What proper subfields does #F_16# have (up to isomorphism)?

1 Answer
Aug 26, 2016

Answer:

The proper subfields of #F_16# are isomorphic to #F_2# and #F_4#.

Explanation:

The set of non-zero elements of a finite field of order #q# form a cyclic group of order #q-1# under multiplication. If #q > 2# then this cyclic group has at least one non-identity element #alpha# that generates it.

Choose #alpha in F_16# that generates all the #15# non-zero elements.

Then #alpha^5# generates the cyclic subgroup of order #3# corresponding to the subfield isomorphic to #F_4#.

#alpha^3# generates the subgroup of order #5#, but #5+1 = 6# is not a power of #2#, so there is no corresponding subfield.

#alpha^15 = 1# generates the subgroup of order #1#, corresponding to the subfield isomorphic to #F_2#.

In general, #F_(p^n)# has subfields isomorphic to #F_(p^k)# where #k# is a divisor of #n#.