# MinuteMochizuki 2

Welcome back to the second episode in the stills-YouTube-channel MinuteMochizuki. Today we will dismantle not just one scheme, but a simple cover of arithmetic schemes and replace it by a

Let $m$ be a square-free number and for simplicity we assume it not to be of the form $m=4l+1$. The ring of integers in the field $\mathbb{Q}(\sqrt{m})$ is then $\mathbb{Z}[\sqrt{m}] = \mathbb{Z}.1 \oplus \mathbb{Z}\sqrt{m}$ and we want to consider the degree two cover of arithmetic schemes

$$\mathbf{Spec}(\mathbb{Z}[\sqrt{m}]) \rightarrow \mathbf{Spec}(\mathbb{Z})$$

So we have to describe all prime ideals of $\mathbb{Z}[\sqrt{m}]$ and see how they lie over the prime numbers $p$ in $\mathbb{Z}$. This all depends on whether or not $\overline{m}$ is a square in $\mathbb{Z}/p\mathbb{Z}$. If $p \not\mid 4m$ then

- if $\overline{m}$ is a square modulo $p$ then there are two prime ideals $P_1$ and $P_2$ of $\mathbb{Z}[\sqrt{m}]$ lying over $(p)$ and $P_1.P_2=\mathbb{Z}[\sqrt{m}]p$. Some say $p$ splits in $\mathbb{Z}[\sqrt{m}]$ and there are infinitely many $p$ for which this happens.
- if $\overline{m}$ is not a square modulo $p$ then there is just one prime $P=\mathbb{Z}[\sqrt{m}]p$ lying over $(p)$. Some say $(p)$ remains prime, or is inert in $\mathbb{Z}[\sqrt{m}]$ and there are infinitely many $p$ for which this happens.

In the remaining (finite number of) cases, when $p \mid 4m$, there is one prime $P$ over $(p)$ but it ramifies meaning that $P^2=\mathbb{Z}[\sqrt{m}]p$. For example, when $m=3$ we get the following picture

and we see that $\mathbf{Spec}(\mathbb{Z}[\sqrt{3}])$ is a two-sheeted cover (some say etale cover) away from the 'bad points', the ramified primes $2$ and $3$.

In the inert cases we have that $\mathbb{Z}[\sqrt{m}]/p\mathbb{Z}[\sqrt{m}] \simeq \mathbb{F}_{p^2}$ and the Frobenius map $\overline{x} \rightarrow \overline{x}^p$ is a non-trivial automorphism of the quotient. We would like to be able to lift all these Frobenius automorphisms from quotients to ring-maps of $\mathbb{Z}[\sqrt{m}]$, but clearly the map $x \rightarrow x^p$ does not behave well for addition in $\mathbb{Z}$.

(Smart-asses will object at this point and argue that as $\mathbb{Q}(\sqrt{m})$ is an Abelian extension we may do all of this, somehow, using $\lambda$-rings. Point being, one would also like to do this in non-Abelian situations and preferable even for $Gal(\overline{\mathbb{Q}}/\mathbb{Q})$.)

In order to include information of these power-maps we will have no choice but to dismantle the two prime-spectra as well as their cover map and to replace this classical setting by the

Frobenioid $C(m)$

We'll start with the recipe of last time, replacing each point (prime ideal) in the two prime spectra by a component $\mathbb{Z}$ into the poset which is this time the disjoint union of two posets, one for each scheme. That is the objects of $C(m)$ are the elements of

$$\bigoplus_{P \in \mathbf{Spec}(\mathbb{Z}[\sqrt{m}])} \mathbb{Z} \bigsqcup \bigoplus_{p~\text{prime}} \mathbb{Z}$$

Last time we saw that we could identify elements of the second factor with positive rational numbers (that is, fractions). We can do something similar with elements of the first factor, but then we have to replace 'fractions' by 'fractional ideals' which are just ideals of $\mathbb{z}[\sqrt{m}]$ divided by an integer.

The poset-structure can again be defined in terms of divisibility if we (again) replace integers by ideals, that is, two fractional ideals $I$ and $J$ satisfy $I \mid J$ if there is an ideal $N \triangleleft \mathbb{Z}[\sqrt{m}]$ such that $I.N = J$.

Morphisms in $C(m)$ come in two flavours (or, as the inimitable Mochizuki would say: "the fundamental dichotomy between types of combinatorial structures, between 'etale-like' structures which are 'indifferent to order' and 'Frobenius-like' structures which are 'order-conscious'." (GeomFrob1, p.9)).

First flavour comes, as last time, from the action-maps of a monoid on each of the components. We already described this action on the component corresponding to $\mathbf{Spec}(\mathbb{Z})$. The action on the other component is similar, apart from having to trow in an extra factor, the Galois group $Gal(\mathbb{Q}(\sqrt{m})/\mathbb{Q})=\mathbb{Z}/2\mathbb{Z}$. So, the action monoid is $\mathbb{N}_{> 0} \times \mathbb{Q}(\sqrt{m})^{\ast} \times Gal$ with product-rule

$$(n,q_1,\sigma_1).(m,q_2,\sigma_2) = (n.m,\sigma_1(q_2)^n.q_1,\sigma_1 \circ \sigma_2)$$

and action $(n,q,\sigma).I = \sigma(I)^n.q^{-1}$. Again, morphisms in $C(m)$ are induced by combining action with poset-maps in both layers.

The second flavour of maps (the 'etale-like' in M-parlance) are substitutes for the cover-maps between the two layers. For every integer $n$ and every $q \in \mathbb{Q}(\sqrt{m})$ we have a map between a fractional ideal $I \in Div(\mathbb{Z}[\sqrt{m}])$ (the upper layer) and a rational number $r \in \mathbb{Q}_{>0}^*$ (the bottom layer) whenever

$$q.(\mathbb{Z}[\sqrt{m}].r).I^{-n}~\text{is an ideal of}~\mathbb{Z}[\sqrt{m}]$$

Again, one shows as last time that this actually is a category, meaning that compositions of maps are well-defined.

Problem for light insomniacs : show that every morphism in $C(m)$ is an epimorphism (that is, if the compositions with two maps are equal, then these two maps have to be equal).

Problem for incurable insomniacs : show that if there is an equivalence of categories between $C(m)$ and $C(n)$, then $\mathbb{Q}(\sqrt{m}) = \mathbb{Q}(\sqrt{n})$.