Ring structure on the $K(1)$-local homotopy of $S^0$ at the prime 2 Let's write $S$ for the $K(1)$-local sphere at the prime 2.  Then there is a cofibre sequence
$$S \to KO \to KO$$
where I'm using $KO$ to denote the $K(1)$-localization of orthogonal K-theory, and the self map of $KO$ is given in terms of Adams operations: $\psi^3-1$.  Drew Heard reminded me that on $\pi_1$, $\psi^3-1=0$ is the trivial endomorphism of $\mathbb{Z} / 2 = \pi_1 KO$; this yields the computation 
$$\pi_0(S) = \mathbb{Z}_2 \oplus \mathbb{Z} / 2,$$
where the 2-adic factor comes from $\pi_0 KO$, and the $\mathbb{Z} / 2$ gets carried around from $\pi_1 KO$ by the connecting map in the cofibre sequence.
My question is: what is the ring structure on $\pi_0(S)$?  
It is certainly a $\mathbb{Z}_2$-algebra, so this describes most of the multiplication.  If I write $x$ for an additive generator of $\mathbb{Z} / 2$ (the image of the connecting map), then the only remaining computation is a formula for $x^2$.  Since $2x=0$, $2x^2=0$, so it must be the case that either $x^2=0$ or $x^2=x$, since these are the only 2-torsion elements of the ring.  Is it known which of these two possibilities holds?
 A: It must be the case that $x^2 = 0$. One can see this in at least two ways:


*

*The action of $\psi^3$ on $KO$ is by a ring map, and so the homotopy fixed-point spectral sequence for its action on $KO$ is multiplicative. The $\Bbb Z/2$ lives on the 1-line of this spectral sequence, and there is nothing on the 2-line or higher, so it must square to zero. I believe that you can also do this calculation for the action of the full group $\Bbb Z_2^\times$ on $KU$. (Note that this also tells you how $x$ acts on other homotopy groups.)

*If we had $x^2 = x$, then $x$ would be idempotent and so we would have a decomposition of the $K(1)$-local sphere into a product of two ring spectra, one $2$-torsion and one not. Idempotents automatically split it as commutative ring spectra (this is in Rognes' memoir -- you can either use obstruction theory or Bousfield localization to prove it), and so that would imply that we had a $K(1)$-local commutative ring spectrum with $2 = 0$. However, $2 = 0$ forces a commutative ring spectrum to be a wedge of Eilenberg-Mac Lane spectra and such an object can't be $K(1)$-local.
Keep warm!
A: Here is another method which is in some sense more direct.  Consider the diagram 
$$ \begin{array}{ccc}
 L_{K(1)}S & \xrightarrow{i} & KO \\
 i \downarrow & & \downarrow (1,\psi^3) \\
 KO & \xrightarrow{(1,1)} & KO\times KO
\end{array} $$
This is a commutative diagram of maps of commutative ring spectra.  Using the fact that $L_{K(1)}S$ is the fibre of $\psi^3-1$, one can check that it is homotopy cartesian.  Whenever you have such a square 
$$ \begin{array}{ccc}
 A & \xrightarrow{i} & B \\
 j \downarrow & & \downarrow k \\
 C & \xrightarrow{l} & D,
\end{array} $$
one can check that $\ker(i).\ker(j)=0$.  In the relevant case we have $i=j$ and $x\in\ker(i)$ so $x^2=0$.
