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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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