$\mathbb{F}_{4096}$,$\mathbb{F}_{16}$ and $\mathbb{F}_{2}$

Question

I'm stuck at these two questions:

i) How many distinct elements $a \in \mathbb{F}_{4096}$ exist, such that $\mathbb{F}_{4096}=\mathbb{F}_2 \left[a\right]$

ii) Find an irreducible $g \in \mathbb{F}_4 \left[x\right]$ such that $\mathbb{F}_{16}=\mathbb{F}_4/(g)$

As for i) I know that the degree of the minimal polynomial of $a$ has to be $12$, since $2$ is prime and $4096=2^{12}$.

As for ii) I'm lost: Does $\deg(g)$ equal to $2$? If so, how would I show that $\mathbb{F}_4/(g)$ indeed is $\mathbb{F}_{16}$? Is $g=x^2+1$ a viable guess?

Answer 1

i) Hint: By a bad choice for $a$, you would end up with $\Bbb F_{2048}$ or a sub-field thereof

ii) Yes, we need $\deg g=2$. However, $x^2+1$ is not irreducible: $x^2+1=(x+1)(x+1)$. In fact, any $x^2+ax+b$ with $a,b\in\Bbb F_2$ will fail to work as it can at most produce a degree 2 extension of $\Bbb F_2$, i.e., it can already be factored in $\Bbb F_4$.

Answer 2

Because $12=2^2\cdot3$, the field $\Bbb{F}_{4096}$ has the following subfields $$ \Bbb{F}_2,\Bbb{F}_4,\Bbb{F}_8,\Bbb{F}_{16},\Bbb{F}_{64}. $$ The exponents $m$ of $2^m$ here are the proper factors of $12$, namely $1,2,3,4$ and $6$.

Of the listed subfields $K_1=\Bbb{F}_{16}$ and $K_2=\Bbb{F}_{64}$ contain the smaller ones. We further observe that $K_1\cap K_2=\Bbb{F}_4$. This lets us answer your first question:

$\Bbb{F}_2[\alpha]=\Bbb{F}_{4096}$ unless $\alpha\in K_1\cup K_2$. We need to exclude $$|K_1\cup K_2|=|K_1|+|K_2|-|K_1\cap K_2|=64+16-4=76$$ elements. Implying that $4096-76=4020$ choices of $\alpha$ will work.

To answer your second question see the bottom quarter of this answer I prepared for referrals. You will find that with $\Bbb{F}_4=\{0,1,\beta,\beta+1=\beta^2\}$ an irreducible quadratic is $g(x)=x^2+x+\beta$. There are a total of six irreducible monic quadratics in $\Bbb{F}_4[x]$. Anyone of them can serve in the role of $g(x)$.