What are the units of the cyclotomic ring $\mathbb Z[X]/(X^n+1)$, with $n$ being a power of $2$?

I am starting to think that the set $\{\pm X^k,k=0,\dots,n-1\}$ contains all units, is that so ?

  • $\begingroup$ Well, I actually stated that $n$ is a power of 2, so it is indeed a cyclotomic polynomial. $\endgroup$ – Tal-Botvinnik Apr 18 '16 at 12:40

It is not true. Dirichlet's theorem says that the unit group of a number field $K$ is finitely generated of rank $r_1+r_2-1$ where $r_1$ and $r_2$ are the number of real (resp. half the number of complex) embeddings of $K$.

The field $\mathbb Q(\zeta)$, $\zeta^n+1=0$, with $n>1$ a power of $2$, has no real embeddings and it has exactly $n$ complex embeddings. Thus the group of units of its ring of integers (which is $\mathbb Z[\zeta]$) has rank $n/2-1$. In particular, it's not a finite group for $n>2$.

What is true is that you described entirely the torsion in the group of units. But there are lots of other units generally.

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  • $\begingroup$ Thanks for the comments. Can anyone give a counterexample ? I.e. a unit that is not of the stated form $\endgroup$ – Tal-Botvinnik Apr 18 '16 at 20:39
  • $\begingroup$ @Tal-Botvinnik Sure. For instance your field contains $\mathbb Q(\sqrt 2)$ for $n>2$, and $1+\sqrt 2$ is a unit in $\mathbb Z[\sqrt 2 ]$. $\endgroup$ – Bruno Joyal Apr 18 '16 at 20:43
  • $\begingroup$ @Tal-Botvinnik Glad to see chess fans hanging out around here BTW. ;) $\endgroup$ – Bruno Joyal Apr 18 '16 at 20:45

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