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Let $S$ be a scheme, $M\to S$ a commutative monoid object in algebraic $S$-spaces, ie. an algebraic $S$-space such that, functorially on $S$-schemes $T$, $M(T)$ is a commutative monoid with neutral element.

Each commutative monoid $M(T)$ has its own group completion $M(T)^{\rm gp}$, given by the classical Grothendieck group construction.

Formation of $M(T)^{\rm gp}$ is functorial in $T$, and we may define the fppf sheaf $M^{\rm gp}$ to be the fppf sheafification of $T\mapsto M(T)^{\rm gp}$.

Is $M^{\rm gp}$ an algebraic $S$-space, hence a commutative group object in algebraic $S$-spaces? Is it at least when $M$ is cancellative?

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Example

Suppose $M$ is, in addition, cancellative (ie. $M(T)$ is a cancellative commutative monoid with zero for all $S$-schemes $T$).

In this case, $M^{\rm gp}$ is the fppf quotient sheaf of $M\times_SM\times_SM\times_SM$ by the following equivalence relation:

$$R := (M\times_SM\times_SM\times_SM)\times_{\mu, M\times_SM,\Delta_{M/S}}M$$

with:

  • $\mu : M\times_SM\times_SM\times_SM\to M\times_SM$ the map defined functorially on $T$-sections by sending $(a_T, b_T, c_T, d_T)$ to $(a_T+d_T, b_T+c_T)$.
  • $\Delta_{M/S} : M\to M\times_SM$ is the diagonal.
  • $s,t : R\to M\times_SM$ are defined to be the pullback along $\mu$ of $\text{pr}_1\circ\Delta_{M/S}$ and $\text{pr}_2\circ\Delta_{M/S}$.

On $T$-sections, $R(T)$ is the equivalence relation that identifies the pairs $((a,b),(c,d))$ with $((a,d),(b,c))$ treating $(a,b)$ and $(c,d)$ as $"a-b"$ and $"c-d"$, and ensuring that $"a-b = c-d"$ if and only if $"a+d = b+c"$. This way, the monoid operation $+$ on $M(T)\times M(T)$ descends to a group operation, where the inverse of $(a,0)$ is $(0,a)$.

$R$ is not an étale equivalence relation, unless $M$ is étale over $S$, as the fiber of $M\times_SM\to M^{\rm gp}$ over $0$ is the diagonal copy of $M$ in $M\times_SM$.

This is always going to be the case, so either the question has negative answer, or one should come up with a better presentation for $M^{\rm gp}$, or verify Artin's axioms.

The case of interest is when $M\to S$ is locally of finite presentation and separated, $S$ is affine.

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    $\begingroup$ The fiber of $R\to M\times_SM$ over $(0,0)$ is $M$, so it's not an étale equivalence relation when $M$ is not étale over $S$. $\endgroup$ Commented Feb 10, 2018 at 23:30
  • $\begingroup$ Yes, fair enough. It may well be possible there's a better presentation, though. Or one may try to verify Artin's axioms, which I haven't tried yet, to settle the general question. $\endgroup$
    – user113393
    Commented Feb 10, 2018 at 23:34

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