On a Special Metric in Cyclotomic Fields

We verify the three conditions of a norm. The triangle inequality follows immediately from the usual triangle inequality in Euclidean spaces $\mathbb{Q}^{p-1}\subseteq\mathbb{R}^{p-1}$.

For any $\alpha\in K$ and $\lambda\in\mathbb{Q}$, we need to prove that $\|\lambda\alpha\|=|\lambda|\|\alpha\|$. This follows from the $\mathbb{Q}$-linearity of trace, since it implies that $v_{\lambda\alpha}=\lambda v_{\alpha}$.

It remains to prove positive-definiteness. Clearly $\|\alpha\|\geq 0$. Suppose $\|\alpha\|=0$. Then $v_{\alpha}\in\mathbb{Q}^{p-1}$ is the zero vector. Hence,

for all $j=1,\ldots,p-1$. Suppose $\alpha\neq 0$. Then we may write

Therefore,

Taking the trace of both sides, and using the fact that trace is $\mathbb{Q}$-linear, we have

which is a contradiction. Hence, $\alpha=0$. ∎

It suffices to show that the norm in Definition 2.1 is invariant under $G$, i.e., $\|\sigma(\alpha)\|=\|\alpha\|$ for all $\alpha\in K$ and $\sigma\in G$. Suppose $\sigma^{-1}(\omega)=\omega^{k}$, where $1\leq k\leq p-1$. Then we have

where the third equality follows from the fact that $\operatorname{Tr}_{K/\mathbb{Q}}$ is invariant under $G$. Since $k$ must be coprime to $p$, it follows that $\{kj:1\leq j\leq p-1\}$ is a permutation of $\{j:1\leq j\leq p-1\}$. Hence, $\|\sigma(\alpha)\|=\|\alpha\|$, as required. ∎

See Lemma 8.1.6 in [7] ∎

Let $G=\operatorname{Gal}(K/\mathbb{Q})$. By Galois theory, $\mathbb{Q}(\alpha)\subseteq\mathbb{Q}(\beta)$ if and only if $\operatorname{Gal}(K/\mathbb{Q}(\beta))\subseteq\operatorname{Gal}(K/\mathbb{Q}(\alpha))$. Since $G$ is a cyclic group of order $p-1$, the preceding condition is equivalent to $\left|\operatorname{Gal}(K/\mathbb{Q}(\beta))\right|$ dividing $\left|\operatorname{Gal}(K/\mathbb{Q}(\alpha))\right|$, which is then equivalent to $[K:\mathbb{Q}(\beta)]$ dividing $[K:\mathbb{Q}(\alpha)]$. By the tower law, this is equivalent to $[\mathbb{Q}(\alpha):\mathbb{Q}]$ dividing $[\mathbb{Q}(\beta):\mathbb{Q}]$.

As in the statement, let $\alpha_{1},\ldots,\alpha_{n}$ be the Galois conjugates of $\alpha$ over $K$, with $\alpha_{1}=\alpha$. Similarly, let $\beta_{1},\ldots,\beta_{m}$ be the Galois conjugates of $\beta$ over $K$, with $\beta_{1}=\beta$. Since $K/\mathbb{Q}$ is Galois, we have $[\mathbb{Q}(\alpha):\mathbb{Q}]=n$ and $[\mathbb{Q}(\beta):\mathbb{Q}]=m$. Therefore, we need to prove that $n$ divides $m$ under the hypothesis that $d(\alpha,\beta)<\frac{1}{2}d(\alpha,\alpha_{i})$ for all $i=2,\ldots,n$.

Let

Then $d(\alpha,\beta)<r$. For any element $x\in K$, denote by $B(x,r)$ the open ball centered at $x$ with radius $r$ under the metric $d$. Observe that if $\alpha_{i},\alpha_{j}$ are two distinct Galois conjugates of $\alpha$, say $\alpha_{i}=\sigma(\alpha)$ and $\alpha_{j}=\tau(\alpha)$, where $\sigma,\tau\in\operatorname{Gal}(K/\mathbb{Q}(\alpha))$, then

It follows that any two distinct conjugates $\alpha_{i}$ and $\alpha_{j}$ are at a distance of at least $2r$ from each other. In particular, the open balls $\left\{B(\alpha_{i},r):1\leq i\leq n\right\}$ are pairwise disjoint.

We claim that for every $\beta_{j}$ there exists an $\alpha_{i}$ such that $\beta_{j}\in B(\alpha_{i},r)$. In other words, the balls contain all conjugates of $\beta$. Indeed, if $\beta_{j}=\sigma(\beta)$, then

by Proposition 3.1. Hence, $\beta_{j}\in B(\sigma(\alpha),r)$.

Furthermore, we claim that for $1\leq i\leq n,$ each ball $B(\alpha_{i},r)$ contains the same number of conjugates of $\beta$. Indeed, suppose $\alpha_{i}=\sigma(\alpha)$. Then by Proposition 3.1 again, we have

Therefore, $\beta_{j}\in B(\alpha_{i},r)$ if and only if $\sigma^{-1}(\beta_{j})\in B(\alpha,r)$. Since $\sigma$ is a bijection, this proves that $B(\alpha,r)$ and $B(\alpha_{i},r)$ contain the same number of Galois conjugates of $\beta$. This number is nonzero because $\beta\in B(\alpha,r)$. Since the balls are disjoint, we conclude that $n$ divides $m$, as desired. ∎

Clearly $\mathbb{Q}(\gamma_{n})\subseteq\mathbb{Q}(\alpha,\beta)$, so it suffices to prove the reverse inclusion. Note that

Thus, when $n$ is sufficiently large, we would have $d(\alpha,\gamma_{n})<\frac{1}{2}d(\alpha,\sigma(\alpha))$ for all $\sigma\in G$. Theorem 3.3 implies that $\mathbb{Q}(\alpha)\subseteq\mathbb{Q}(\gamma_{n})$. In particular, $\alpha\in\mathbb{Q}(\gamma_{n})$, and so $\beta\in\mathbb{Q}(\gamma_{n})$, as desired. ∎

When $r$ is odd, the sum is zero because $a^{r}+(-a)^{r}=0$. When $r$ is even, this follows from the well-known Faulhaber’s formula of sums of powers (see [3] for example). ∎

By Lemma 4.1, we have

We break this sum into two pieces by linearity. The first piece equals

The second piece equals

We now simplify the second piece. If $i\neq j$, then the terms $a_{i}-b_{i}$ and $a_{j}-b_{j}$ are independent, in which case the sum is zero because

If $i=j$, then $(a_{i}-b_{i})(a_{j}-b_{j})=(a_{i}-b_{i})^{2}$, in which case the sum becomes

which is exactly the same as (3), up to a difference in the coefficient. It follows that

Again, we break the above sum by linearity, and noting that

and

where the value of $S_{2}(N)$ is computed in Lemma 6.2. Therefore, we obtain

and the result follows from dividing the above quantity by $\#B(p,N)^{2}=(2N+1)^{2p-2}$. ∎

We break (8) into two pieces according to whether $i$ equals $j$. The $i=j$ piece equals:

Since $(a_{1}-b_{1})^{4}=(a_{1}^{4}+b_{1}^{4})+4(a_{1}^{3}b+a_{1}b_{1}^{3})+6a_{1}^{2}b_{1}^{2}$, we can further rewrite the sum in (9) as

since all term with odd powers vanish. The above quantity can be computed directly using Lemma 6.2.

On the other hand, the $i\neq j$ piece of (8) equals

The innermost sum inside the square has been previously calculated in (5) and (6). The result now follows from combining the $i=j$ piece and the $i\neq j$ piece. We omit the details of the tedious calculation. ∎