Number of elements $\alpha \in \mathbb F_{2^4}$ such that $\mathbb F_2(\alpha) = \mathbb F_{2^4}.$

Question

What is the number of elements $\alpha \in \mathbb F_{2^4}$ such that $\mathbb F_2(\alpha) = \mathbb F_{2^4}\ $?

If $\alpha \in \mathbb F_{2^4}$ is such an element then the minimal polynomial of $\alpha$ over $\mathbb F_2$ has degree $4.$ Since any monic irreducible polynomial is a minimal polynomial of each of it's roots it follows that there will be $4k$ many such $\alpha \in \mathbb F_{2^4}$ such that $\mathbb F_2 (\alpha) = \mathbb F_{2^4}.$ Now how to determine $k\ $? Since $\left \lvert \mathbb F_{2^4} \right \rvert = 2^4 = 16$ it follows that $k \leq 4.$ Also $\{0,1\} = \mathbb F_2 \subseteq \mathbb F_{2^4}.$ So $k \leq 3.$ Also I am eager to know that $:$ Is there any way to generalize this result for $\mathbb F_p \hookrightarrow \mathbb F_{p^n}$ for any prime $p$ and for any positive integer $n\ $? Any help in this regard will be warmly appreciated.

Thanks for your time.

$\Bbb F_{2^4}$ has $\phi(2^4-1)=8$ primitive elements $\alpha$ with $\Bbb F_2(\alpha)=\Bbb F_{2^4}$. In general the number is $\phi(p^n-1)$. — Dietrich Burde, Apr 03 '21 at 08:28
But @Dietrich Burde the answer given is $12.$ Are the primitive elements generate the multiplicative group of a field? — Anacardium, Apr 03 '21 at 08:29
For a proof see for example here, for a proof in Hagen's answer. — Dietrich Burde, Apr 03 '21 at 08:35
@Dietrich Burde see here $:$ https://math.stackexchange.com/q/2958747/512080 See Angina Seng's answer for example. Primitive elements doesn't necessarily generate the multiplicative group of the field. — Anacardium, Apr 03 '21 at 08:41
"Depends on the definitions. This correctly answers the question in the body, but is wrong for the question in the title. Caveat reader." OK, so they don't always generate the multiplicative group. — Dietrich Burde, Apr 03 '21 at 08:43
So how to proceed now @Dietrich Burde? You have misguided me at first. Please don't mark it as a duplicate question. — Anacardium, Apr 03 '21 at 08:45
Look for the number of irreducible monic polynomials over a finite field. There is a formula. — Dietrich Burde, Apr 03 '21 at 08:58
You see the formula in the answer below. This holds in general for all finite fields. See this duplicate. — Dietrich Burde, Apr 03 '21 at 09:30

score 1 · Accepted Answer · answered Apr 03 '21 at 09:36

My other answer finishes it in the way you were trying and also answers your generalisation.

However, here's another way of doing that for this simple case:
Note that we have the tower of subfields $$\Bbb F_2 \subset \Bbb F_{2^2} \subset \Bbb F_{2^4}$$ and there are no other subfields of $\Bbb F_{2^4}$. (Recall that $\Bbb F_{p^m} \subset \Bbb F_{p^n}$ iff $m \mid n$.)

Thus, if we pick any element of $\Bbb F_{2^4}$ outside $\Bbb F_{2^2}$, it would be a primitive element. Conversely, any primitive element would necessarily have to be outside of $\Bbb F_{2^2}.$

Thus, there are $|\Bbb F_{2^4}| - |\Bbb F_{2^2}| = 16 - 4 = \boxed{12}$ such elements.

In this example, we had a nice situation because the poset of subfields was actually a chain. In general, you'd face the same if you have an extension of the form $\Bbb F_{p} \hookrightarrow \Bbb F_{p^{q^n}}$ where $p$ and $q$ are primes. In that case, you'd get the answer as $p^{q^{n}} - p^{q^{n - 1}}.$
(Our case was $p = q = n = 2$.)

In more complicated examples, trying to use the above strategy would result in you ending up with the necklace polynomial as in the other answer.

Many many thanks for the detailed answer. I will upvote the other answer what you gave after I look into the necklace formula and understand it clearly. — Anacardium, Apr 03 '21 at 10:35

score 0 · Answer 2 · answered Apr 03 '21 at 09:26

You are looking for all those elements which whose (monic) irreducible polynomial over $\Bbb F_2$ has degree exactly $4$.

We have an explicit formula to get the number. This is the Necklace polynomial (see the fourth point under Applications). Derivation can be found in Lecture 11 here.

In this case, we have the number as $$\frac{1}{4}(1\cdot2^4 - 1\cdot2^2 + 0\cdot2^1) = 3.$$

Since all these polynomials are separable, they will all have four distinct roots (and two distinct polynomials won't share any root). Thus, the total number is $3 \times 4 = \boxed{12}$.

In general, the necklace polynomial will let you conclude in the same way.

Number of elements $\alpha \in \mathbb F_{2^4}$ such that $\mathbb F_2(\alpha) = \mathbb F_{2^4}.$

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