Ring structure in the Serre spectral sequence I've tried to understand what's going on in Example 1.5 on page 27-28 in Hatcher's notes on spectral sequences. There is one part in the reasoning that I can't understand here. He writes down a table with the $E^2$-page of a spectral sequence looking like (arrows omitted)
$$
\begin{array}{ccccccc}
\mathbb{Z}a & 0 & \mathbb{Z}ax_2 & 0 & \mathbb{Z}ax_4 & 0 & \ldots\\
\mathbb{Z}1 & 0 & \mathbb{Z}x_2 & 0 & \mathbb{Z}x_4 & 0 &\ldots
\end{array}
$$
What I'm trying to figure out is how do we know that $ax_2$ is the generator of $E^{2,1}_2$? Hatcher just writes:

The generators for the $\mathbb{Z}$'s
  in the upper row are $a$ times the
  generators in the lower row, because
  the product $E_2^{0,q}\times E_2^{s,t}\to E_2^{s,t+q}$ is just
  multiplication of coefficients.

Can someone explain to me what's going on here?

I'll attempt to add to Dylan Wilson's excellent comment. 
The OP refers to the Serre spectral sequence in cohomology for the path-loop fibration $K(\mathbb{Z},1) \to P \to K(\mathbb{Z},2)$. In general for a fibration $F \to E \to B$, the $E^2$ page is given by $E_2^{pq} = H^p(B,H^q(F))$, where we view $H^q(F)$ as a local system under the monodromy action. But in this case, $K(\mathbb{Z},2)$ is simply connected, so there is no monodromy. Even better, as $K(\mathbb{Z},1) \simeq S^1$, the fiber $K(\mathbb{Z},1)$ has free cohomology groups, namely
$$ H^q(K(\mathbb{Z},1)) = \begin{cases} \mathbb{Z} & q = 0,1 \\ 0 & \text{else.}\end{cases}$$
Thus the universal coefficient theorem gives
$$ E_2^{pq} = H^p(K(\mathbb{Z},2)) \otimes H^q(K(\mathbb{Z},1)).$$
Hence, once $E_2^{2,0} = H^p(K(\mathbb{Z},2))\otimes \mathbb{Z}$ is determined, by arguing that $E_3 = E_\infty$, we immediately know $E_2^{2,1}$ as well. 
The extra magic of this computation is that the differentials in the Serre spectral sequence are derivations with respect to the ring structure on $E_2^{pq}$, which itself is the tensor product of the ring structures on the cohomology of base and fiber (when there is no monodromy). 
