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?
 A: 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). 
