TITLE: Construction of $p^n$ field
QUESTION [5 upvotes]: I heard that one can construct field based on $p^n$ elements where $p$ is prime. I tried with $p = n = 2$. It seemed easy as there are 2 groups which $\{0,1,2,3\}$ can form and 1 group formed by $\{1,2,3\}$. However each time I get into contradiction. Is there anything I missed or it is not possible to form $p^n$ element group?

REPLY [3 votes]: The very nice answer by Arturo Magidin describes explicit constructions of finite fields with $p^n$ elements. Let me work out a couple of explicit examples here. 
For instance, you can construct a field $\mathbb{F}_4$ with $4$ elements by considering the quotient  $\mathbb{F}_4=\mathbb{F}_2[x]/(x^2+x+1)$. Here $\mathbb{F}_2 = \mathbb{Z}/2\mathbb{Z}$ is a field with two elements, and notice that $x^2+x+1$ is irreducible over $\mathbb{F}_2$, because neither $0$ or $1$ are roots. Indeed, you can check that the $4$ polynomials $\{0, 1, x, 1+x\}$ form a complete set of representatives for the quotient of $\mathbb{F}_2$ modulo $(x^2+x+1)$, because any polynomial $q(x)$ of degree $2$ or higher is congruent to one of $0$, $1$, $x$ or $1+x$ modulo $x^2+x+1$ (just use long division of $q(x)$ by $x^2+x+1$). Here is the addition table for the field $\mathbb{F}_4$:
$$\begin{array}{c|cccc}

+ & 0 & 1 & x & 1+x \\
\hline
0 & 0 & 1 & x & 1+x \\

1 & 1 &  0 & 1+x & x \\

x & x &  1+x & 0 & 1 \\

1+x & 1+x &  x & 1 & 0 \\

\end{array}  $$
Notice that the shape of the multiplication table agrees with the one given by Robin Chapman. Here is the multiplication table for $\mathbb{F}_4$:
$$\begin{array}{c|cccc}

\times & 0 & 1 & x & 1+x \\
\hline
0 & 0 & 0 & 0 & 0 \\

1 & 0 &  1 & x & 1+x \\

x & 0 &  x & 1+x & 1 \\

1+x & 0 &  1+x & 1 & x \\

\end{array}  $$
For example, $x(1+x)=x+x^2\equiv 1+(1+x+x^2)\equiv 1 \bmod (1+x+x^2)$.
Finally, let's construct a field with $9$ elements. This can be done by considering $\mathbb{F}_3/(x^2+1)$, since $x^2+1$ is irreducible over $\mathbb{F}_3$. Notice that $2\equiv -1\bmod 3$ is not a quadratic residue modulo $3$, thus $x^2+1\equiv x^2-2$ must be irreducible. Similarly, if $p$ is any prime with $p\equiv 3 \bmod 4$, then $x^2+1$ is irreducible and $\mathbb{F}_p/(x^2+1)$ is a field with $p^2$ elements.