Can you give me one example where the characteristic of a field is not equal to the number of elements in the field?
I know that the characteristic of a prime field $GF(p)$ is $p$, but even for a non-prime $q$, say $q = 6$, the characteristic is equal to 6.
So when the characteristic and the number of elements in a fields will be unequal?
On
The only possible characteristics for a field are $0$ (but these are all infinite so not in the scope of the question) and primes. Here's a rough idea as to why: if a field $F$ had characteristic $n = ab$ with $1 < a, b < n$, then $a, b \neq 0$ in $F$ but $ab = 0$. But of course, a field cannot have zero divisors. Here, when I refer to $a, b$ as elements of $F$, I mean that as $a * 1_{F}, b * 1_{F}$, i.e. the image of $a, b$ under the unique ring homomorphism $\mathbb Z \longrightarrow F$ which defines the characteristic.
Furthermore, as indicated in the comments, any finite field must have prime power order. Here's the sketch of this proof: For any finite field $F$ of characteristic $p$, we can consider the subset $\{1, 2, \dots p - 1\} \subseteq F$. This subset is isomorphic to $GF(p)$ and subsequently, $F$ inherits the structure of a finite dimensional vector space over $GF(p)$. Hence, its order must be a power of $|GF(p)| = p$.
As @AlekosRobotis indicated, there are many examples of non-prime order fields - such as the splitting field of $x^q - x \in GF(p)[x]$. Remarkably, there is only one finite field of each prime power order, and a finite field $GF(p^n)$ contains a unique copy of $GF(p^d)$ for any $d \mid n$. I've sort of gone off the rails here, and I cannot include a whole proof of this classification of finite fields, but it all comes down to the polynomials $x^n - x$.
The field $\Bbb{F}_{q}$ for $q=p^k$ is a field of characteristic $p$, but contains $p^k$ elements. To construct this you take $\Bbb{F}_p[x]/(m(x))$ for $m(x)$ a monic irreducible polynomial of degree $k$. Alternatively, this can be seen as the splitting field of $x^q-x\in \Bbb{F}_p[x]$.