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 […]