# Problem Involving Finite Fields

I've arrived at a Theorem in text that I'm confused about:

Note: My question below is about the statement of this theorem, not about a proof for it. (The proof is supplied in the text)

Theorem: Let $E$ be a field of $p^{n}$ elements contained in an algebraic closure $\tilde{\mathbb{Z}_{p}}$ of $\mathbb{Z}_{p}$. The elements of $E$ are precisely the zeros in $\tilde{\mathbb{Z}_{p}}$ of the polynomial $x^{p^{n}} - x$ in $\mathbb{Z}_{p}[x]$.

The first line startles me somewhat. So far in this book we have never considered the algebraic closure of any structure which wasn't a field. And for $\mathbb{Z}_{p}$ to be a field, we must have that $p$ is prime. This is not given in the theorem, and there was no blanket statement at the beginning of the section as there sometimes is.

My Question: Have I missed some key fact regarding the orders of a finite field needing to be prime powers?

To give perspective to my background and where this chapter fits into development, the purpose of the chapter I am reading is to build the Galois Field of order $p^{n}$, with which I am not yet familiar.

• Well, I guess the book should have said $p$ prime, since after all there is a field with $(5^3)^{17}$ elements, and $5^3$ is not prime. But the letter $p$ in field context is pretty much reserved for primes. – André Nicolas Aug 1 '12 at 21:38
• Assuming that you refer to Theorem $33.3$ in Fraleigh's A First Course in Abstract Algebra, then note that he writes in the first paragraph of Section $33$ "The purpose of this section is to determine the structure of all finite fields. We shall show that for every prime $p$ and positive integer $n$, there is exactly one finite field (up to isomorphism) of order $p^n\,\ldots$" (my emphasis). – Bill Dubuque Aug 1 '12 at 21:40
• Sorry for taking so long to reply. Yes I am reading exactly that book and referring to exact that theorem. I didn't draw the same conclusion from that introductory paragraph. As someone who didn't know the fact that almost every comment refers to, for all I know it may have been required to prove some facts about fields of non-prime power order before accomplishing such a goal. But with all these in mind, thanks for reminding me of that paragraph because now I cannot interpret it any other way. :) – roo Aug 1 '12 at 22:29
• "...prove some facts about fields of non-prime power order..." Does a finite field whose order is composite but not a power of any prime exist? – Dilip Sarwate Aug 2 '12 at 1:31

The theorem states that $E$ is a field. For a field to have $p^n$ elements, $p$ must be prime (unless the author is a psycho, see comments); since all finite fields have order $p^n$ for $p$ prime. So the statement that $E$ is a field of order $p^n$ already determines the fact that $p$ is prime, and so (provided this result is known) there is no need to mention that $p$ is prime in the statement of the theorem. If this is not a known result, then I refer you to almost any undergraduate algebra textbook which covers fields for a proof.

• A field can have $p^n$ elements for $p=4$, which is not prime. (I think it may be implicit in the statement of the theorem that $p$ must be prime, simply due to the variable letter chosen for it). – hmakholm left over Monica Aug 1 '12 at 21:36
• @HenningMakholm: I guess that's right, I think I'm just too used to the convention! – Clive Newstead Aug 1 '12 at 21:41
• That seems totally reasonable, yet we haven't developed this fact yet. I'll try to verify it. – roo Aug 1 '12 at 21:46

I'm writing this answer both to satisfy myself and to address the key problem in my question.

Fact: A finite field must have a prime-power order

The proof is an easy application of the immediately preceding Corollary to this theorem which I have already asked about here: Question about a corollary about Finite Fields

Proof: Since the preceding corollary states that if a field has characteristic $p$, then it must have $p^n$ elements for some $n\geq 1$, all that needs to be shown is that a field cannot have a non-prime characteristic.

If $F$ has characteristic $p$, and $p = xy$ for some $1 < x,y < p$, then $x = x\cdot 1\neq 0$ and $y = y\cdot 1\neq 0$ by definition of characteristic. But then $xy = xy\cdot 1 = p\cdot 1 = 0$ by assumption. Therefore $x$ and $y$ are zero divisors contained in the field $F$ which is a contradiction.

Thanks to everyone as usual for their seemingly infinite patience with my questions.