Degree of Field Extension over $\mathbb{F}_2$

Question

While studying for quals, this question came up:

Let $K = \mathbb{F}_2$, and $\alpha^{17} = 1$ where $\alpha \in \overline{K}$. If $\alpha \neq 1$ then $K(\alpha)/K$ has degree $8$.

This essentially asserts that all factors of the polynomial $$x^{16} + x^{15} + \ldots + x + 1$$ are of degree eight. I know I could factor this into$$ (x^8 + x^5 + x^4 + x^3 + 1)(x^8 + x^7 + x^6 + x^ 4 + x^2 + x + 1)$$ and show that each factor is irreducible, but that's awfully time consuming to do during a timed test.

Question: Is there a quick and slick way to prove this? (Or, more generally, is there a technique I could have used but didn't?)

Answer 1

For any field and natural number $n$, a field has no more than one multiplicative group of order $n$ (it's cyclic and collects together all elements satisfying $\alpha^n=1$). Suppose $\alpha\in\overline{\Bbb F}_p^\times$ has order $n$, so that $\langle\alpha\rangle$ is the unique group of order $n$ in $\overline{\Bbb F}_p^\times$. The field $\Bbb F_p(\alpha)$ is the smallest one containing $\alpha$, which is equivalent to the smallest field containing $\langle\alpha\rangle$, which is the smallest field containing a multiplicative group of order $n$. A finite field $K$ contains a multiplicative group of order $n$ if and only if ${\rm char}(K)\nmid n$ and $n\mid\#K^\times$.

In this case, $\Bbb F_2(\alpha)\subseteq\Bbb F_{2^r}$ if and only if $17\mid(2^r-1)$. Since $2^4=16\equiv-1$ mod $17$, the order of $2$ mod $17$ is $8$, hence $r=8$ is minimal for which this holds, hence $\Bbb F_2(\alpha)=\Bbb F_{2^{\large 8}}$.

Answer 2

I fully endorse Blue's approach for its simplicity. Since this is for quals you can also use the known Galois theory of finite fields. The conjugates of $\alpha$ are gotten by applying the Frobenius repeatedly so they are $\alpha,\alpha^2,\alpha^4,\alpha^8,\alpha^{16}=\alpha^{-1}$, $\alpha^{-2}$, $\alpha^{-4}$, $\alpha^{-8}$ but then the cycle closes as $\alpha^{-16}=\alpha$. Eight conjugates total, so that settles it.

Mind you, above we saw that $\alpha$ and $\alpha^{-1}$ are Galois conjugates. This bit of information manifests itself in your factorization. Namely it explains why the irreducible factors of the seventeenth cyclotomic polynomial are palindromic polynomials. After all the reciprocal of the minimal polynomial of $\alpha$ is the minimal polynomial of $\alpha^{-1}$, which now must coincide.