Is $\mathbb{F}_5$ a subfield of $\mathbb{F}_7$? I can think of the
answer 'yes' because they have the same set op operations $+ \cdot$
and the answer 'no' because in $\mathbb{F}_5: 2\cdot3=1$ and in
$\mathbb{F}_7: 2\cdot3=6$.When I consider the finite field with four elements $\mathbb{F}_4$:
$\{0,1,\omega,\omega^2=\omega+1\}$ as being $\mathbb{F}_2 \times
\mathbb{F}_2$ how do I prove or know that in this field $1+1=0$ like
in $\mathbb{F}_2$?
EDIT: by $\mathbb{F}_2 \times \mathbb{F}_2$ I mean that the product may be defined in a complicated way, e.g. $(a,b)\cdot(c,d)=(ac+bd,ad+bc+bd)$. Unfortunately I don't know the correct notation.Can it be proved also for the field with 8 elements $\mathbb{F}_8 =
\mathbb{F}_2 \times \mathbb{F}_2\times \mathbb{F}_2$?Is it possible to enumerate the elements of $\mathbb{F}_8$ like an
extension of the elements of $\mathbb{F}_4$:
$\{0,1,\omega,\omega^2=\omega+1, \gamma, \gamma^2, \ldots, \delta, \ldots\}
$
Sunday, 7 February 2016
Four questions about finite fields
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