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

Aug 26, 2016

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 $n$.