Solid rings and Tor

Question

A solid ring is a ring $R$ such that the multiplication $R\otimes_{\mathbb{Z}} R \to R$ is an isomorphism.
These were classified by Bousfield and Kan; they are

subrings of $\mathbb{Q}$,
$\mathbb{Z}/n$,
products $R\times \mathbb{Z}/n$ with $R\subseteq \mathbb{Q}$ and every divisor of $n$ invertible in $R$
colimits of these.

I wonder how small the list gets if I put the additional constraint that $\mathrm{Tor}_{\mathbb{Z}}(R,R) = 0$.

REFERENCE: Bousfield, A. K.; Kan, D. M. The core of a ring. J. Pure Appl. Algebra 2 (1972), 73–81.

It seems R must be {\mathbb Q}. Also, I think you must mean colimits, not limits. — Steven Landsburg, Apr 25 '12 at 14:44
Your summary of Bousfield and Kan's results is inaccurate in a number of ways. You should probably start by reviewing their paper. I think it works out that the only solid rings with $\text{Tor}_{\mathbb{Z}}(R,R)=1$ are the localisations $\mathbb{Z}[J^{-1}]$ (for any set of primes $J$). — Neil Strickland, Apr 25 '12 at 14:55
I apologize for the mangling of the classification of solid rings; fixed now, I think. — Jeff Strom, Apr 26 '12 at 14:06

Steven Landsburg · Accepted Answer · 2012-04-25T20:55:57.227

Let $R^t$ be the torsion submodule and consider the exact sequence

$$0\rightarrow R^t\rightarrow R \rightarrow R/R^t\rightarrow 0$$

Bousfield and Kan show that the ring on the right is a localization of ${\mathbb Z}$, hence flat over ${\mathbb Z}$, so its $Tor$ with $R$ vanishes. Thus if we $Tor$ the above with $R$, we get $Tor(R^t,R)=Tor(R,R)$.

Now tensor the exact sequence with $R^t$ instead of $R$. This gives $Tor(R^t,R^t)=Tor(R^t,R)$.

Thus $Tor(R,R)=Tor(R^t,R^t)$. But if $R^t$ is nonzero then (see Bousfield and Kan) it contains some ${\mathbb Z}/p{\mathbb Z}$ as a direct summand and hence $Tor(R^t,R^t)$ does not vanish. Thus $Tor(R,R)=0$ implies $R^t=0$. It follows (B/K 3.7) that $R$ is a localization of ${\mathbb Z}$.

Solid rings and Tor

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