Fields - Finite Fields

With groups, it’s easy for us to find finite ones that have the same order, but are not the same (up to isomorphism). The simplest example I can give is $\mathbb{Z}_4$ and $\mathbb{Z}_2\times\mathbb{Z}_2$, which both have order $2$, but they are not isomorphic. The former is a cyclic group, but the latter is not!

However, with fields, I hope to show in this post that fields are the exactly the same as each other (up to isomorphism) if they have the same finite order. This means when I tell you I have a field of order $4$, I really mean the field of order $4$–there is only one.

Splitting Fields

I do not want to spend too much time focusing on splitting fields, but I will mention some important takeaways that are useful for this proof.

Definition 1. A splitting field is a field such that for a polynomial $f(x)$ over a field $K$, all roots are contained in the field.

For example, the splitting field for $r(x)=x^2-2$ over $\mathbb{Q}$ is $\mathbb{Q}(\sqrt{2})$, since the roots of this function are $\pm\sqrt2$, meaning we must attach $\sqrt2$ to $\mathbb{Q}$.

Theorem 2. Let $f(x)$ be a polynomial over $K$. If the extensions fields $F$ and $E$ are both splitting fields over $K$, then they are isomorphic.

Proof. An entire chapter of the book. See these notes for a full proof. $\blacksquare$

Size of Finite Fields

Definition 3. The characteristic of a field $F$ is the smallest positive number $n$ such that $1\cdot n =0$. If one does not exist, the characteric is $0$. Denoted $\mathrm{char}(F)$.

All integral domains have characteristic $0$ or $p$, where $p$ is a prime. Since all fields are integral domains, we know that fields can only have characteristic $0$ or $p$.

Theorem 4. All finite fields with characteristic $p$ have order $p^n$, where $p$ is a prime number, and $n$ is a positive integer.

Proof. First note that if a field $F$ has characteristic $p$, then the homomorphism $\phi: \mathbb{Z}\rightarrow F$ such that $\phi(n)=n\cdot1$ has a kernel $p\mathbb{Z}$. From the ring isomorphism theorem, we can say that the image of $\phi(\mathbb{Z})$, which we will call $K$, is isomorphic to $\mathbb{Z}/\ker(\phi) = \mathbb{Z}/p\mathbb{Z} \cong \mathbb{Z}_p$.

Now, since $F$ is a vector space over $K$, and both $F$ and $K$ are finite, we let $[F:K]=n$. Every element $f \in F$ has the unique form

\[f=a_1v_1+a_2v_2+\cdots+a_nv_n\]

for a basis $v_i$ and coefficients $a_i$, with $a_i \in K$. To count the number of possible elements we can form, simply use combinatorics. There are $p$ choices for each $a_i$, because there are $p$ elements in $K$. So the number of possible elements is $p^n$, for the $n$ coefficients. $\blacksquare$

We call $K$ the prime subfield of $F$. Note that $1 \in K$. This theorem gives us an idea about how some fields are structured.

Useful Theorems/Lemmas

Theorem 5. Let $F$ be a finite field with $p^n$ elements. Then $F$ is a splitting field for $x^{p^n}-x$ over its prime subfield $K$.

Corollary 6. Two finite fields are isomorphic if and only if they have the same number of elements.

Proof. Let $F$ and $E$ both have $p^n$ elements, and have prime subfields $K$ and $L$. $K\cong L\cong \mathbb{Z}_p$. Since the polynomial $x^{p^n}-x$ splits over $F$ and $E$ in $K$ and $L$, $F$ and $E$ are both splitting fields, and by Theorem 2, they are isomorphic. $\blacksquare$

Lemma 7. Let $F$ be a field with characteristic $p$, and let $n$ be a positive integer.

\[\{ a \in F \mid a^{p^n}-a=0\}\]

is a subfield of $F$.

Lemma 8. Let $F$ be a field of characteristic $p$. If $n$ is a positive integer not divisible by $p$, then $x^n-1$ has no repeated roots in any extension field of $F$.

The Main Theorem

Theorem 9. For prime $p$ and positive integer $n$, there is a field with $p^n$ elements.

Proof. We want to create a field $F$ with $p^n$ elements. So let $F$ be the splitting field for $f(x)=x^{p^n}-x$ over $\mathbb{Z}_p$. Since

\[f(x)=x^{p^n}-x = x(x^{p^n-1}-1),\]

and $p$ does not divide $p^n-1$, we know from Lemma 8 that there are no repeated roots in the polynomial. Lemma 7 tells us that all these roots form their own field. Therefore $F$ is a field that contains all roots of $f(x)$, of which there are $p^n$ unique ones. $\blacksquare$

Since any finite field of the same order is isomorphic, and we can create fields of any possible order $p^n$, we can call these finite fields the field of order $9,27,100,729$ etc. In fact, the field with $p^n$ elements is called the Galois field of order $p^n$, denoted $\mathrm{GF}(p^n)$.

The main point in all of this is that we have a fairly good understanding of how fields can behave when they are finite–something that we can’t say for even finite groups (see Monster group). This makes finite fields a good candidate for studying cryptography, where encryption needs to be hard to crack, but easy to implement.