abstract algebra - Two elements of $mathbb F_{q^m}$ that share the same minimal polynomial over $mathbb F_q$ have the same multiplicative order

Let $\mathbb F_{q^m}$ be the field with $q^m$ elements, where $q$ is a prime number (constructed as the quotient field $\mathbb F_q[x]/(p)$, where $p$ is an irreducible, monic polynomial over $\mathbb F_q$). Let $\alpha, \beta \in \mathbb F_{q^m}$, so that $\alpha$ and $\beta$ share the same the same minimal polynomial $m_\alpha$ over $\mathbb F_q$.

I now want to show that $\alpha, \beta$ have the same order in $\mathbb F_{q^m}^\times$ (the multiplicative group of $\mathbb F_{q^m}$). Using that, I'm then supposed to conclude that $m \mid \phi(q^m - 1)$, where $\phi$ is Euler's totient function.

Now my idea was to maybe look at the multiplicative, cyclic subgroup $\langle \alpha \rangle$ of $\mathbb F_{q^m}^\times$ (the subgroup that's generated by $\alpha$). Now we have $\mid \langle \alpha \rangle \mid = \text{ order of } \alpha \text{ in } \mathbb F_{q^m}^\times =: ord_{\mathbb F_{q^m}^\times}(\alpha)$.

Now I suspect that it's true that $\langle \alpha \rangle = \langle \beta \rangle$, from which the first statement would follow right away. But I'm not sure how I could derive that from only the fact that $\alpha, \beta$ are roots of the same minimal polynomial in $\mathbb F_q$.

(I'm also open to any other approaches – this was just the first that came to mind.)

Answer

The multiplicative order of $\alpha$ is the minimum $k$ such that $\alpha$ is a root of $x^{k} - 1$.

Now if $F$ is a field and $f \in F[x]$, the following are equivalent

• $\alpha$ is a root of $f$, and



• the minimal polynomial of $\alpha$ over $F$ divides $f$.

---

So the multiplicative order of $\alpha$ is the minimum $k$ such that $m_{\alpha} \mid x^{k} -1$. And the multiplicative order of $\beta$ is the minimum $h$ such that $m_{\beta} \mid x^{h} -1$. But $m_{\alpha} = m_{\beta}$, thus...

---

As to the second part, $\varphi(q^m - 1)$ is the number of generators of the cyclic group of non-zero elements of the field with $q^{m}$ elements. Every such generator generates the field with $q^{m}$ elements over the field with $q$ elements, and thus its minimal polynomial over the field with $q$ elements has degree $m$. But by the first part all the $m$ roots of such a minimal polynomial have the same multiplicative order, hence the number of generators of the cyclic group of non-zero elements of the field with $q^{m}$ elements is a multiple of $m$.