At the end of last lecture we arrived at the question of whether every finite field is a for some prime
In this lecture we show that this is not the case, by exhibiting a field of 4 elements. We also find some general properties of finite fields. Finite fields have many interesting applications (in cryptography, for example), but we will not deal much with them as our focus through the course is on number fields, that we will begin discussing next lecture.
We begin by proving the following result:
Lemma 13. Suppose that
is a finite field. Then there is some natural number
such that the sum of
ones vanishes,
The least such
is a prime that divides the size of the field.
Proof. Suppose Consider the following elements of
There are
elements in this list and only
elements in
so necessarily two of these expressions must coincide, i.e., for some two distinct numbers
with
since one of the field axioms tells us that
Suppose towards a contradiction that
is not a prime, so we can write
where
and
Then
as can be easily checked from the field axioms. We now have a product
in a field
As shown last lecture, immediately after Lemma 12, it must be the case that either
or
i.e., either
or
But this contradicts the minimality of
and we are forced to conclude that
is a prime.
We are left to argue that (recall that
). This is a particular case of an argument we will encounter again later on, Lagrange's theorem. Letting
then:
since otherwise we must have two elements of
that are equal, which (reasoning as in the first paragraph of the proof) gives us a number strictly smaller than
in
is closed under addition and under additive inverses, i.e., if
then also
and
This is because, essentially,
with addition behaves like
with addition mod
For write
for the set
and note:
- For any
- For any
either
or else
Both these properties are easily checked using the two properties of we pointed out right before.
It follows that we have partitioned into disjoint sets, all of them of the same size
It must therefore be the case that
as we wanted to show.
Definition 14. If is a finite field, the prime number
from Lemma 13 is called the characteristic of
Let us now return to Example 7. Can we have a field of 4 elements? Say Then
because
is a prime that divides 4. This means that
which implies that
as well, since for example
Clearly,
cannot be any of
since
and
Thus
Similar reasoning shows us that there is only one way to complete the addition table of
and
for any
and
It is easy to check that is commutative and associative, that there is an additive identity and every element has an additive inverse.
Now we need to see if there is a way of constructing the multiplication table of Notice that
It cannot be 0 since
cannot have zero divisors. It cannot be 1, since
implies
which in turn implies that
or
It cannot be
since
implies that
so
or
Thus
Similar reasoning shows us that there is only one way to complete the multiplication table of
for all
for all
It is easy to check that is commutative and associative, that there is a multiplicative identity and that every nonzero element has a multiplicative inverse. It is also easy to check that
distributes over
It follows that
is indeed a field.
This is an interesting object: It is a field of 4 elements, so it is not a Notice that (up to renaming of its elements) there is only one such field, we did not have any freedom in the construction of the tables for addition or multiplication. Notice also that
for any
so
This is an interesting phenomenon that always holds in finite fields.
I list without proof the following general results:
- If
is a finite field, then
for
and some
- For any prime
and any positive integer
there is a field of size
this field is unique up to renaming of the elements. (The technical term for “renaming” is isomorphism, a notion we will soon study in detail.)
- In the field
of size
the equation
is true for all
Next time we will see several examples of infinite fields.
[…] 1. Show directly that there is no field of elements. (“Directly” means, among other things, that we cannot use the facts mentioned without proof at the end of lecture 4.3.) […]
[…] This polynomial has no roots in since Recall that in lecture 4.3 we constructed a field of 4 elements. Using the same notation we used there, where (this easily […]
[…] 1. Show directly that there is no field of elements. (“Directly” means, among other things, that you cannot use the facts mentioned without proof at the end of lecture 4.3.) […]
[…] This polynomial has no roots in since Recall that in lecture 4.3 we constructed a field of 4 elements. Using the same notation we used there, where (this easily […]