Primitive roots in $\mathbb{F}_7[x]/(x^2+1)$

Question

Are $2+x$ and $1+x$ primitive roots in $\mathbb{F}_7[x]/(x^2+1)$? I'm having trouble with the arithmetic associated with finding primitive roots.

Answer 1

The field ${\Bbb F}_7[x]/\langle x^2+1\rangle$ has $48$ non-zero elements. We have $$\eqalign{ (1+x)^2&=1+2x+x^2=2x\cr (1+x)^3&=2x+2x^2=5+2x\cr (1+x)^4&=5+7x+2x^2=3\cr}$$ and so on. Sooner or later you will find $(1+x)^n=1$. If the first time this happens is when $n=48$, then $1+x$ is a primitive root.

There are some short cuts: the first value of $n$ must be a factor of $48$, so if you show that it is not $1,2,3,4,6,8,12,16$ or $24$ then it must be $48$. And with a bit of thought you can see that it is only necessary to confirm that $$(1+x)^{16}\ne1\quad\hbox{and}\quad (1+x)^{24}\ne1\ .$$