Are the units of the quotient ring $\mathbb{F}_2[x]/\langle x^k+1 \rangle$ known in general, where $\mathbb{F}_2$ is the finite field with two elements? I'm specifically interested in the case where $k$ is divisible by two, such as say $k=8$, or $k=12$.
If so, can they be constructed easily? I'd be grateful if you could provide a few examples.
As Jyrki Lahtonen points out in the comments, one way to determine whether or not a particular polynomial $p(x)$ is a unit in the quotient is to compute the greatest common divisor of $p(x)$ and $x^k+1$. If it's $1$, then it's a unit. Otherwise it will be something other than $1$, and will not be a unit.
That can be computationally intense though, so I'm also going to recommend a quick trick that reliably produces units.
When $2$ divides $k$, we can be sure that $(x^k+1)=(x^{\frac{k}{2}}+1)^2$, and this means that the ring has a nonzero Jacobson radical. (The radical must contain at least $I=(x^{\frac{k}{2}}+1)$.) Then you can be assured that everything in $\{1+i\mid i\in I\}$ is a unit.
To retrieve the full Jacobson radical and maximize the number of units produced this way, you'd have to factor $x^k+1$ completely over $F_2$ and then look at the ideal generated by the product of prime divisors of $x^k+1$.
So for example in $F_2[x]/(x^2+1)$, you have a ring of four elements whose radical is $(x+1)$. Then $x+1$ and $0$ are nonunits, while $x$ and $1$ are units.
In $F_2[x]/(x^4+1)$, you have $x^4+1=(x+1)^4$, and in fact the radical is again $(x+1)$ and that everything inside the radical is nilpotent, and everything outside is a unit.
In $F_2[x]/(x^6+1)$ we have a first factorization that isn't trivial. We have $x^6+1=(x^3+1)^2=(x+1)^2(x^2+x+1)^2$. The radical is then $(x^3+1)$. Nothing in the radical is a unit, and $1+x$ for any $x$ in the radical is a unit. There can be other units and nonunits outside of these.
