Let be the sphere spectrum. Then is, in particular, a commutative ring spectrum (i.e. a commutative monoid in the stable homotopy category), using the canonical identification . Let be a prime. We seek the answer the following question:
When is a ring spectrum?
To turn 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 , and observe that there is an extension to 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 is zero, or equivalently, that . This can be calculated via a sequence of homotopy long exact sequence calculations using , or we can do it more systematically via the Adams spectral sequence.
Since is the cofiber of , we can compute its (co)homology using the cellular chain complex. We find that
is in degrees and , and vanishes otherwise. The Bockstein homomorphism acts non-trivially between the degrees.
Dualizing, we get
As a Steenrod comodule, we have
Note that when
, the comodule
is not a subalgebra of the Steenrod algebra, where
is not nilpotent. This immediately lets us conclude
does not have the structure of a comodule algebra. Hence does not admit a ring structure.
for the comodule action. If
had an algebra structure, then
However, the image of
is given by linear combinations of
Alternatively, we can compute using the Adams spectral sequence and show that it has order elements, hence is non-zero.
For , we shall show that the spectrum does admit a commutative ring spectrum structure.
Noting that is already -local, the Adams spectral sequence gives us
By the change of rings theorem, we can write
where , since is a trivial comodule.
Now the terms of lowest degree in are generated by and have degree . So in the cobar complex, we see that apart from , all terms have . In particular,
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
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 . Then for . More precisely, the maps
are bijections for .
We can directly calculate the value of
. To see that the map we wrote down in particular is a bijection, we note that both sides are
, and the map is non-zero since the unit map
is in the image.
In particular, the case of tells us there is a unique choice of up to homotopy, and in particular, since the map is symmetric in the 's, the same is true of . Similarly, the case tells us the multiplication is associative, since that is true for .
does not admit the structure of a ring spectrum.
has a multiplication but is not (necessarily) associative.
is a (homotopy) commutative ring spectrum for .