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No credit card required 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