# Abstract Nonsense

## Finite Fields

Point of Post: In this post we discuss the classification of finite fields, and discuss some other incidental properties relating to such.

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Motivation

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There are very few quick, clean, and satisfying classifications in mathematics. Among these though is the classification of finite fields. Now, while this may not seem like a very important class of fields, finite fields have stupendous importance in many areas of mathematics, perhaps most prominently in number theory. That said, for now we shall view them mostly as curiosities to test our knowledge against.

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Ok, so why are finite fields so “curious”? What’s so interesting about them is that even though we know entirely what their additive and multiplicative structures look like up to isomorphism it is very difficult to construct them explicitly. For example, we shall show that there is a field $\mathbb{F}_4$ with four elements. Now, clearly $\mathbb{F}_4\cong\mathbb{Z}_2^2$ as groups (if this isn’t obvious, we shall prove this below) and also that $\mathbb{F}_4^\times\cong\mathbb{Z}_3$.  That said, try to sit down and construct a field with four elements just by writing out a multiplication table–I bet you will have a hard time.

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In fact, there is only one surefire technique for constructing such a field. Indeed, suppose that we wanted a field of order $p^n$ where $p$ is some prime (as we shall see these are the only orders of fields), then to construct a field $k$ of order $p^n$ it seems sensible to construct a degree $n$ extension of the field $\mathbb{Z}_p$. Moreover, we have a good way to do this. Namely, if we were able to find a polynomial $f(x)\in\mathbb{Z}_p[x]$ which is irreducible and of degree $n$ then we know that $\mathbb{Z}_p[x]/(f)$ is a field extension of $\mathbb{Z}_p$ of degree $n$. That said, it’s difficult to actually find such a polynomial. For example, the problem of finding fields of order $4$ is simple because $f(x)=1+x+x^2$ is an irreducible (it is identically $1$ as a function) polynomial in $\mathbb{Z}_2[x]$ and so $\mathbb{Z}_2/(f)$ is such a field–but, these are really small numbers.

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No, instead of trying to construct such finite fields explicitly we shall construct them as splitting fields of certain polynomials in $\mathbb{Z}_p[x]$.

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Finite Fields

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Let’s begin by figuring out first what any supposed finite field “should” look like. So, let’s assume that $k$ is such a field.  The first obvious question one might ask about $k$ is what its prime subfield $k_0$ may look like? Well, we know from first principles that $k_0$ is isomorphic to either $\mathbb{Z}_p$ for some prime $p$ or $\mathbb{Q}$. Now, clearly since $k_0\not\cong\mathbb{Q}$ we may conclude that $k_0\cong\mathbb{Z}_p$ for some prime $p$. Now, since $k$ is an extension of $k_0$ we know that $k$ is a $k_0$-algebra and thus isomorphic, as a $k_0$-vector space, to $k_0^n$ where $n=[k:k_0]$ (which of course is necessarily finite). While this may seem somewhat unimpressive note that it tells us that $|k|=p^n$. Thus, we find that the only possible orders of finite fields are prime powers (we shall, in fact, show that all such orders exist).

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Here’s another somewhat novel argument for why a finite field must have prime power order, using group theory opposed to vector space theory. Indeed, suppose that $k$ is a finite field of characteristic $p$ and $q\ne p$ is prime such that $q\mid |k|$. By Bezout’s lemma we can find integers $n,m$ such that $nq+mp=1$. Now, by Sylows’ theorems we can find $x\in (k,+)$ with $|x|=q$. Note then $x=(nq+mp)x=nqx+mpx=0$ (since $qx=0$ and $px=0$ since the field is characteristic $0$), which is a contradiction.

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Very interestingly we know more than $|k|=p^n$ we actually know what the two structures $(k,+)$ and $(k,\cdot)$ actually look like (up to isomorphism). Indeed, since $k\cong \mathbb{Z}_p^n$ as vector spaces they are obviously isomorphic as abelian groups and so we find that additive $(k,+)$ is just the elementary abelian group $\mathbb{Z}_p^n$. Multiplicatively we know that $k^\times$ looks like $\mathbb{Z}_{p^n-1}$ since, as we have proven before, finite subgroups of the multiplicative group of a field are cyclic. Now, note that this was regardless of what field $k$ was–with all these similarities between all fields of order $p^n$ we should get a strong inkling that, perhaps, all fields of order $p^n$ are isomorphic.

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Ok, so enough with the screwing around, let’s actually discuss how one can actually produce these finite fields.

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Suppose that we wanted to make a field $k$ of order $p^n$. Let’s consider the polynomial $x^{p^n}-x\in \mathbb{Z}_p[x]$. We know that this polynomial has a splitting field, call it $K$. What we’d like to claim is that $|K|=p^n$. Indeed, let $\alpha_1,\cdots,\alpha_{p^n}$ be the roots of $x^{p^n}-x$ in $K$ (we shall eventually show that all of these roots are distinct, but for now we just take this as true). What we know then is that $K=\mathbb{Z}_p(\alpha_1,\cdots,\alpha_{p^n})$. Now, we merely know that any $x\in K$ can be written in the form $\displaystyle \sum_i c_i\alpha_i^{n_i}$ with $c_i\in\mathbb{Z}_p$. We see then that

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$\displaystyle x^{p^n}=\left(\sum_i c_i\alpha_i^{n_i}\right)^{p^n}=\sum_i c_i^{p^n}\left(\alpha_i^{p^n}\right)^{n_i}=\sum_i c_i\alpha_i^{n_i}=x$

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where we used the fact that since $K$ has characteristic $p$ that $(x+y)^{p^n}=x^{p^n}+y^{p^n}$ and by Fermat’s little theorem $z^p=z$ for all $z\in\mathbb{Z}_p$ and so $z^{p^n}=z$ for all $z\in\mathbb{Z}_p$. From this we can see that $|K|\leqslant p^n$ but since $\alpha_1,\cdots,\alpha_{p^n}$ are distinct we may conclude that $|K|=p^n$ as desired.

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So, now we know that finite fields of orders $p^n$ for $n\in\mathbb{N}$ and $p$ prime exist. From this we actually get a pretty cool theorem. Namely, let $K$ be a field of order $p^n$. We know then that $K^\times$ is cyclic of order $p^n-1$ and so there exists $\alpha\in K^\times$ which generates $K^\times$ and so clearly $K=\mathbb{Z}_p(\alpha)$. We see then that if we consider the minimal polynomial $m_{\alpha,\mathbb{Z}_p}$ that $K\cong \mathbb{Z}_p[t]/(m_{\alpha,\mathbb{Z}_p}(t))$ and thus $m_{\alpha,\mathbb{Z}_p}$ is irreducible and $\deg m_{\alpha,\mathbb{Z}_p}=[K:\mathbb{Z}_p]=n$. Thus, we find that

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Theorem: Let $p$ be a prime. Then, for every $n\in\mathbb{N}$ there exists an irreducible polynomial of degree $n$ in $\mathbb{Z}_p[x]$.

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Which, at least to me, is not a priori an obvious fact (I personally don’t know an example of an easy way to construct such polynomials–I’m sure there exists such a method though).

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Anyway, up until this point all we know is that the only possible orders of a finite field are $p^n$ where $p$ is a prime, and moreover, there exists at least one such field for every one of these admissible orders. Thus, the next order of business is to show that the order of a finite field determines it up to isomorphism–or that there is, up to isomorphism, only one field of each order $p^n$.

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The idea behind this is simple. Namely, we constructed a field of each order $p^n$ as the splitting field of $x^{p^n}-x\in\mathbb{Z}_p[x]$ and so using circular logic we know that every field of that order is going to have to be a splitting field for $x^{p^n}-x$. Thus, turning circular logic into the good kind we actually prove that every field $K$ of order $p^n$ really is a splitting field for $x^{p^n}-x$.  To do this we merely note that clearly $0$ satisfies $x^{p^n}-x$, and if $\alpha\in K^\times$ then $1=\alpha^{|K^\times|}=\alpha^{p^n-1}$ and thus $\alpha^{p^n}=\alpha$. Thus, we see that every element of $K$ is a root of $x^{p^n}-x$ and since $|K|=p^n$ we see that it constitutes a full set of roots of $x^{p^n}-x$. Since these roots (being all of the field) generate the field we may conclude that $K$ really is a splitting field for $p^n$. Using the fact that splitting fields are unique up to isomorphism we may finally state the full classification of finite fields:

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Theorem(Classification of Finite Fields): Every finite field has order $p^n$ for some prime $p$ and some $n\in\mathbb{N}$. Moreover, for each such $p$ and $n$ there exists precisely one, up to isomorphism, field of order $p^n$.

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We denote the unique field of order $p^n$ as $\mathbb{F}_{p^n}$. It is customary to just write $\mathbb{F}_q$ where one assumes that $q=p^n$ for some prime $p$.

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While there is quite a bit to say about finite fields, we only say a little bit more for now. Namely, we’d like to discuss how the inclusions of finite fields inside other finite fields works. For example, let’s suppose that $\mathbb{F}_{p^m}\subseteq\mathbb{F}_{p^n}$. We see then that $\mathbb{F}_{p^m}^k\cong\mathbb{F}_{p^n}$ for some $k$ and so $(p^m)^k=p^n$ or $n=mk$. Thus, $m\mid n$. What we’d like to show is that, conversely, if $m\mid n$ then $\mathbb{F}_{p^m}\subseteq\mathbb{F}_{p^n}$. To do this it suffices to show that if $m\mid n$ then any root of $x^{p^m}-x$ is a root of $x^{p^n}-x$. But, if $m\mid n$ then $n=mk$ for some $k$ and we see that if $\alpha$ is a root of $x^{p^m}-x$ then

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$\alpha^{p^n}=\alpha^{p^{mk}}=(\alpha^{p^m})^{p^{m(k-1)}}=\alpha^{p^{m(k-1)}}=\cdots=\alpha^{p^m}=\alpha$

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from where everything follows. Thus, we may conclusively deduce that

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Theorem: The inclusion $\mathbb{F}_{p^m}\subseteq\mathbb{F}_{p^n}$ holds if and only if $m\mid n$

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References:

[1] Morandi, Patrick. Field and Galois Theory. New York: Springer, 1996. Print.

[2] Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 2004. Print.

[3] Lang, Serge. Algebra. New York: Springer, 2002. Print.

[4] Conrad, Keith. Collected Notes on Field and Galois Theory. Web. <http://www.math.uconn.edu/~kconrad/blurbs/&gt;.

[5] Clark, Pete. Field Theory. Web. <http://math.uga.edu/~pete/FieldTheory.print

April 12, 2012 -

1. When you write $\mathbb{F}_4\cong \mathbb{Z}_2^2$ you must remark this is an isomorphism of groups.

Tonio

Comment by Tonio P. | April 12, 2012 | Reply

• Haha! Tonio–thank you for pointing that out–but I do not thank you for making me nervous knowing that someone with whom I hold such respect is watching out for my mistakes.

It’s good to see you man.

Best,
Alex

Comment by Alex Youcis | April 13, 2012 | Reply

2. Some typos to fix.
Motivation
Paragraph 2 “we sill show” to “we shall
show”

Paragraph 3 “then construct a field” to “then to construct a field”

Comment by Stephen Tashiro | April 13, 2012 | Reply

• Stephen,

Thank you very much!

Best,
Alex

Comment by Alex Youcis | April 14, 2012 | Reply

3. […] (inseparable, if you will) to the Galois group of the extension. For example, consider the finite field and the rational function field . Then, we claim that is a simple extension of degree . Indeed, […]

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