TITLE: Are all finite fields isomorphic to $\mathbb{F}_p$?
QUESTION [16 upvotes]: I've recently started taking some algebra courses and I was wondering whether or not every finite field is isomorphic to $\mathbb{F}_p$, where $p$ is prime.

REPLY [20 votes]: You're close, every finite field is isomorphic to $\mathbb{F}_{p^n}$, which is a quotient of $\mathbb{F}_p[x]$ by an irreducible polynomial of degree $n$.  It's often difficult to see that things like this can exist without a concrete example.  The easiest is $\mathbb{F}_4$:
$$\begin{array}{c|cccc}
+       &   0       &   1       &   \alpha  &   1+\alpha\\ \hline
0       &   0       &   1       &   \alpha  &   1+\alpha\\
1       &   1       &   0       &   1+\alpha&   \alpha  \\
\alpha  &   \alpha  &   1+\alpha&   0       &   1       \\
1+\alpha&   1+\alpha&   \alpha  &   1       &   0       \\
\end{array}$$
$$\begin{array}{c|cccc}
\times  &   \color{grey}{0}       &   1       &   \alpha  &   1+\alpha\\ \hline
\color{grey}{0}       &   \color{grey}{0}       &   \color{grey}{0}       &   \color{grey}{0}       &   \color{grey}{0}       \\
1       &   \color{grey}{0}       &   1       &   \alpha  &   1+\alpha\\
\alpha  &   \color{grey}{0}       &   \alpha  &   1+\alpha&   1       \\
1+\alpha&   \color{grey}{0}       &   1+\alpha&   1       &   \alpha  \\
\end{array}$$
You can see that $\mathbb{F}_4$ is a group under addition and $\mathbb{F}_4^\times=\mathbb{F}_4\setminus \{0\}$ is a group under multiplication.
In this case, I made $\mathbb{F}_4=\mathbb{F}_{2^2}$ out of the polynomial $x^2+x+1$, which is irreducible over $\mathbb{F}_2$.  I get the multiplication table by taking all polynomials of degree $1$ or less (i.e. below the degree of the irreducible polynomial minus one), then multiplying them together and reducing them mod $x^2+x+1$. (For calculation, it doesn't matter whether you use $x$ or $\alpha$.)