={(1, 0), (6, 0),

(0, 1), (0, 6), (2, 2), (2, 5), (5, 2), (5, 5)}.

We can state Fermat’s Last Theorem in the language of varieties.

Fermat’s Last Theorem is the following assertion:

For any positive integer n, let the variety V

n

be deﬁned by

x

n

+ y

n

= z

n

.

Then if n > 2, V

n

(Z) contains only solutions where one or

more of the variables is 0.

One reason to state Fe rmat’s Last Theorem in this way is that

it is easier to study V

n

(R), V

n

(C), or V

n

(F

p

)thanV

n

(Z). Then

some information about V

n

(Z) (the solution set we are really

interested in) can be derived from the preceding three sets, using

advanced theorems in number theory and algebraic geometry. The

complete proof of Fermat’s Last Theorem, however, required Galois

representations and many other additional ideas, as we shall see.

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