Past exam of the Probability, statistics and options II course of the University of Oxford / 2008 / B (Introduction to Fields)

The characteristic is the smallest positive integer $p$ with $p{1}_F=0$. Such an integer exists because the additive group is finite. If it factored as $p=ab$ with $1<a,b<p$, then $(a{1}_F)(b{1}_F)=0$ with both factors nonzero, contradicting the absence of zero divisors. Hence it is prime. The elements generated by $1_F$ form a subfield isomorphic to ${\mathbb{F}}_p$. The finite set $F$ is a finite-dimensional vector space of some dimension $a\ge 1$ over that subfield, so $\boxed{|F|=p^a}$.

A nonzero polynomial of polynomial degree $n$ over a field has at most $n$ distinct roots. This follows by induction: a root $r$ gives a factor $X-r$, and every different root must be a root of the remaining factor by cancellation. Thus $X^n-1$ has at most $n$ roots. In the finite multiplicative group, the elements whose order divides $n$ are exactly those satisfying $x^n=1$. The supplied cyclicity criterion therefore proves that the multiplicative group is cyclic.

For the prime of the specified form write $m=2^{r}s$ with $s$ odd. If $s>1$, then $2^{2^r}+1$ is a proper factor of $(2^{2^r}{)}^s+1$, impossible. Thus $m$ is a power of two. Since $2^m\equiv -1(\mathrm{mod}p)$, the order of two divides $2m$ but does not divide $m$. All divisors of the power of two $2m$ other than $2m$ itself divide $m$. ConsequentlyIt generates the multiplicative group precisely when $2m=p-1=2^m$. For positive integers $m$ this holds exactly at $m=1,2$: from $m=3$ onward $2^m>2m$, and the inequality persists on increasing $m$. These give $\boxed{p=3\text{ or }5}$, equivalently the stated range among Fermat primes.