# Finding a cyclotomic field implied by Kronecker-Weber

I was wondering what are the most efficient ways to find a cyclotomic field s.t. given $K/\mathbb{Q}$ we have $K\leq \mathbb{Q}(\zeta_n)$? For quadratic fields this is easy by just considering factors of $d$ for the adjoined square root $\sqrt{d}$. How about cubic extensions and higher?

I guess such a formula can't be solely based on the primes that ramify, because we could construct fields with discriminant of the form $p^aq^b$ for rather large $a$ and $b$, and this would require $n$ be large too.

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The smallest such integer is the conductor, so you might google information about the conductors of abelian fields. For instance, you can find on the wikipedia page for conductors the fact that the conductor of your extension can be computed from the local conductors, and these are related to the unit groups in the completions at ramified primes. I don't know how practical that is for concrete computation, but for instance it shows that for a tamely ramified prime p, the power of p dividing the conductor will be 1. –  Barry Smith May 3 '11 at 0:34
@Barry: You can relate the conductor and discriminant of your extension using the conductor-discriminant formula. Regards, –  Matt E May 3 '11 at 0:37
I thought about mentioning that, but I think of that as a formula for computing the discriminant from conductors and not vice-versa. Can this viewpoint be reversed in a practical setting? –  Barry Smith May 3 '11 at 1:02
Since I can't edit the previous comment...to avoid confusion, I should have said for a tamely ramified p, the exact power of p dividing the conductor will be $p^1$ –  Barry Smith May 3 '11 at 1:20
@Barry: Dear Barry, With regard to your question "Can this viewpoint be reversed ... ?", see my answer below for an indication as to how. Best wishes, –  Matt E May 3 '11 at 14:11

Suppose that $K$ is ramified at $p_1,\ldots,p_n$. Then certainly $K \subset \mathbb Q(\zeta_{p_1^{m_1}},\ldots,\zeta_{p_n^{m_n}})$ for some $m_1,\ldots,m_n$.

Now you could try to pin down the field precisely by using some combination of information about the discriminant and the Galois group of $K/\mathbb Q$.

For example, suppose that our abelian extension $K$ is ramified at a single odd prime $p$. Then $Gal(\mathbb Q_{\zeta_{p^m}})/\mathbb Q) = (\mathbb Z/p^m\mathbb Z)^{\times}$ is cyclic; in fact it is a product of the cyclic groups $(1 + p\mathbb Z_p)/(1+p^m\mathbb Z_p)$ (of order $p^{m-1}$) and $(\mathbb Z/p\mathbb Z)^{\times}$ (of order $p-1$).

Any quotient is then equal to $(1+p\mathbb Z_p)/(1+p^{m'}\mathbb Z_p)$ (for $1 \leq m'\leq m$) and some quotient $H$ of $(\mathbb Z/p\mathbb Z)^{\times}$.

Thus if we choose $m$ minimally in the first place, the Galois group of $K$ over $\mathbb Q$ (which is a quotient of $Gal(\mathbb Q_{\zeta_{p^m}}/ \mathbb Q)$) is isomorphic to $(1 + p\mathbb Z_p)/(1+p^m\mathbb Z_p)\times H$, for a quotient $H$ as above.

The problem that you asked about is to compute $m$ given $K$. In this case we see that it is just a matter of computing the power of $p$ dividing the order of $Gal(K/\mathbb Q)$: the number $m$ is one more than the power of $p$ dividing this order.

As a slightly more complicated example, suppose that $K$ is ramified at two odd primes $p < q$, so that $K \subset \mathbb Q(\zeta_{p^m},\zeta_{q^n})$. Again, let's choose $m$ and $n$ minimally. Then $Gal(K/\mathbb Q)$ is a quotient of $$(\mathbb Z/p^m\mathbb Z)^{\times} \times (\mathbb Z/q^n\mathbb Z)^{\times}$$ $$= (\mathbb Z/p\mathbb Z)^{\times} \times (1 + p\mathbb Z_p)/(1+p^m\mathbb Z_p) \times (\mathbb Z/q\mathbb Z)^{\times} \times (1+q\mathbb Z_q)/(1+q^n\mathbb Z_q)^{\times} .$$

Now we see that it makes a difference whether or not $p | q - 1$. More precisely, suppose that $p^e$ is the precise power of $p$ that divides $q-1$. Then (thinking about the possible Sylow subgroup structures of a quotient of the above product, and the fact that $m$ and $n$ were chosen minimally), we see that $n$ is one more than the power of $q$ dividing the order of $Gal(K/\mathbb Q)$. On the other hand, if $m'$ is the power of $p$ dividing this order, then what we see is that $m-1 \leq m' \leq m-1+e$, so if $e \geq 1$, then we have not pinned down $m$ precisely just by knowing $m'$.

However, if we now apply the conductor-discriminant formula, one will find that knowledge of the discriminant of $K$ should allow us to determine $m$. The point is that the contribution to $m'$ that is coming from the $(\mathbb Z/q\mathbb Z)^{\times}$ factor will lead to additional powers of $q$ in the discriminant.

As a more precise example, suppose that $e = 1$, and write $(\mathbb Z/q\mathbb Z)^{\times} = H \times H',$ where $H$ has order prime-to-$p$ and $H'$ has order $p$. Consider two possibilities for $Gal(K/Q)$. In the first case, suppose that $Gal(K/Q) = (\mathbb Z/p^2\mathbb Z)^{\times} \times H$, while in the second case suppose that $Gal(K/Q) = (\mathbb Z/p\mathbb Z)^{\times}\times (\mathbb Z/q\mathbb Z)^{\times}$. In each case $Gal(K/Q)$ is the product of a cyclic group of order $(p-1)$ and a cyclic group of order $(q-1)$.

But in the first case the discriminant (up to sign) will be $p^{2p^2 - 3p} q^{(q-p-1)/p}$, while in the second case it will be $p^{p-2}q^{q-2}$.

And in the first case we have that $K \subset \mathbb Q(\zeta_{p^2},\zeta_q)$ (but no smaller cyclotomic field), while in the second case $K = \mathbb Q(\zeta_p,\zeta_q)$.

I didn't try to work this method out more systematically, but presumably one can, and certainly it shouldn't be so bad to apply by hand to any particular small example that you have in mind.

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