Roots of polynomials of the form $x^q-x-\alpha$ over $\mathbb F_{q^m}$

Question

Let $q$ be a prime power and $\mathbb F_q$ a finite field with $q$ elements. Moreover let $\mathbb F_{q^m}$ be an extension field. For every $\alpha \in \mathbb F_{q^m}$ define the polynomial $$ g_{\alpha}(x)=x^q-x-\alpha \in \mathbb F_{q^m}[x].$$ My questions:

1. For which $\alpha$ (and $m$) the polynomial $g_{\alpha}(x)$ has at least one root in $\mathbb F_{q^m}$?

2. Which is the splitting field of $g_{\alpha}$?

Here my thoughts.

1. If $\beta$ is a root of $g_{\alpha}$, then the set of roots of $g_{\alpha}$ is $$ \beta+\mathbb F_q:=\{\beta+\lambda \,|\, \lambda \in \mathbb F_q \}.$$ This means that if $g_{\alpha}$ has one root in $\mathbb F_{q^m}$ then all the roots belongs to $\mathbb F_{q^m}$.

2. If $\alpha \neq 0$ then $\beta \notin \mathbb F_q$.

Any help or reference will be appreciated. Thanks.

Alessandro

EDIT

It seems by computation with Macaulay2, for every $\alpha \in \mathbb F_{q^m}$ there are only 2 possibilities for the polynomial $g_{\alpha}$.

1. $g_{\alpha}$ splits into linear factors.

2. There exist $\lambda \in \mathbb F_q$, $\gamma_1,\ldots, \gamma_{\frac{q^m}{p}}\in \mathbb F_{q^m}$ (where $p$ is the characteritic of the field), s.t. $$g_{\alpha}(x)=\prod_{i=1}^{\frac{q^m}{p}} (x^p-\lambda x-\gamma_i). $$ But I don't know how to prove it.

Answer 1

Partial answer and/or something that is too long to fit into a comment.

Consider the relative trace function $$ tr^{m}_1:\Bbb{F}_{q^m}\to\Bbb{F}_q,x\mapsto x+x^q+x^{q^2}+\cdots+x^{q^{m-1}}. $$ We easily see (or know) that $tr^m_1$ is $\Bbb{F}_q$-linear, and surjective. Furthermore, $$ tr^m_1(z^q)=tr^m_1(z) $$ for all $z\in\Bbb{F}_{q^m}$. Therefore, again for all $z\in\Bbb{F}_{q^m}$, we have $tr^m_1(z^q-z)=0$. The mapping $$P:\Bbb{F}_{q^m}\to\Bbb{F}_{q^m},x\mapsto x^q-x$$ is also $\Bbb{F}_q$-linear. Clearly $\operatorname{ker}(P)=\Bbb{F}_q$, so rank-nullity theorem implies that $\operatorname{im}(P)$ has dimension $m-1$. But we just saw that $\operatorname{im}(P)\subseteq \operatorname{ker}(tr^m_1)$. Therefore we get equality

$$\operatorname{im}(P)=\operatorname{ker}(tr^m_1).$$ Together with your observation this implies that

$g_\alpha(x)$ has a zero in the field $\Bbb{F}_{q^m}$ if and only if $tr^m_1(\alpha)=0$.

As you observed, when this holds all the $q$ zeros of $g_\alpha(x)$ will be in the field $\Bbb{F}_{q^m}$.

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The factorization you observed can be described as follows. Assume that $tr^m_1(\alpha)\neq0$. Let $z$ be a zero of $g_\alpha(x)$ in some extension field of $\Bbb{F}_{q^m}$. Let's consider the Galois conjugates of $z$ over $\Bbb{F}_{q^m}$. These are gotten by iterating the Frobenius automorphism $F:z\mapsto z^{q^m}$. But we know that $F(\alpha)=\alpha$ and that $$ F(z)=z^{q^m}=z^{q^m}-g_\alpha(z)=z+\alpha. $$ An easy induction then proves that for all $i=1,2,\ldots,p$ we have $$ F^i(z)=z+i\cdot\alpha. $$ In particular the lowest power of $F$ that maps $z$ to itself is $F^p$. Therefore the minimal polynomial of $z$ over $\Bbb{F}_{q^m}$ is $$ m_z(x):=\prod_{i=0}^{p-1}(x-z-i\alpha). $$ Let's denote $T(x)=x^p-x=\prod_{i=0}^{p-1}(x-i)\in\Bbb{F}_p[x]$. We see that $$ T(\frac{x-z}\alpha)=\prod_{i=0}^{p-1}(\frac{x-z}{\alpha}-i)=\alpha^{-p}m_z(x). $$ But $T$ is additive, i.e. $T(x+y)=T(x)+T(y)$ for all $x,y\in\overline{\Bbb{F}_q}$, so $$ m_z(x)=\alpha^p T(\frac{x-z}\alpha)=\alpha^p T(x/\alpha)-\alpha^p T(z/\alpha) =x^p-\alpha^{p-1}x-\alpha^pT(z/\alpha). $$ In particular $T(z/\alpha)$ must be an element of $\Bbb{F}_{q^m}$. These are the factors that you see in your Macaulay output. In particular your $\lambda=\alpha^{p-1}$.