## Solid Rings and Tor

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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

1. subrings of $R\subseteq\mathbb{Q}$,

2. $\mathbb{Z}/n$,

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

4. 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.

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Did you mean Tor_i = 0 for i > 0? – Jason Polak Apr 25 2012 at 14:22
Is R supposed to be $\mathbb{Q}$ on the second line? – Sean Tilson Apr 25 2012 at 14:31

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It seems R must be {\mathbb Q}. Also, I think you must mean colimits, not limits. – Steven Landsburg Apr 25 2012 at 14:44

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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 2012 at 14:55
I apologize for the mangling of the classification of solid rings; fixed now, I think. – Jeff Strom Apr 26 2012 at 14:06

## 1 Answer

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 ${\mathbb Z}$.

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This is perfect! – Jeff Strom Apr 26 2012 at 14:06
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