# The Morm is Dead! Long Live the Morm!

Let $K$ be a number field of degree $d$ over $\mathbb Q$. Let $\mathfrak F$ be the group of fractional ideals of $K$, and let $\mathcal N$ and $\mathfrak N$ be respectively the elements of absolute norm 1 in $K$ and the fractional ideals of norm 1 of $K$. We will use primes to denote principal ideals, so that for instance $\mathfrak N’$ is the subgroup of principal ideals of norm 1. The ring of integers of $K$ will be denote $\mathcal O = \mathfrak o$ depending on whether we want to view it as a set of elements, or as a fractional ideal. The group of units will be denoted $\mathcal U$ and the group of roots of unity (which we will prompty ignore) will be denoted $\mathcal W$.

Claim: If $\xi \in \mathcal N$ there exists a unit $u \in \mathcal U$, an algebraic integer $\alpha$ and rational $r > 0$ such that $r \xi = u \alpha$.

Proof. Set $\mathfrak x = \xi \mathcal O$ and write $\mathfrak x = \prod_p \mathfrak x_p$, where $\mathfrak x_p$ consists of the prime factors of $\mathfrak x$ that lie over the same rational prime $p$. In this case, $$\mathfrak x_p = \prod_{\mathfrak p | p} \mathfrak p^{r_{\mathfrak p}}$$ where $\mathbf r_p = (r_{\mathfrak p})_{\mathfrak p | p}$ is a vector of integers. It should also be remarked that $\mathbf r_p \neq \boldsymbol 0$ for only finitely many rational primes. The norm-1 condition implies that the $\mathbf r_p$ dotted with the vector of inertial degrees $\mathbf f_p = (f_{\mathfrak p})_{\mathfrak p | p}$ is equal to 0. Write $\mathbf s_p = ( f_{\mathfrak p} e_{\mathfrak p} )_{\mathfrak p | p}$ where $e_{\mathfrak p}$ is the ramification degree of $\mathfrak p | p$. Then, consider $$\{\mathbf r_p + \ell \mathbf s_p : \ell \in \mathbb Z, r_{\mathfrak p} + s_{\mathfrak p} \geq 0 \}.$$ Let $\ell_p$ be the minimum of this set. In which case, $p^{\ell_p} \mathfrak x_p$ is an integral ideal in $\mathcal O$. Set $r = \prod_p p^{\ell p} \in \mathbb Q$. Then $r \mathfrak x$ is an integral ideal in the same ideal class as $\mathfrak x$. Finally, $\mathfrak x = \xi \mathcal O$ is principal, in which case there exists $\alpha \in \mathfrak O$ such that $r \xi \mathcal O = \alpha \mathcal O$ and the claim follows by choosing the appropriate unit. $\square$.

Definition: With $\xi \in \mathcal N$ if $\alpha \in \mathcal O$ is as in the proof of the previous claim, then we call $\alpha$ a visible integer for $\xi$. We denote the set of visible integers by $\mathcal V$. A principal ideal $\mathfrak a = \alpha \mathcal O$ generated by a visible integer is called a visible ideal. Every generator of $\mathfrak x = \xi \mathcal O$ has the same visible ideal, and so we say $\mathfrak a$ is the visible ideal for $\mathfrak x$. We define the morm of $\mathfrak x$, $\mathbb M\mathfrak x$ to be the (ideal) norm of $\mathfrak a$, and we define the morm of $\xi$, $M(\xi)$ to be the (field) norm of $\xi$ over $\mathbb Q$. That is $$\mathbb M \mathfrak x = \mathbb N \mathfrak a; \qquad M(\xi) = N_{K/\mathbb Q}(\alpha).$$ Given $r \in [0, \infty)$ we define $\mathfrak N'(r) = \{ \mathfrak x \in \mathfrak N’ : \mathbb M \mathfrak x \leq r \}$ to be the morm-exhaustion of $\mathfrak N’$.

Claim: If $\alpha$ is a visible integer, then the norm of $\alpha$ is the $d$th power of a rational integer.

Proof. With $\alpha, \xi$ and $r$ as in the previous claim, $N_{K/\mathbb Q}(\alpha) = N_{K/\mathbb Q}(r \xi) = r^d.$ The claim follows because $N_{K/\mathbb Q}$ is a rational integer.

Definition: If $n > 1$ is a rational integer not equal to $0, 1, -1$, and $\alpha$ is a visible integer, then $n \alpha$ is called a hidden integer. The corresponding integral ideal $n \mathfrak a = n \alpha \mathcal O$ is called a hidden ideal. An integer or integral ideal which is not visible and not hidden is said to be invisible. Non-invisible ideals can be extended to the group of non-invisible fractional ideals $\mathfrak R$. The quotient group $\mathrm{Inv} = \mathfrak F/\mathfrak R$ is called the invisibility group of $K$.

The local invisibility group over $p$ is the subgroup of $\mathbb Z/d \mathbb Z$ given by $$\mathrm{Inv}_p = \{ \mathbf a \cdot \mathbf f_p \bmod d : \mathbf a \in \mathbb Z^{J_p} \}.$$ Here $J_p$ is the number or primes $\mathfrak p | p$. $p$ is inert exactly when $J_p = 1$, in which case $\mathrm{Inv}_p = \{ 0 \}.$ In simple situations, for instance when $K$ is Galois, the $\mathrm{Inv}_p$ can be explicitly determined in terms of the splitting classes of $p$.

Claim: $\mathrm{Inv}$ is the restricted direct sum of the $\mathrm{Inv}_p$ over the rational primes, with respect to the identity subgroups $\{0\} \in \mathrm{Inv}_p$. That is, given any fractional ideal $\mathfrak c$, there exists a rational $r$ and a vector of non-negative integers indexed by the rational primes $(m_p)$, with $0 \leq m_p < d$ and $m_p = 0$ for all but finitely many $p$, such that $\mathbb N (r \mathfrak c) = \prod_p p^{m_p}$.

For each $p$ and $r \in \mathrm{Inv}_p$ we select a representative $\mathbf n_p(r)$ such that $\mathbf f \cdot \mathbf n_p(r) \equiv r \bmod d$. We write $\mathfrak n_p(r) = \prod_{\mathfrak p | p} \mathfrak p^{n_{\mathfrak p}(r)}$. If $\mathbf r = (r_p) \in \mathrm{Inv}$, then we define $$\mathfrak n(\mathbf r) = \prod_p \mathfrak n_p(r_p),$$ and call $\mathfrak n(\mathbf r)$ a representative ideal for the invisibility class $\mathbf r$. We take the representative ideal for the trivial invisibility class $\mathbf 0$ to be the ring of integers $\mathfrak 0$.