# Roots of unity in rings of algebraic integers

Context: Let $$K$$ be an algebraic number field. Let $$O_K$$ be the ring of integers of $$K$$. Let $$O_K^\times$$ denote the group of units of $$O_K$$. By Dirichlet's unit theorem, $$O_K^\times$$ is always a finitely generated abelian group. The torsion subgroup of $$O_K^\times$$ is always a finite cyclic group and is exactly all the roots of unity that are in $$K$$.

Question 1: Are there any well known sufficient conditions for $$O_K^\times$$ to have free rank $$0$$ (in other words, conditions implying $$O_K^\times$$ finite) (Edit: By Dirichlet's unit theorem $$O_K^\times$$ is finite if and only if $$K=\mathbb{Q}$$ or $$K$$ is an imaginary quadratic field, see comment from Lukas Heger)

Question 2: Are there any well known formulas for determining the largest $$d$$ for which the root of unity $$\zeta_d$$ exists in $$K$$? (in other words, finding the order of $$Tor(O_K^\times)$$)

Edit (just fleshing out the comment from Bart Michels): Since roots of unity are algebraic integers then all the roots of unity in an algebraic number field $$K$$ are also in $$O_K$$. Thus the $$w_K$$ appearing in the class number formula https://en.wikipedia.org/wiki/Class_number_formula denotes both the number of roots of unity in $$K$$ and the number of roots of unity in $$O_K$$.

• The group of units is finite exactly for $\Bbb Q$ and imaginary quadratic fields Jul 7, 2023 at 16:39
• @LukasHeger Oh great! Does that follow easily from Dirichlet's unit theorem or something like that? Jul 7, 2023 at 16:41
• Yes, exactly it follows from Dirichlet's unit theorem Jul 7, 2023 at 16:42
• Ok I'll edit the question to reflect that, thanks Jul 7, 2023 at 16:43
• The class number formula involves the size of the torsion group. Jul 7, 2023 at 16:49

Let $$w$$ be the number of roots of unity in $$K$$, so $$w$$ is even. If $$p$$ is a prime factor of $$w$$ then $$\mathbf Q(\zeta_p) \subset K$$. Since $$(p) = (1-\zeta_p)^{p-1}$$ in $$\mathbf Z[\zeta_p]$$, every prime $$\mathfrak p$$ over $$p$$ in $$K$$ ramifies when $$p > 2$$. Hence when a prime $$\mathfrak p$$ in $$K$$ is unramified and $$\mathfrak p$$ doesn't lie over $$2$$, $$\mathfrak p \nmid (w)$$. Then $$x^w - 1 \bmod \mathfrak p$$ is separable and splits completely, so $$w \mid ({\rm N}(\mathfrak p)-1)$$.

There is a converse result: if $$d \mid ({\rm N}(\mathfrak p)-1)$$ for all but finitely many unramified $$\mathfrak p$$, then $$d \mid w$$, so $$w$$ is the gcd of the integers $${\rm N}(\mathfrak p)-1$$ as $$\mathfrak p$$ runs over any set of all but finitely many unramified primes in $$K$$. This may not look like a practical way of computing $$w$$, but if you can compute norms of prime ideals in $$K$$ then you can look at a large finite set of such numbers $${\rm N}(\mathfrak p) - 1$$ to make a plausible guess at the gcd of all such numbers (with $$\mathfrak p$$ unramified and not lying over $$2$$) in order to guess a value for $$w$$.

Example. Take $$K = \mathbf Q(i)$$, for which $$w = 4$$. When $$(\pi)$$ is a prime in $$\mathbf Q(i)$$ other than $$(1+i)$$ then $${\rm N}(\pi)-1$$ is divisible by $$4$$, either by a direct calculation (since $${\rm N}(\pi)$$ is a prime that's $$1 \bmod 4$$ or is $$p^2$$ where $$p \equiv 3 \bmod 4$$) or because $$\mathbf Z[i]/(\pi)$$ is a field containing a primitive $$4$$th root of unity. If $$(\pi)$$ runs over all but finitely many primes in $$\mathbf Q(i)$$ other than $$(1+i)$$ then it can be proved that the numbers $${\rm N}(\pi)-1$$ have gcd $$4$$. This generalizes the fact that if $$m \geq 2$$ then and $$p$$ runs over all but finitely many primes that are $$1 \bmod m$$, then the numbers $$p-1$$ have gcd $$m$$ (proved by Dirichlet's theorem).

• $(p) = ((1-\zeta_p))^{p-1} = ((1-\zeta_p)^{p-1}) \implies (1-\zeta_p)^{p-1} \times (1+\zeta^*_p)^{p-1} \in ((1-\zeta_p)^{p-1})$ $\implies i^{p-1} 2^{p-1} sin^{p-1}(2 \pi/p) \in ((1-\zeta_p)^{p-1})$. But i am not sure what multiple of $p$ is actually equal to $i ^{p-1} 2^{p-1} sin(2 \pi/p)$. Overall for $(p) = ((1-\zeta_p))^{p-1} = ((1-\zeta_p)^{p-1})$ there must be factorization of $p$ in terms $(1-\zeta_p)^{p-1}$ and a unit. Can you clarify the factorization by explicitly giving a unit which multiplies with $(1-\zeta_p)^{p-1}$ and gives $p$ ? Jul 8, 2023 at 13:50
• $\mathbb{Z}[\zeta_w]/(p) = \mathbb{Z}_p[\zeta_w]$ which is a finite field of order $p^w -1$ and $w$ divides $p^w - 1 = N(p) - 1$. Is this a more straight forward argument to prove $w$ divides $N(p)-1$? I think what you are telling is a more general result. Can you please clarify ? Jul 8, 2023 at 14:27
• When $\mathfrak p | (p)$, $\mathcal O_K/\mathfrak p$ has characterisic $p$, so when $p \nmid w$ the polynomial $x^w - 1$ has distinct roots in characteristic $p$. When $\mathbf Q(\zeta_w) \subset K$, the polynomial $x^w - 1$ has $w$ roots in $\mathcal O_K$ and thus $w$ distinct roots in $\mathcal O_K/\mathfrak p$ if $p \nmid w$. Hence the group $(\mathcal O_K/\mathfrak p)^\times$ has a subgroup of order $w$, so $w \mid ({\rm N}(\mathfrak p)-1)$.
– KCd
Jul 8, 2023 at 14:44
• @Balajisb Your 2nd comment has an error at the start: the ring $\mathbf Z[\zeta_w]/(p)$ has characteristic $p$ but it is not usually a field because $(p)$ need not be a prime ideal in $\mathbf Z[\zeta_w]$, e.g., whenever $p \equiv 1 \bmod w$ (think about $w = 4$)
– KCd
Jul 8, 2023 at 14:47
• @Balajisb That $(p) = (1-\zeta_p)^{p-1}$ as ideals in $\mathbf Z[\zeta_p]$ is discussed in lots of books on algebraic number theory (see chapters on cyclotomic fields). An explicit unit factor is $p/(1-\zeta_p)^{p-1}$, which admittedly is perhaps not a satisfying answer, but look at how algebraic number theory books talk about cyclotomic fields to see more details on this topic.
– KCd
Jul 8, 2023 at 14:48