Mandelbrot Set and Fermat’s Last Theorem
111
(ii) Let x, y ∈ F. ⇒ x
−1
.y = 0 ⇒ X − {x
−1
.y} is a neighbourhood of 0. ⇒ There
is a symmetric neighbourhood V ⊂ X − x
−1
.y ⇒ x.V and y.V are nbds of x and
y. ⇒ x. V ∩ y.V = φ. So there exist v 1 , v 2 ∈ V such tha tv 1 v
−1
2 ∈ V , which is a
contradiction. Hence F is Hausdorff.
2.2 Finite Fields
Denote F[P
n
], where P is a prime n is a positive integer F[ p
n
] has P
n elements. For
n=1, fields are of the form F( p) = {0, 1, 2, . . . , p − 1} multiplication and addition
are usual operations, except multiples of p should be left out of the set (modulo p).
F(5) = {0, 1, 2, 3, 4}
All of the multiplications in the example are mod 5 because P = 5 addition on F[4]
is addition of polynomials. (a + bx) + (c + dx) = (a + c) + x(b + d). What is the
additive identity?
(a + bx) + (c + dx) = (a + bx) = (a + c) + x(b + d)
a + c = 0, c = 0 b + d = b, d = 0. Therefore 0 is still the additive identity. (a +
bx) has the additive inverse −a − bx. Usual multiplication of polynomials except
need to use x
2
+ x + 1 = 0
(a + bx) ∗ (c + dx) = ac = x(bc + ad) + x
2
(bd)
This seems to be a problem because F[4] only has linear polynomials but we ended
up with a quadratic one. When constructing the elements of F[2] remember had
multiplication modulo p = 2, so F[4] as the “constant” polynomials. ⇒ Remember
when making these tables that each element will only show up once in any column or
row. This is because we want to show that each element has a multiplicative inverse
and that there are no zero divisors
F[9] = F[3
2
] = {a + bx|a, b F[9]}
. As a vector space, to find the multiplication table we need a monic-quadratic that
has no zeros in F[9]. A monic-quadratic will have a coefficient of 1 on the highest
degree term.
Try x
2
+ 0x + 1, f (0) = 1, f (1) = 2, f (2) = 2 (no zeros!).
Note: The constant term must to be non-zero, because otherwise 0 is a zero. Also
note that we only need one polynomial that works. From the above explanation, it is
clear that in F( p), FLT is true only if n = ( p − 1) or n =
p−1
2
.
If we change the metric on R
2 , its geometry is different and hence FLT is no more
true. For example, if we consider the taxi cab metric d on R
2
111
(ii) Let x, y ∈ F. ⇒ x
−1
.y = 0 ⇒ X − {x
−1
.y} is a neighbourhood of 0. ⇒ There
is a symmetric neighbourhood V ⊂ X − x
−1
.y ⇒ x.V and y.V are nbds of x and
y. ⇒ x. V ∩ y.V = φ. So there exist v 1 , v 2 ∈ V such tha tv 1 v
−1
2 ∈ V , which is a
contradiction. Hence F is Hausdorff.
2.2 Finite Fields
Denote F[P
n
], where P is a prime n is a positive integer F[ p
n
] has P
n elements. For
n=1, fields are of the form F( p) = {0, 1, 2, . . . , p − 1} multiplication and addition
are usual operations, except multiples of p should be left out of the set (modulo p).
F(5) = {0, 1, 2, 3, 4}
All of the multiplications in the example are mod 5 because P = 5 addition on F[4]
is addition of polynomials. (a + bx) + (c + dx) = (a + c) + x(b + d). What is the
additive identity?
(a + bx) + (c + dx) = (a + bx) = (a + c) + x(b + d)
a + c = 0, c = 0 b + d = b, d = 0. Therefore 0 is still the additive identity. (a +
bx) has the additive inverse −a − bx. Usual multiplication of polynomials except
need to use x
2
+ x + 1 = 0
(a + bx) ∗ (c + dx) = ac = x(bc + ad) + x
2
(bd)
This seems to be a problem because F[4] only has linear polynomials but we ended
up with a quadratic one. When constructing the elements of F[2] remember had
multiplication modulo p = 2, so F[4] as the “constant” polynomials. ⇒ Remember
when making these tables that each element will only show up once in any column or
row. This is because we want to show that each element has a multiplicative inverse
and that there are no zero divisors
F[9] = F[3
2
] = {a + bx|a, b F[9]}
. As a vector space, to find the multiplication table we need a monic-quadratic that
has no zeros in F[9]. A monic-quadratic will have a coefficient of 1 on the highest
degree term.
Try x
2
+ 0x + 1, f (0) = 1, f (1) = 2, f (2) = 2 (no zeros!).
Note: The constant term must to be non-zero, because otherwise 0 is a zero. Also
note that we only need one polynomial that works. From the above explanation, it is
clear that in F( p), FLT is true only if n = ( p − 1) or n =
p−1
2
.
If we change the metric on R
2 , its geometry is different and hence FLT is no more
true. For example, if we consider the taxi cab metric d on R
2