1

I want to explicitly determine the subfield of GF$(16)$=$\mathbb{F}_{16}$ with 4 elements. I know this to be $\mathbb{F}_4$ (or isomorphic to it) since a field of prime power (in this case $2^4$) has a unique subfield of order $2^m$ iff $m$ divides $4$. The method I used seemed quite cumbersome.

I determined the roots of $Y^4-Y$ in $\mathbb{F}_{16}$ since $\mathbb{F}_4$ is the decomposition field of the polynomial $Y^4-Y$. I found that $$Y^4-Y=Y(Y-1)(Y^2+Y+1),$$ and so $0,1$ are already in $\mathbb{F}_4$ as a subfield of $\mathbb{F}_{16}$. Also $$\mathbb{F}_4\cong\dfrac{\mathbb{Z}_2[X]}{\langle X^2+X+1\rangle},$$ which means that $X^2=X+1$ in $\mathbb{F}_4$ and $$\mathbb{F}_{16}\cong \dfrac{\mathbb{Z}_2[X]}{\langle X^4+X+1\rangle},$$ which means $X^4=X+1$ in $\mathbb{F}_{16}$ and so elements of $\mathbb{F}_{16}$ are of the form \begin{equation} a_3X^3+a_2X^2+a_1X+a_0 \end{equation} with $a_i \in \mathbb{Z}_2$ for $i=0,1,2,3$. Next, I substituted $a_3X^3+a_2X^2+a_1X+a_0$ into $Y^2+Y+1$ and used the properties $X^4=X+1$, $X^2=X+1$ and found, after a long calculation, something of the form $$b_1X+b_0,$$ with $b_1,b_0\in\mathbb{Z}_2$ some sum of $a_i$'s and powers of them. This means the two remaining elements of $\mathbb{F}_4$ are $X+1$ and $X$ and so I found that $\mathbb{F}_4=\{0,1,X,X+1\}$ as a subfield of $\mathbb{F}_{16}.$

First of all, is this method correct? I believe it is since I found a correct answer for $\mathbb{F}_4$.

Second, is there a more simple solution to this? I was hoping for a more generic way of finding such subfields.

TheHunter
  • 305
  • 2
    Error. You don't have $X^2=X+1$ IN THE FIELD $\Bbb{F}{16}$. You only have $X^4=X+1$. To avoid confusions resulting from the many roles played by $X$ I write elements of $\Bbb{F}{16}$ in terms of the constant $\gamma=X+(X^4+X+1)$, when $\gamma^4=\gamma+1$ becomes the only relation I need to play with. Also, it is cumbersome to keep those different ideals in mind. – Jyrki Lahtonen Dec 06 '20 at 12:01
  • Also, see the last two thirds of this answer I prepared for referrals like this back in the day :-) – Jyrki Lahtonen Dec 06 '20 at 12:04
  • @JyrkiLahtonen Isn't it true that for elements of $\mathbb{F}4$ as a subfield of $\mathbb{F}{16}$ both the properties $X^4=X+1$ and $X^2=X+1$ hold? Since every element will be in $\mathbb{F}_{16}$ (and so the former holds) and every element will also be in the subfield with 4 elements, which is isomorphic to a field for which the latter holds. Correct me if I'm wrong because I don't immediately see why this wouldn't be the case. – TheHunter Dec 06 '20 at 12:18
  • 1
    You haven't taken on board the comment about confusing different uses of "$X$". The cube root of $1$, call it $\omega$ (a solution of the poly $X^2+X+1$), is the fifth power of a primitive $15$-th root of $1$,call it $\eta$ (where $\eta$ is a solution of $X^4+X+1$.) We have $\eta=X+\langle X^4+X+1\rangle$, and $\omega=X+\langle X^4 + X +1\rangle$. – ancient mathematician Dec 06 '20 at 12:44
  • 1
    Compare. In the field $K_1=\Bbb{Q}[X]/\langle X^2-2\rangle$ the coset of $X$ can be seen as $\sqrt2$. In the field $K_2=\Bbb{Q}[X]/\langle X^4-2\rangle$ the coset of $X$ can be seen as $\root4\of2$. There also $\Bbb{Q}(\sqrt2)\subset \Bbb{Q}(\root4\of2)$. If $X$ meant the same thing in both fields, that would mean that $\sqrt2=\root4\of2$. Surely you see how absurd that would be. The $X$ in $\Bbb{F}_2[X]/\langle X^2+X+1\rangle$ has nothing whatsoever to do with the $X$ in $\Bbb{F}_2[X]/\langle X^4+X+1\rangle$ for the same reason. – Jyrki Lahtonen Dec 06 '20 at 13:02
  • 1
    Just show that $X^2+X+\langle X^4+X+1\rangle$ is a zero of $Y^2+Y+1$ and hence an element of the subfield $\Bbb{F}_4$. – Jyrki Lahtonen Dec 06 '20 at 13:04

0 Answers0