Irreducibility of $f(x)=x^{q+1}+x+1$ over $\mathbb{F}_q$

Question

I want to show that $\mathbb{F}_{q^3} = \mathbb{F}(\gamma)$ when $\gamma ^{q+1}+\gamma +1 = 0$. I showed that if $\alpha$ is a root of $f$ then $\alpha \in \mathbb{F}_{q^3}$, one thing in my mind is to show that $\alpha$ is a primitive element of $\mathbb{F}_{q^3}$.

Answer 1

I guess that the question is about the relation between the polynomial $f(x)$ and the minimal polynomials of its roots. You see, there is no reason for $f(x)$ to be irreducible. In fact, when $q>2$ the calculations show that it cannot be!!

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As a simple example consider the case $q=3$. We are in characteristic three so $f(1)=1+1+1=0$, and therefore $f(x)$ factors as $$ f(x)=x^4+x+1=(x-1)(x^3+x^2+x-1). $$ Here the cubic factor is irreducible, because otherwise it would have a linear factor and hence also a zero in $\Bbb{F}_3$.

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As a way of confirming the OP's findings let me include the following calculation. If $f(\alpha)=0$, then $$ \alpha^q=\frac{f(\alpha)-\alpha-1}\alpha=-\frac{\alpha+1}\alpha=-1-\frac1\alpha. $$ Consequently $1/\alpha^q=-\alpha/(\alpha+1)$ and $$ \alpha^{q^2}=-\frac{\alpha^q+1}{\alpha^q}=-1-\frac1{\alpha^q}=-\frac1{\alpha+1}. $$ Repeating the dose we then get $$ \alpha^{q^3}=-1-\frac1{\alpha^{q^2}}=-1+(\alpha+1)=\alpha. $$ The conclusion is that $\alpha\in\Bbb{F}_{q^3}$. This means that the minimal polynomial of $\alpha$ over $\Bbb{F}_q$ is either cubic or linear.

The polynomial $f(x)$ is a product of linear and cubic factors from $\Bbb{F}_q[x]$. In particular it is reducible whenever $q>2$.

In other words, we are working with the familiar cyclic group of three fractional linear transformations.

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The other question seems to be whether the zeros of $f(x)$ are primitive elements of the cubic extension. It may happen that no zero of $f(x)$ would be a primitive element of $\Bbb{F}_{q^3}$. As an example let me proffer the case of $q=4$. In this case we have the factorization $$ f(x)=x^5+x+1=(x^2+x+1)(x^3+x+1) $$ over $\Bbb{F}_2$. The quadratic factor $x^2+x+1$ splits into a product of irreducibles over $\Bbb{F}_4$, but zeros of the cubic factor $x^3+x+1$ are elements of $\Bbb{F}_8$, thus all of multiplicative order seven. Adjoining a seventh root of unity to $\Bbb{F}_4$ gives $\Bbb{F}_{4^3}$ as promised, but none of the roots of $f(x)$ have order $63$.

In general we can do the following. If $\alpha$ is a zero of $f(x)$ then $$ \begin{aligned} &\alpha^{q+1}=-1-\alpha\\ \implies &\alpha^{q(q+1)}=-1-\alpha^q\\ \implies &\alpha^{q^2+q+1}=-\alpha-\alpha^{q+1}\\ \implies &\alpha^{q^2+q+1}=1. \end{aligned} $$ So we can conclude that the order of $\alpha$ is a factor of $q^2+q+1$. This is a proper factor of $q^3-1=(q-1)(q^2+q+1)$ unless $q=2$.

If $q>2$ then none of the zeros of $f(x)$ are primitive elements of $\Bbb{F}_{q^3}$.