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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?

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    $\begingroup$ What if I take an algebraic reflexive relation which is not an eq. relation on the affine line? For instance, the relation (x,x) and (x,0) is algebraic, and the corresponding quotient algebra of $\mathbb{C}[x,y]$ is $\mathbb{C}[x,y]/(xy-y^2)$. $\endgroup$ Jan 10, 2011 at 2:53
  • $\begingroup$ That would be a counterexample to the claim that Aff is Mal'cev. I'm not sure why you're taking a quotient of $\mathbb{C}[x,y]$, though. $\endgroup$
    – David Roberts
    Jan 10, 2011 at 3:36
  • $\begingroup$ Quotients of $\mathbb{C}[x,y]$ are closed subschemes of $A^1\times A^1$, where $A^1$ is the complex affine line. The quotient I gave corresponds to the relation I mention. $\endgroup$ Jan 10, 2011 at 4:10
  • $\begingroup$ Ah, I see. Thanks. Would you like to post this as an answer? $\endgroup$
    – David Roberts
    Jan 10, 2011 at 4:36

2 Answers 2

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(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).

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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 $t=1$.

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