# Roots of unity in a characteristic 2 integral domain

Let $$A$$ be an integral domain of characteristic $$2$$, and let $$T=\{ a \in A \ |\ \exists n \geq 1 \text{ s.t. } a^n=1 \}$$ be finite. I need to prove that $$T$$ is a cyclic group of odd order.

## 4 Answers

Let $$R$$ be the subring of $$A$$ generated by $$T$$. It has a finite generating family over $$\mathbb F_2$$ which is $$\{1\}\cup \{t^k, t\in T, k\leq n(t)\}$$ where $$n(t)$$ is the least integer $$\geq 1$$ such that $$t^n = 1$$; therefore it is a finite dimensional $$\mathbb F_2$$ vector space in particular it is finite.

Therefore it is a finite integral domain : this implies it is a field of characteristic $$2$$. But now we have :

A finite field of characteristic $$p$$ has size $$p^n$$ for some $$n$$

And

Let $$K$$ be a field and $$G\subset K^\times$$ a finite subgroup. Then $$G$$ is cyclic.

One easily checks that $$T\subset R^\times$$ and that the converse inclusion holds (by Lagrange's theorem) so $$T=R^\times$$ is cyclic, and has order $$|R|-1 = 2^n-1$$ for some $$n$$.

One can even prove that $$T = \{1\}$$.

Indeed, let $$t \in T$$ and $$n$$ such that $$t^n=1$$. then $$(1+t)(1+t+t^2+\ldots+t^{n-1}) =$$ $$1+t+t^2+\ldots+t^{n-1}+$$ $$t+t^2+\ldots+t^{n-1}+1$$ $$=0$$ Now $$\ t+1\$$ is a zero divisor unless $$\ t=1$$ or $$1+t+t^2+\ldots+t^{n-1} = 0$$. The latter is not the case: Let R be the subring generated by $$t$$, then $$R$$ is an algebra over the field $$\mathbb{F}_2$$ with (as a vector space) basis $$\{1, t, t^2, \ldots, t^{n-1}\}$$ so the factor under consideration has coefficients $$(1,1,\ldots,1)$$.

• That is definitiely not true. Take $A=\mathbb F_4$, then $A^\times$ is of cardinal $3$ and of course by Lagrange's theorem any $x\in A^\times$ has $x^3=1$. You haven't proved that $\{1,t...,t^{n-1}\}$ was a basis, it's just a generating set – Max Jun 22 at 13:37
• @Max Indeed you are right. I tried to delete the answer, but I'm not allowed to do that since it was already accepted. – Marc Bogaerts Jun 22 at 13:57

Here is another way:

Suppose $$T$$ is finite. Then, recall the following lemma:

Let $$G$$ be a finite abelian group. Then, $$G$$ is cyclic iff $$G$$ contains at most most one subgroup for each divisor of $$|G|$$.

In fact, the hypothesis that $$G$$ be abelian can be dropped. See Dummit and Foote Proposition 5 in section 6.1. Although, the proof is slightly easier when $$G$$ is abelian.

Let $$|T| =n$$ and $$k \mid n$$. Since $$A$$ is an integral domain there are at most $$k$$ roots of the polynomial $$x^k - 1$$. Hence, if there were 2 distinct subgroups of order $$k$$, say $$H,K$$, then every element of $$H \cup K$$ would be a root of $$x^k - 1$$ by lagranges theorem. But, $$|H \cup K| > k$$ since these subgroups are distinct. This contradicts there being at most $$k$$ roots of the polynomial $$x^k - 1$$. Thus, $$T$$ has at most 1 subgroup for every order dividing $$n$$ and by the lemma above $$T$$ must be cyclic. (Note: this proof is very similar to one used to show that the multiplicative group of units in a field is cyclic. This relies on the fact that a polynomial of degree $$n$$ over a field has at most $$n$$ roots in a field. This is also true in an integral domain as you can embed your domain in its fraction field.)

Now, we have shown that $$T$$ is cyclic, we need to show that $$|T|$$ is odd. If $$|T|$$ were even, then there would exists a subgroup of $$T$$ of order 2, specifically an element of order 2. Let $$x \in T$$ such that $$|x| = 2$$. Then, $$x^2 = 1 \implies x^2-1 = 0 \implies (x-1)(x+1)=0$$. Since $$A$$ is an integral domain either $$x = 1$$ or $$x = -1$$. But, $$A$$ has characteristic 2 so that $$-1 = 1$$ in $$A$$. Thus, $$x = 1$$, contrary to $$|x| = 2$$. As a result, $$|T|$$ must be odd.

Hint: Consider the $$\mathbb{F}_2$$-span of $$T$$.

• Could you be more explicit? Thanks – Christa Wolf Jun 18 at 9:54
• Note that $T$ is closed under multiplication. Hence the $\mathbb{F}_2$-span of $T$ is a finite sub-domain of $A$, hence a field $K$. $T$ is a finite multiplicative subgroup of $K^\times$, so is cyclic. – user10354138 Jun 22 at 14:08