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[Skip down to the bottom for a correction] Let's work over a field k, assume it is as nice as you need it to be.. Suppose I have an ordinary (edit: commutative) affine group scheme G = Spec(A) over k, I may produce from G the following objects:

1.) A simplicial ring k(A,n) having homotopy groups equal to A in degree n and k elsewhere with all (co)degeneracy maps being k-linear. Probably the most obvious choice for a notion of an Eilenberg-MacLane space. Let Spec(k(A,n)) be the associated derived scheme.

2.) We can also associate to X the (iterated...) classifying stack K(G,n). I'm not fully sure how such a thing is constructed for n>1. (is this even a remotely nice object? or are we talking (n,infinity)-cats here).

To what extent are these objects known to be 'the same'. To be concrete, let n = 1 and let's focus on the following case

1'.) k(A,1) is a simplicial ring with non-trivial homotopy concentrated in degree 1. You can very directly construct it by starting with the constant simplicial ring associated to k, then adding and killing off generators in higher degree.

2'.) K(G,1)(X) is of course, the classifying stack of principle G-bundles on a scheme X. In any case, that's what I wish K(G,1) is... (though, morally, I should probably be looking at the classifying space of principle G-bundles on the derived scheme Spec(A*) where A* is the constant simplicial ring associated to A.

Second edit: Qiaochu Yuan's comment below shows that the first construction above does not make sense as stated. I guess what I really want to know is this: Is there a derived scheme playing the role of the n-th Eilenberg MacLane space to the group scheme associated to the multiplicative group of a field and does it homtopically classify the objects one hopes it does (n = 0 characters on the scheme, n = 1 line bundles, n =2 azumaya algebras maybe??).

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  • $\begingroup$ But what is $X$? $\endgroup$ Mar 7, 2015 at 19:57
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    $\begingroup$ X is just G, sorry. Corrected that. I guess maybe I'm misunderstanding something fundamental? Shouldn't K(A,0) be the constant simplicial ring on A. $\endgroup$
    – ahar
    Mar 7, 2015 at 20:13
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    $\begingroup$ You're right, I tend to shift dimensions by $+1$ in my mind by how the equivalence between simplicial sets and spaces works. $\endgroup$ Mar 7, 2015 at 20:18
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    $\begingroup$ I don't understand your first construction. What is the multiplication on the simplicial ring when $n \neq 0$? I think you're trying to deloop on the wrong side, as it were; whatever a simplicial model for $B^n G$ looks like, it should crucially involve the multiplication on $G$, or equivalently the comultiplication on $A$, which it doesn't look like your first construction does. Also, for $n \ge 2$ you need $A$ to be commutative. $\endgroup$ Mar 7, 2015 at 20:25
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    $\begingroup$ To be more specific, in the first construction you're trying to deloop the underlying abelian group of $A$, which is not what you want. You should be trying to write down a simplicial scheme (essentially the bar construction), not a simplicial ring. $\endgroup$ Mar 7, 2015 at 20:40

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