Let S\mathbb {S} be the sphere spectrum. Then S\mathbb {S} is, in particular, a commutative ring spectrum (i.e. a commutative monoid in the stable homotopy category), using the canonical identification SSS\mathbb {S}\wedge \mathbb {S}\overset {\sim }{\to } \mathbb {S}. Let pp be a prime. We seek the answer the following question:

Question

When is S/p\mathbb {S}/p a ring spectrum?

To turn S/p\mathbb {S}/p into a ring spectrum, we have to solve the extension problem

\begin{useimager} 
\[
  \begin{tikzcd}
    \S \wedge \S \ar[r, "\sim"] \ar[d] & \S \ar[d]\\
    \S/p \wedge \S/p \ar[r, dashed] & \S/p
  \end{tikzcd}
\]
\end{useimager}

and then check associativity of the multiplication map (unitality is clear).

To solve the extension problem, we factor the left vertical map as SSSS/pS/pS/p\mathbb {S}\wedge \mathbb {S}\to \mathbb {S}\wedge \mathbb {S}/p \to \mathbb {S}/p \wedge \mathbb {S}/p, and observe that there is an extension to SS/p\mathbb {S}\wedge \mathbb {S}/p given essentially by the identity map. Thus we have to solve

\begin{useimager} 
\[
  \begin{tikzcd}
    \S \wedge \S/p \ar[d, "p"] \\
    \S \wedge \S/p \ar[r, "\sim"] \ar[d] & \S/p\\
    \S/p \wedge \S/p \ar[ur, dashed]
  \end{tikzcd}
\]
\end{useimager}

Here the vertical maps form a cofibration sequence, so solving the lifting problem is equivalent to showing that p:S/pS/pp: \mathbb {S}/p \to \mathbb {S}/p is zero, or equivalently, that [S/p,S/p]=Z/pZ[\mathbb {S}/p, \mathbb {S}/p] = \mathbb {Z}/p\mathbb {Z}. This can be calculated via a sequence of homotopy long exact sequence calculations using SSS/p\mathbb {S}\to \mathbb {S}\to \mathbb {S}/p, or we can do it more systematically via the Adams spectral sequence.

Since S/p\mathbb {S}/p is the cofiber of p:SSp: \mathbb {S}\to \mathbb {S}, we can compute its (co)homology using the cellular chain complex. We find that

Lemma

HFp(S/p)H\mathbb {F}_ p^*(\mathbb {S}/p) is Fp\mathbb {F}_ p in degrees 00 and 11, and vanishes otherwise. The Bockstein homomorphism acts non-trivially between the degrees.

Dualizing, we get

Corollary

As a Steenrod comodule, we have

(HFp)(S/p)={E(ξ1)p=2E(τ0)p>2. (H\mathbb {F}_ p)_*(\mathbb {S}/p) = \begin{cases} E(\xi _1) & p = 2\\ E(\tau _0) & p > 2 \end{cases}.\fakeqed

Note that when p=2p = 2, the comodule E(ξ1)E(\xi _1) is not a subalgebra of the Steenrod algebra, where ξ1\xi _1 is not nilpotent. This immediately lets us conclude

Proposition

E(ξ1)E(\xi _1) does not have the structure of a comodule algebra. Hence S/2\mathbb {S}/2 does not admit a ring structure.

Proof
Write ψ\psi for the comodule action. If E(ξ1)E(\xi _1) had an algebra structure, then ψ(ξ12)=ψ(ξ1)2=ξ121+1ξ12. \psi (\xi _1^2) = \psi (\xi _1)^2 = \xi _1^2 \otimes 1 + 1 \otimes \xi _1^2. However, the image of ψ\psi in AE(ξ1)\mathcal{A} \otimes E(\xi _1) is given by linear combinations of 111 \otimes 1 and ξ11+1ξ1\xi _1 \otimes 1 + 1 \otimes \xi _1.

Alternatively, we can compute π2(S/2)\pi _2(\mathbb {S}/2) using the Adams spectral sequence and show that it has order 44 elements, hence 2:S/2S/22: \mathbb {S}/2 \to \mathbb {S}/2 is non-zero.

For p>3p > 3, we shall show that the spectrum S/p\mathbb {S}/p does admit a commutative ring spectrum structure.

Noting that S/p\mathbb {S}/p is already pp-local, the Adams spectral sequence gives us

E2s,t=ExtAs,t(E(τ0),E(τ0))[ΣtsS/p,S/p]. E_2^{s, t} = \operatorname{Ext}^{s, t}_{\mathcal{A}_*} (E(\tau _0), E(\tau _0)) \Rightarrow [\Sigma ^{t - s} \mathbb {S}/p, \mathbb {S}/p].

By the change of rings theorem, we can write

E2s,t=ExtA//E(τ0)s,t(E(τ0),F2)=ExtA//E(τ0)s,t(F2,F2)F2{1,x} E_2^{s, t} = \operatorname{Ext}^{s, t}_{\mathcal{A}_*/\! \! /E(\tau _0)} (E(\tau _0), \mathbb {F}_2) = \operatorname{Ext}^{s, t}_{\mathcal{A}_*/\! \! /E(\tau _0)} (\mathbb {F}_2, \mathbb {F}_2) \otimes \mathbb {F}_2\{ 1, x\}

where xE20,1x \in E_2^{0, -1}, since E(τ0)E(\tau _0) is a trivial A//E(τ0)\mathcal{A}_*/\! \! /E(\tau _0) comodule.

Now the terms of lowest degree in A//E(τ0)\overline{\mathcal{A}_*/\! \! /E(\tau _0)} are generated by ξ1\xi _1 and have degree 2p22p - 2. So in the cobar complex, we see that apart from F2{1,x}\mathbb {F}_2\{ 1, x\} , all terms have ts2p3t - s \geq 2p - 3. In particular,

Lemma

[S/p,S/p]=Z/pZ[\mathbb {S}/p, \mathbb {S}/p] = \mathbb {Z}/p\mathbb {Z}.

This tells us we can solve our original extension problem. We can actually go further and understand the set of possible lifts. This amounts to understanding the kernel of the map

[S/pS/p,S/p][SS,S/p]. [\mathbb {S}/p\wedge \mathbb {S}/p, \mathbb {S}/p] \to [\mathbb {S}\wedge \mathbb {S}, \mathbb {S}/p].

This is not difficult because we can use the same technique to compute these groups explicitly. Using Künneth's formula and the same calculation, we find that

Lemma

Let p>3p > 3. Then [S/pk,S/p]=Z/pZ[\mathbb {S}/p^{\wedge k}, \mathbb {S}/p] = \mathbb {Z}/p\mathbb {Z} for k=1,2,3k = 1, 2, 3. More precisely, the maps

[S/pk,S/p][Sk,S/p] [\mathbb {S}/p^{\wedge k}, \mathbb {S}/p] \to [\mathbb {S}^{\wedge k}, \mathbb {S}/p]

are bijections for k=1,2,3k = 1, 2, 3.

Proof
We can directly calculate the value of [S/pk,S/p][\mathbb {S}/p^{\wedge k}, \mathbb {S}/p]. To see that the map we wrote down in particular is a bijection, we note that both sides are Z/pZ\mathbb {Z}/p\mathbb {Z}, and the map is non-zero since the unit map SS/p\mathbb {S}\to \mathbb {S}/p is in the image.

In particular, the case of k=2k = 2 tells us there is a unique choice of μ:S/pS/pS/p\mu : \mathbb {S}/p \wedge \mathbb {S}/p \to \mathbb {S}/p up to homotopy, and in particular, since the map SSS/p\mathbb {S}\wedge \mathbb {S}\to \mathbb {S}/p is symmetric in the S\mathbb {S}'s, the same is true of μ\mu . Similarly, the k=3k = 3 case tells us the multiplication is associative, since that is true for S\mathbb {S}.

Answer
  • S/2\mathbb {S}/2 does not admit the structure of a ring spectrum.

  • S/3\mathbb {S}/3 has a multiplication but is not (necessarily) associative.

  • S/p\mathbb {S}/p is a (homotopy) commutative ring spectrum for p>3p > 3.