How to guarantee existence of a finite field Existence of a finite field:
Solution:
I can understand that if I  have a finite field $F$ of characteristic $p$ where $p$ is prime then I can consider $\mathbb Z_p$ as its prime field and hence $F$ will have $p^n$ elements
But how to find such a field $F$ ?
 A: For any $n$ one can always find an irreducible degree $n$ polynomial $q(x)$ over $\mathbb{Z}_p$ (edit see below). In particular $(q(x))$ is maximal, and so $\mathbb{Z}_p / (q(x))$ is a field. The elements of this field can be written uniquely as $$a_0 + a_1 x + \cdots + a_{n - 1} x^{n - 1} + q(x), \qquad a_0, \ldots, a_{n - 1} \in \mathbb{Z}_p,$$
so this field has precisely $p^n$ elements.
Edit By request, here's an easy way to see why such a polynomial always exists: By construction, $x^{p^n} - x$ is the product of all irreducible polynomials over $\mathbb{F}_p$ of degree dividing $n$. On the other hand, the degree of the product of all irreducible polynomials over $\mathbb{F}_p$ of degree strictly dividing $n$ (that is, dividing $n$ and less than $n$) is
$$\sum_{d \,|\, n, \, d < n} p^d \leq \sum_{k < n} p^d = \frac{p^n - 1}{p - 1} < p^n = \deg (x^{p^n} - x).$$
Thus, $x^{p^n} - x$ has a factor irreducible over $\mathbb{F}_p$ of degree $n$.
A: You can construct an algebraic closure $\overline{\mathbf{F}}_p$ of $\mathbf{F}_p := \mathbf{Z}/p\mathbf{Z}$ and show that the set of elements $x$ in it such that $x^{p^n}-x$ is a field (because the Frobenius $y\mapsto y^{p^n}$ is additive, classic), and as it has $p^n$ elements...
A: The cleanest general solution for this that I know of is to construct the said field as the splitting field of the polynomial $x^{p^n}-x$. In practice you want to construct it as the quotient ring $\Bbb{Z}_p[x]/\langle p(x)\rangle$, where $p(x)\in\Bbb{Z}_p[x]$ is an irreducible polynomial of degree $n$.
This practical approach begs the question: Can you show me such an irreducible polynomial? That's when the fun begins. Proving their existence is easier than actually finding one. Deriving a formula telling us exactly how many there are is also easier than actually finding one. I use tables and/or ad hoc techniques. 
