MinuteMochizuki 3

Surely there must be a more elegant formulation than that of Mochizuki of an arithemetic Frobenioid, some commenters on google+ pointed out this week. Here, a feeble attempt on my part.

An arithmetic Frobenioid is a category associated to a Galois extension $L$ of the rational numbers $\mathbb{Q}$ and consists of two parts. First, one associates a 'layer of objects' $Frob(K)$ to every subfield $\mathbb{Q} \subset K \subset L$ and next one links these different layers by 'pull-back' or Galois morphisms.

Let $R$ be the ring of integers in $K$ (that is, $R$ is the integral closure of $\mathbb{Z}$ in $K$), then $R$ is a Dedekind domain implying that all non-zero ideals of $R$ are maximal and that any non-zero ideal $I$ of $R$ can be written uniquely as a product of prime-ideals

$$I=P_{i_1}^{e_1}.P_{i_2}^{e_2}. \cdots . P_{i_k}^{e_k}$$

and therefore the non-zero ideals of $R$ form an abelian monoid

$$Div(R) = \bigoplus_{P \in \mathbf{Spec}(R)} \mathbb{N}$$

the ideal $I$ above corresponding the the element having $e_l$ in the factors determined by $P_{i_l}$ and zeroes elsewhere. The group generated by it is the divisor group $Div(K)$ and consists of all fractional ideals of $K$. Such a fractional ideal has again a unique decomposition in powers of prime ideals, this time allowing for negative exponents where one defines for a prime ideal $P \in \mathbf{Spec}(R)$

$$P^{-1} = \{ k \in K~|~P.k \subset R \}$$

There's a natural equivalence relation on fractional ideals by declaring $I \sim J$ if and only if there is a $k \in K$ such that $I.k=J$.

To some category theorists, sets with an equivalence relation are easy examples of groupoids which are very special categories in which every morphism is an isomorphism. The objects are the elements of the set (in our case the fractional ideals) and there is exactly one isomorphism between any pair of elements which are equivalent to each other.

These isomorphisms are precisely the isomorphisms in the layer $Frob(K)$. From number theory one gets that the number of isomorphism classes (or connected components in the groupoid) is finite and that the classes are labeled by the elements of a finite Abelian group $Cl(R)$ which is called the class group of $R$. This group is also the quotient monoid

$$Cl(R) \simeq Div(R)/Prin(R)$$

where $Prin(R)$ is the sub-monoid of all principal ideals $Rr$ of $R$.

The remaining morphisms in $Frob(R)$ are given by the action-map of a certain noncommutative monoid on $Div(K)$ (or on its groupoid). This monoid is generated by the operations

$[p](I) = I^p$ for any prime number $p$, and

$P(I) = P.I$ for any non-zero prime ideal $P$ of $R$.

One easily verifies that the only relations between these are (for prime numbers $p$ and $q$ and prime ideals $P$ and $Q$)

$[p] \circ [q] = [q] \circ [p]$, $P \circ Q = Q \circ P$ and $[p] \circ Q = \underbrace{Q \circ Q \circ \cdots \circ Q}_p \circ [p]$

and so the acting monoid is the twisted monoid $\mathbb{N}^{\times}_{>0} \ast Div(R)$. These morphisms turn $Frob(K)$ into a category and it is not that difficult to see that the endomorphism-monoid of any element $I \in Div(K)$ is isomorphic (as abstract monoid) to the twisted monoid $\mathbb{N}^{\times}_{>0} \ast Prin(R)$.

The remaining part in the definition of the arithmetic Frobenioid are the morphisms between the different layers. So, assume $K \subset K' \subset L$ and let $\sigma$ be a $\mathbb{Q}$-automorphism of $K'$ fixing $K$. then there is a ringmorphism between the rings of integers $\sigma~:~R \rightarrow R'$.

If $I \in Div(K)$ there is for each $\sigma$ a pull-back morphism

$\sigma~:~R'.\phi(I) \rightarrow I$

and one verifies that composition of pull-backs are again a pull-back morphism, so that we have indeed associated a category to any Galois extension $L/\mathbb{Q}$ : the arithmetic Frobenioid of $L$.

Next time we will try to reconstruct $L$ from the category...