2
$\begingroup$

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 $.

$\endgroup$
6
  • 2
    $\begingroup$ The group of units is finite exactly for $\Bbb Q$ and imaginary quadratic fields $\endgroup$ Jul 7, 2023 at 16:39
  • $\begingroup$ @LukasHeger Oh great! Does that follow easily from Dirichlet's unit theorem or something like that? $\endgroup$ Jul 7, 2023 at 16:41
  • $\begingroup$ Yes, exactly it follows from Dirichlet's unit theorem $\endgroup$ Jul 7, 2023 at 16:42
  • $\begingroup$ Ok I'll edit the question to reflect that, thanks $\endgroup$ Jul 7, 2023 at 16:43
  • 2
    $\begingroup$ The class number formula involves the size of the torsion group. $\endgroup$ Jul 7, 2023 at 16:49

1 Answer 1

2
$\begingroup$

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).

$\endgroup$
11
  • $\begingroup$ $(p) = ((1-\zeta_p))^{p-1} = ((1-\zeta_p)^{p-1}) \implies (1-\zeta_p)^{p-1} \times (1+\zeta^*_p)^{p-1} \in ((1-\zeta_p)^{p-1})$ $ \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 $p$ ? $\endgroup$
    – Balaji sb
    Jul 8, 2023 at 13:50
  • $\begingroup$ $\mathbb{Z}[\zeta_w]/(p) = \mathbb{Z}_p[\zeta_w]$ 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 ? $\endgroup$
    – Balaji sb
    Jul 8, 2023 at 14:27
  • $\begingroup$ 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 $w \mid ({\rm N}(\mathfrak p)-1)$. $\endgroup$
    – KCd
    Jul 8, 2023 at 14:44
  • 1
    $\begingroup$ @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 $w = 4$) $\endgroup$
    – KCd
    Jul 8, 2023 at 14:47
  • 1
    $\begingroup$ @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. $\endgroup$
    – KCd
    Jul 8, 2023 at 14:48

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .