Fermat’s Last Theorem - Elliptic Curves: Number Theory and Cryptography - page 462

By counting the streets emanating from the station, you can immediately
determine which town you are in. The reason is that you have a base point. If
you didn't, then you might be on any of the vertices of I or II. You would not
be able to count streets and identify the town. The configuration of streets at
the station is the analogue of the tangent space at the base point. Of course,
it is possible that two towns could have the same tangent spaces, but Wiles
shows that this does not happen in his situation.

Tangent spaces

We now want to translate the notion of a tangent space into a useful alge-
braic formulation. Let R [ x, y ] be the ring of polynomials in two variables and
let f ( x, y ) R [ x, y ]. We can regard f as a function from the xy -plane to R .

Restricting f to the parabola y = x 2

∈

− 6 x , we obtain a function

f : parabola −→ R .

If g ( x, y ) ∈ R [ x, y ], then f and g give the same function on the parabola if
and only if f − g is a multiple of y +6 x − x 2 . For example, let f = x 3 − y and

g =6 x + xy +5 x 2 .Then

f − g = − ( x +1)( y +6 x − x 2 ) .

a 2 =0,so

If we choose a point ( a, b ) on the parabola, then b +6 a −

f ( a, b )= g ( a, b ) − ( a +1)( b +6 a − a 2 )= g ( a, b ) .

Therefore, there is a one-to-one correspondence

x 2 ) . polynomial functions on the parabola ←→ R [ x, y ] / ( y +6 x −

The ring on the right consists of congruence classes of polynomials, where
we say that two polynomials are congruent if their difference is a multiple of
y +6 x−x 2 . In this way, we have represented a geometric object, the parabola,
by an algebraic object, the ring R [ x, y ] / ( y +6 x x 2 ).

Now let's consider the tangent line y +6 x =0at(0 , 0). It is obtained by
taking the degree 1 terms in y +6 x

−

x 2 . We can represent it by the set −

{ax + by | a, b ∈ R } mod ( y +6 x ) ,

where we are taking all linear functions and regarding two of them as congru-
ent if they differ by a multiple of y +6 x . Of course, we could have represented
the tangent line by the ring R [ x, y ] / ( y +6 x ), but, since we already know that
the tangent line is defined by a linear equation, we do not lose any information
by replacing R [ x, y ] by the linear polynomials ax + by .