Show that a finite field must have exactly \(p^n\) elements for some prime \(p\) and some positive integer \(n\).
List all the subfields of the field of size \(3^{12}\).
Give an example of an infinite field of characteristic \(3\).
Solution
1. In part (a), a finite field \(F\) has \(p^n\) elements for a prime \(p\) and an integer \(n\ge1\).
The characteristic of a field is \(0\) or a prime. In characteristic \(0\) the prime subfield is isomorphic to \(\QQ\), which is infinite, so \(F\) has prime characteristic \(p\) and prime subfield \(\FF_p=\ZZ/p\ZZ\). Field addition and multiplication by elements of \(\FF_p\) make \(F\) an \(\FF_p\)-vector space, which is finite-dimensional because \(F\) is finite. If \(n=\dim_{\FF_p}F\), then \(n\ge1\) and \(F\cong\FF_p^n\) as vector spaces, so \(\abs{F}=p^n\).
2. In part (b), the subfields of \(\FF_{3^{12}}\) are its unique subfields of orders\[ 3,\quad 3^2=9,\quad 3^3=27,\quad 3^4=81,\quad 3^6=729,\quad 3^{12}=531441. \]
A subfield \(K\) of \(\FF_{p^m}\) contains \(\FF_p\), and by the tower law \(d=[K:\FF_p]\) divides \(m=[\FF_{p^m}:\FF_p]\). Conversely, for each positive divisor \(d\) of \(m\), the polynomial \(x^{p^d}-x\) divides \(x^{p^m}-x\) in \(\FF_p[x]\), and its roots in \(\FF_{p^m}\) form the unique subfield of order \(p^d\) [@DF04]. Here \(p=3\), \(m=12\), and the positive divisors of \(12\) are \(1,2,3,4,6,12\).
3. In part (c), \(\FF_3(t)\), the field of rational functions in one indeterminate \(t\) over \(\FF_3\), is an infinite field of characteristic \(3\).
\(\FF_3(t)\) is the field of fractions of the integral domain \(\FF_3[t]\), so it is a field. Its characteristic is \(3\), since \(1+1+1=0\) in \(\FF_3\subseteq\FF_3(t)\). The monomials \(t^k\), \(k\ge0\), are pairwise distinct elements, so \(\FF_3(t)\) is infinite. An algebraic closure \(\overline{\FF}_3=\bigcup_{n\ge1}\FF_{3^n}\) is another infinite field of characteristic \(3\).
4. Q.E.D.
Steps 1, 2, and 3 answer parts (a), (b), and (c).