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Given a simplicial commutative ring $A$ and a simplicial set $K$ does $K \otimes A$ make sense as a (commutative) simplicial ring?

I'm asking as I've seen the expression $S^n \otimes A$ written down and (say A is just an ordinary (commutative) ring seen as a constant simplicial ring) this should correspond the nth suspension of A (-nth shift if you view A as a cdga).

This seems counterintuitive to me as simplicial $S^n$ certainly has complicated homotopy groups while $S^n \otimes \mathbb{Z}$ should have a fairly simple homotopy algebra (it's almost as if the latter $S^n$ should be interpreted as a homological $S^n$)

EDIT: OK, I still don't understand why $S^n \otimes A$ should correspond to a shift. I tried an example, take simplicial $S^1$ and a simplicial abelian group (or $\mathbb{Z}$-module) M and consider $S^1 \otimes M$. If it corresponded to a shift then under the Dold-Kan the complex we obtain should be concentrated in one degree.

But the complex $N(S^1 \otimes M)$ has $M \oplus M \oplus M$ in degrees one and zero and vanishes elsewhere. The matrix which gives the differential is $(-1,1,0), (-1,0,1), (0,-1,1)$ (these are the columns), which has both kernel and cokernel. Something went wrong there, is the problem that it's a simplicial module and not an algebra? But it should work with algebras as well no?

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    ah, thanks for the tip2012-02-17

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