Let \(F\) be a finite field. Give a complete proof that \[ |F|=p^r \] for some prime \(p\) and integer \(r\ge1\).
Solution
1. The characteristic of \(F\) is a prime number \(p\).
Because \(F\) is finite, the elements \[ 0,\ 1,\ 1+1,\ 1+1+1,\ \ldots \] cannot all be distinct. Hence there is a least positive integer \(m\) such that \[ m\cdot1_F=0. \] Thus \(\operatorname{char}F=m>0\).
If \(m=uv\) with \(1<u<m\) and \(1<v<m\), then \[ (u\cdot1_F)(v\cdot1_F) = m\cdot1_F =0. \] By minimality of \(m\), neither factor is zero. This contradicts the fact that a field has no zero divisors. Therefore \(m\) is prime; write \(m=p\).
2. The prime subfield of \(F\) is isomorphic to \(\FF_p\).
1. Hence its image is a subfield of \(F\)isomorphic to\[ \ZZ/p\ZZ=\FF_p. \]We identify this image with \(\FF_p\).
3. The field \(F\) is a finite-dimensional vector space over \(\FF_p\).
Since \(\FF_p\subset F\), the field \(F\) is an \(\FF_p\)-vector space. Any linearly independent subset of \(F\) is, in particular, a subset of the finite set \(F\), so it is finite. Thus a basis of \(F\) over \(\FF_p\) is finite. Write \[ \dim_{\FF_p}F=r. \] Because \(1_F\neq0\), this vector space is nonzero, so \(r\geq1\).
4. The field \(F\) has exactly \(p^r\) elements.
Choose a basis \(e_1,\ldots,e_r\) of \(F\) over \(\FF_p\). Every element of \(F\) has a unique expression \[ c_1e_1+\cdots+c_re_r, \qquad c_j\in\FF_p. \] There are \(p\) independent choices for each of the \(r\) coefficients, so there are exactly \(p^r\) such coefficient tuples. Hence \[ \boxed{\abs{F}=p^r}. \]
5. Q.E.D.
Step 1 gives a prime \(p\), step 3 gives an integer \(r\geq1\), and step 4 gives the required cardinality.