Size of a field with field extensions is hopefully equal to $p^n-1$ I've been given some terminology that I'm not sure is correct.
I have a finite field of prime multiplicative order $p$.  Then I "adjoin" square roots of numbers, whose square does not exist modulo $p$.  For example, modulo $5$, we have
$$1^2 \equiv 1, 2^2 \equiv 4, 3^2 \equiv 4, 4^2 \equiv 1$$
My understanding is that I can create a second value, say $\sqrt{2}$, that doesn't exist modulo $5$, and use it like I would use the imaginary number $i$ in the complexes modulo $5$.
So what I'm trying to do is use a set of $n$ square roots to extend the primes to $p^n - 1$ numbers plus a zero value, just like with complexes there are $p^2 - 1$ unique values besides $0$.  I believe that I can assume the field is closed with respect to addition, subtraction, and multiplication if the product of any of these square roots does not create an additional, "new" sqauare root.
My question is, do these fields have $p^n$ total elements, including zero, and what can I call them?  I guess I'm sort of looking for a crash course in "field extensions".
 A: Let us try to add $\sqrt 2$ to the field $K = \mathbb Z / 5 \mathbb Z$, as in your example. Note that you want to have an element $x$ in this extension that satisfies $x^2 -2 = 0$. Now it turns out that the minimal way to accomplish this is to consider the field
$$L = K[x]/(x^2-2),$$
where $K[x]$ is the polynomial ring over $K$ in one variable and $x$ is an indeterminate. Now $L$ is indeed a field since $(x^2-2)$ is a maximal ideal. You can think about this field as follows. An element is an equivalence class of polynomials, and the polynomial $x$ satisfies $x^2 = 2$. This should show you how to do arithmetic in this thing.
Alternatively, if you haven't done any ring theory, a possibly more instructive way of viewing this field (also pointed out by @GreginGre) is to have it be $L = \{a + b\sqrt{2} \mid a,b\in K\}$ (which is isomorphic to the previous $L$), which maybe makes it more obvious that $L$ is a vector space over $K$.
More generally, any field is a vector space over a subfield, so if you add things to $\mathbb Z/p \mathbb Z$ (for $p$ prime) you will get a field of $p^n$ elements, where $n$ depends on what you added. In our case we have $n = 2$ (because the degree of $x^2-2$ is $2$). But you see, if $\sqrt 3$ doesn't by chance happen to be an element of $L$ (it won't I guess), you will have to extend $L$ by $\sqrt 3$, yielding a field of $(5^2)^2$ elements.

As it turns out, I guessed wrong (see Jyrki Lahtonen's comment). The fact than all finite fields of order $p^n$ are isomorphic implies that whenever you add one square root to $\mathbb Z/p \mathbb Z$, you in fact add all the square roots.
