be a finite field (that means,
a field which has only a finite number of elements.)
Then:
such that
holds in
.
contains
as a subfield.
is a power of
.
, we have
.
is
a cyclic group of order
.
The next task is to construct such fields. An important tool is the following lemma.
and for any non zero polynomial
,
there exists a field
containing
such that
is decomposed into linear factors in
.To prove it we use the following lemma.
and for any irreducible polynomial
of
degree
, we have the following.
is a field.
be the class of
in
. Then
satisfies
.
Then we have the following lemma.
be a prime number. Let
be a power of
.
Let
be a field extension of
such that
is
decomposed into polynomials of degree
.
Then
containing
.
elements.
Finally we have the following lemma.
be a prime number. Let
. Then we have the following facts.
elements.
of degree
.
is divisible by the polynomial
as above.
which has exactly
-elements, there exists an element
.
In conclusion, we obtain:
of
, there exists a field which has exactly
elements.
It is unique up to an isomorphism. (We denote it by
.)
The relation between various
's is described in the following lemma.
to
if and only if
is a power of
.