Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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
2  
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. :) –  Kyle Schlitt Aug 1 '12 at 22:29
1  
"...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
add comment

2 Answers

up vote 2 down vote accepted

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.

share|improve this answer
3  
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). –  Henning Makholm 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. –  Kyle Schlitt Aug 1 '12 at 21:46
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.