MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
Did you mean Tor_i = 0 for i > 0? – Jason Polak Apr 25 '12 at 14:22
Is R supposed to be $\mathbb{Q}$ on the second line? – Sean Tilson Apr 25 '12 at 14:31
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
up vote 8 down vote accepted

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 ${\mathbb Z}$.

share|cite|improve this answer
This is perfect! – Jeff Strom Apr 26 '12 at 14:06

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.