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

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

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

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

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

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

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

$H\mathbb {F}_p^*(\mathbb {S}/p)$ is $\mathbb {F}_p$ in degrees $0$ and $1$, and vanishes otherwise. The Bockstein homomorphism acts non-trivially between the degrees.

Dualizing, we get

As a Steenrod comodule, we have

$(H\mathbb {F}_p)_*(\mathbb {S}/p) = \begin{cases} E(\xi _1) & p = 2\\ E(\tau _0) & p > 2 \end{cases}.$$E(\xi _1)$ does not have the structure of a comodule algebra. Hence $\mathbb {S}/2$ does not admit a ring structure.

However, the image of $\psi$ in $\mathcal{A} \otimes E(\xi _1)$ is given by linear combinations of $1 \otimes 1$ and $\xi _1 \otimes 1 + 1 \otimes \xi _1$.

Alternatively, we can compute $\pi _2(\mathbb {S}/2)$ using the Adams spectral sequence and show that it has order $4$ elements, hence $2: \mathbb {S}/2 \to \mathbb {S}/2$ is non-zero.

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

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

$E_2^{s, t} = \operatorname{Ext}^{s, t}_{\mathcal{A}_*} (E(\tau _0), E(\tau _0)) \Rightarrow [\Sigma ^{t - s} \mathbb {S}/p, \mathbb {S}/p].$By the change of rings theorem, we can write

$E_2^{s, t} = \operatorname{Ext}^{s, t}_{\mathcal{A}_*/\! \! /E(\tau _0)} (E(\tau _0), \mathbb {F}_2) = \operatorname{Ext}^{s, t}_{\mathcal{A}_*/\! \! /E(\tau _0)} (\mathbb {F}_2, \mathbb {F}_2) \otimes \mathbb {F}_2\{ 1, x\}$where $x \in E_2^{0, -1}$, since $E(\tau _0)$ is a trivial $\mathcal{A}_*/\! \! /E(\tau _0)$ comodule.

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

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

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

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

$[\mathbb {S}/p^{\wedge k}, \mathbb {S}/p] \to [\mathbb {S}^{\wedge k}, \mathbb {S}/p]$are bijections for $k = 1, 2, 3$.

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

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

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