Groupification and perfection of a commutative monoid


(1) If $M \to M^{pf}$ is not injective, there exist $x \neq y$ and $n\in \mathbb{N}_{\geq 1}$ such that $f_{n1}(x) = f_{n1}(y)$. (The fact that elements are equal in the colimit iff they are equal in some term requires some thought. It is not a general property; it follows here from the structure of the diagram defining the colimit.) So let us assume that $x,y\in M$ satisfy $nx = ny$.

Let $Z = [x] - [y]$ in $M^{gp}$. We have $nZ = [nx] - [ny] = 0$, which lies in the image $\psi(M)$, so saturation implies that $Z \in \psi(M)$. Say $Z=\psi(z)$ for $z\in M$. Since $Z$ satisfies $Z + [y] = [x]$, integrality implies that $z + y = x$. The element $z$ satisfies $\psi(nz)=nZ=0\in M^{gp}$, so integrality implies $nz = 0\in M$. Since $M$ is torsion-free this implies $z = 0$, so $z+y=x$ becomes $y = x$. This shows that $M\to M^{pf}$ is injective.

(2) Since $M^{pf}$ is perfect (an inverse to $x\mapsto nx$ is the map induced by sending $x\in M_a$ to $x\in M_{na}$), an isomorphism $M\to M^{pf}$ implies that $M$ is perfect. Conversely, if $M$ is perfect, we can change coordinates on $M_a$ by $x \mapsto \frac{1}{n}x$ (the inverse of $x\mapsto nx$, which is automatically a morphism). Now the direct system is constant and so $M\to M_1\to M^{pf}$ is an isomorphism.


Related videos on Youtube

Makoto Kato
Author by

Makoto Kato

Updated on August 01, 2022


  • Makoto Kato
    Makoto Kato over 1 year

    The following definitions and proposition are taken from the paper The geometry of Frobenioids I by S. Mochizuki.

    Let $\mathbb{N}_{\ge 1}$ be the set of positive integers. $\mathbb{N}_{\ge 1}$ is a directed set by the order relation $a|b$. Let $M$, $N$ be commutative monoids(the binary operations are written additively). A map $f\colon M \rightarrow N$ is called a morphism if $f(x + y) = f(x) + f(y)$ for any $x, y$ and $f(0) = 0$.

    For every $n \in \mathbb{N}_{\ge 1}$, the multiplication map $x \rightarrow nx$ on $M$ is a morphism.

    $M$ is said to be torsion-free if $nx = 0$ implies $x = 0$ for every $n \in \mathbb{N}_{\ge 1}$.

    Let $M^{gp}$ be the groupification of $M$, i.e. the grothendieck group of $M$. Let $\psi\colon M \rightarrow M^{gp}$ be the canonical map. If $\psi$ is injective, $M$ is said to be integral.

    $M$ is said to be saturated if $na \in \psi(M)$ for some $n \in \mathbb{N}_{\ge 1}$ and $a \in M^{gp}$, then $a \in \psi(M)$.

    $M$ is said to be perfect if the multiplication map $x \rightarrow nx$ is bijective for every $n \in \mathbb{N}_{\ge 1}$.

    The perfection $M^{pf}$ of $M$ is defined as follows. Let $M_a = M$ for every $a \in \mathbb{N}_{\ge 1}$. if $a|b$, we define a morphism $f_{ba}\colon M_a \rightarrow M_b$ by $f_{ba}(x) = (b/a)x$. Then $(M_a)_{a\in \mathbb{N}_{\ge 1}}$ and $(f_{ba})$ consitute a direct(or inductive) system. We define $M^{pf} = colim_{a\in \mathbb{N}_{\ge 1}} M_a$. There exists the canonical morphism $\phi\colon M = M_1 \rightarrow M^{pf}$.

    My question: How do we prove the following proposition?


    (1) $\phi\colon M \rightarrow M^{pf}$ is injective if $M$ is torsion-free, integral, and saturated.

    (2) $M$ is perfect if and only if $\phi$ is an isomorhism.

    Motivation Recently(August, 2012), S. Mochizuki submitted a series of papers(Inter-universal Teichmuller Theory I,II,III,IV) which develops his new theory. As an application of his theory, he wrote a proof of ABC conjecture. I think the validity of the proof has not yet been confirmed by other mathematicians. However, considering his track record, I think it's worthwhile to read the papers. He referred to The geometry of Frobenioids I for the notation and terminolgy concerning monoids and categories.

    • Makoto Kato
      Makoto Kato almost 10 years
      @user18921 I'm afraid I don't understand what you mean. Could you rephrase it?
  • Berci
    Berci about 11 years
    I got stucked, too.. sorry, Turn back on that later.
  • Makoto Kato
    Makoto Kato about 11 years
    I think (1) is correct. However I don't understand (2) fully. Could you elaborate on it?
  • Tom Church
    Tom Church about 11 years
    If $M$ is perfect, then each map $f_{ba}\colon M_a\to M_b$ is an isomorphism of monoids. Therefore the colimit is boring: it is isomorphic to $M_1$. (To see this, check that for any $a\in \mathbb{N}_{\geq 1}$ and any $x\in M_a$, there exists some $y\in M_1$ so that $f_{a1}(y)=x$. This is worth working out.)