Is there a lot of ring spectrum which are idempotent in the sense that the multiplication map $R \wedge R \rightarrow R$ is an equivalence ?


The sphere spectrum $\mathbb{S}$ and the $0$ spectrum are clearly examples.


I believe that $\mathbb{S}[n^{-1}]$ i.e. the ring spectrum obtained by starting from $\mathbb{S}$ and formally inverting multiplication by a sets of integer, by taking the inductive limits of $\mathbb{S} \overset{\times n}{\rightarrow} \mathbb{S} \overset{\times n}{\rightarrow} \mathbb{S} \dots$ gives us an other example. And one can generalize by localizing at a set of integer.


Is there other type of such example ? Can we classify them ?


In case there is indeed more examples and they cannot be classified, I have an additional question:


Is there any example other than $\mathbb{S}$ and $0$ which are connective and such that the fiber of the unit map $\mathbb{S} \rightarrow R$ is also connective (which if I'm correct amount to saying that $R$ is connective and the map $\mathbb{Z} = \pi\_0(\mathbb{S}) \rightarrow \pi\_0(R)$ is surjective, and it rules out all the localization mentioned above).



Idempotent ring spectrum
There are actually quite a number of other examples. In particular, this is necessary and sufficient for $R$ to be a so-called *smashing localization* of the sphere $\Bbb S$, and there are several prominent examples called the $E(n)$-local spheres as we range over primes $p$ and natural numbers $n > 0$.


However, there are not many examples when $R$ is connective: all of them are obtained from the sphere spectrum $\Bbb S$ by inverting some set of primes. Here is a proof.


1. The left unit map $R = R \wedge \Bbb S \to R \wedge R$ is an equivalence, and so is the right unit map.
2. Suppose $E$ is a homology theory with a perfect Kunneth formula:
$$E\_\*(X \wedge Y) \cong E\_\*(X) \otimes\_{E\_\*} E\_\*(Y).$$
Then the maps
$$
E\_\* (R) \to E\_\*(R) \otimes\_{E\_\*} E\_\*(R) \to E\_\*(R)
$$
induced by the left unit and the multiplication are isomorphisms.
3. In particular, this is true for homology with coefficients in any field $\Bbb F$. Therefore, $H\_\*(R; \Bbb F)$ is an $\Bbb F$-algebra $A$ such that the left unit $A \to A \otimes\_{\Bbb F} A$ is an isomorphism. By choosing a basis, we can conclude that either the unit $\Bbb F \to A$ is an isomorphism or $A = 0$. This gives us a set $S$ of primes such that the mod-$p$ homology of $R$ is trivial, and in the other cases the map $\Bbb S \to R$ is a mod-$p$ homology isomorphism.
4. We can then use the universal coefficient theorem relating integral homology with mod-p homology. For any prime in $S$, both $H\_\*(R;\Bbb Z) \otimes \Bbb Z/p$ and $Tor(H\_\*(R;\Bbb Z), \Bbb Z/p)$ are trivial, so multiplication-by-$p$ is an isomorphism on $H\_\*(R;\Bbb Z)$.
5. Similarly, for primes *not* in $S$, the map $\Bbb S \to R$ being a mod-$p$ homology isomorphism and naturality of the universal coefficient theorem imply that $Tor(H\_\*(R;\Bbb Z), \Bbb Z/p) = 0$. Thus $H\_\*(R; \Bbb Z)$ is torsion-free, hence flat. This means $H\_\*(R; \Bbb Z)$ is a subring of the rational homology.
6. Now applied to the rational homology, we find that $H\_\*(R; \Bbb Q)$ is either trivial or $\Bbb Q$, so $H\_\*(R; \Bbb Z)$ is either $0$ or a subring of $\Bbb Q$. In the latter case, this forces $H\_\*(R;\Bbb Z)$ to be the localization of $\Bbb Z$ at the set of primes in $S$.
7. If we finally assume that $R$ is connective, then $\pi\_0(R) \cong H\_0(R;\Bbb Z) = \Bbb Z[S^{-1}]$ by the Hurewicz theorem. This means that we get a factorization $\Bbb S \to \Bbb S[S^{-1}] \to R$ which is an isomorphism on integral homology. Since both sides are connective, by the Whitehead theorem we find that $\Bbb S[S^{-1}] \to R$ is a weak equivalence.


