Do affine schemes form a Mal'cev category? - MathOverflow

Do affine schemes form a Mal'cev category?

2011-01-10T02:29:57Z

This may be a silly question, but I have no intuition in this direction. Every category internal to a Mal'cev category is a groupoid (this is why categories internal to $Grp$ are groupoids). If this was true it would put restrictions on generalising algebraic stacks (which are stacks of groupoids by default) to algebraic stacks of categories.

So my question is:

Is the opposite of the category of commutative rings (= category of affine schemes) Mal'cev?

Answer by Greg Muller for Do affine schemes form a Mal'cev category?

2011-01-10T14:15:06Z

(This is a repeat of an above comment.) The category of affine schemes is not Mal'cev. This can be disproven by producing an reflexive, non-symmetric relation on an affine scheme $X$ whose graph is a closed subscheme of $X\times X$.

Take $X=\mathbb{A}^1=Spec(\mathbb{C}[x])$. The relation $(x,x)$ and $(x,0)$ (as $x$ runs over all points) is reflexive and asymmetric. If we identify $X\times X$ with $$Spec(\mathbb{C}[x]\otimes \mathbb{C}[y])=Spec(\mathbb{C}[x,y]),$$ then the graph of the quotient is the union of the lines $x=y$ and $y=0$. This is a closed subscheme. In terms of rings, this looks like the quotient $\mathbb{C}[x,y]\rightarrow \mathbb{C}[x,y]/(xy-y^2)$.

There's also a simpler and less-satisfying example. Since the category of affine schemes contains the category of finite sets (with $[n]\rightarrow Spec(\mathbb{C}^n)$, one can choose any reflexive, non-symmetric relation on a finite set (since they are all algebraic).

Answer by Dmitry Roytenberg for Do affine schemes form a Mal'cev category?

2011-10-28T08:52:23Z

Dear David,

There's also a less direct way to see that the category of affine schemes is not Mal'cev, but one that is more in line with your motivation -- namely, by exhibiting an internal category which is not a groupoid: for instance, take the multiplicative monoid. As a functor from commutative rings to monoids, it just forgets addition, and is co-represented by the ring $\mathbb{Z}[t]$, with co-multiplication sending $t$ to $xy\in\mathbb{Z}[x,y]$ and the co-unit given by evaluating at $t=1$.