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.
 A: 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: 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.
