multiplicative structure of Ext Basically, I am trying to compute something with the Adams spectral sequence (as a toy example). The $E^2$ page reduced to computing $Ext^{s,t}_{\Gamma} (\mathbb{F}_2, \mathbb{F}_2)$, where $\Gamma = \mathbb{F}_2 x/ x^2$ and $deg(x) =1$. Keeping very close track of degree, I found that $Ext^{s,t}_{\Gamma} (\mathbb{F}_2, \mathbb{F}_2)$ is only nonzero in degree $s=t$. In this case, we have $\mathbb{F}_2 \{ \phi_t \} $ where $\phi_t \in Hom^t_{\mathbb{F}_2 x/x^2} (\Sigma^t \mathbb{F}_2 x/x^2, \mathbb{F}_2)$ sends $1$ to $1$ and $x$ to $0$. (the superscript $t$ denotes morphisms that lower degree by $t$.)
So I have come to the point that I need to understand the multiplicative structure of $Ext$. I know I am supposed to get that $\phi_1^2 = \phi_2$ (because I know what the spectral sequence is supposed to converge to--the 0th column should have a copy of $\mathbb{F}_2$ in each degree, but all of it generated by $\phi_1$ ). However, I am not really sure how to make sense of $\phi_1^2$.  So my questions are 
1) how to make sense of $\phi_1^2$? or have I confused things somewhere along the way
2) in order to work with the multiplicative structure in the second page of Adams spectral sequence, I need to understand the multiplicative structure of $Ext$. I would like to work directly with objects in $Ext$, but how does one multiply? and is it easier to do it this way or does one typically use the isomorphism between objects of $Ext$ and extensions, multiply the extensions (which is simple), and then translate back? But the isomorphism for higher Ext does not seem easy to work with. 
Note: I specifically am trying to do with without the cobar complex. 
Edit: I should add that the resolution I took was $\ldots \Sigma^2 \mathbb{F}_2 x/x^2 \xrightarrow{d} \Sigma \mathbb{F}_2 x/x^2 \xrightarrow{d} \mathbb{F}_2 x/x^2 \to \mathbb{F}_2$, where $d$ sends $\Sigma^{t} 1$ to $\Sigma^{t-1} x$ and everything else to $0$. I'm adding the $\Sigma$ to help keep track of degree. 
 A: So, maybe I should format this as an answer.
Consider any pair of objects $A, B$ in an abelian category with, say, enough projective objects. Then, the extension group $Ext^i(A,B)$ is defined as $H^i(Hom(A^{\bullet}, B^{\bullet}))$, where $A^{\bullet}, B^{\bullet}$ are resolutions of objects $A, B$ respectively, and $A^{\bullet}$ needs to be projective, and no requirement on $B^{\bullet}$ but we will take it to be projective too to define composition.
Composition for Exts is then induced by composition $Hom(A^{\bullet}, B^{\bullet}) \otimes Hom(B^{\bullet}, C^{\bullet}) \rightarrow Hom(A^{\bullet}, C^{\bullet})$.
Let us denote $R = \mathbb{F}_2[x]/x^2$. Then, $\mathbb{F}_2$ admits the following resolution: $A^{\bullet} = ... \rightarrow \Sigma^2 R \rightarrow \Sigma R \rightarrow R \rightarrow 0$, the maps being multiplication by $x$.
The hom-complex $Hom(A^{\bullet},A^{\bullet})$ is then quasiisomorphic to $Hom(A^{\bullet}, \mathbb{F}_2)$ (it is always true for any pair of complexes such that left one is a complex of projective modules and the right one is switched to the quasiisomorphic one). Concretely, it means that any closed map $\phi: A^{\bullet} \rightarrow \mathbb{F}_2$ can be lifted to the map $\phi: A^{\bullet} \rightarrow A^{\bullet}$ uniquely up to homotopy. So, the map we lift is your $\phi_t$, and the lift is defined as $\phi_t: A^k \rightarrow A^{k-t} = Id$ for $k \geq t$ and $0$ otherwise.
It is now clear that $\phi_t \phi_s = \phi_{t+s}$
$\blacksquare$
the rule is basically "switch everything to resolutions and then map complexes instead of objects". it works very well if you have a category with enough projective or injective objects
A: The composition product $Ext(N,P) \otimes Ext(M,N) \to Ext(M,P)$ can be computed as follows.    Given cocycles $x : N_{s_1} \to \Sigma^{t_1} P$  and $y :   M_{s_2} \to \Sigma^{t_2} N$, let $\{ y_s : M_{s+s_2} \to \Sigma^{t_2} N_s\}$ be a chain map lifting $y$.    Then $xy$ is represented by the cocycle $x y_{s_1} : M_{s_1+s_2} \to \Sigma^{t_2} N_{s_1} \to \Sigma^{t_1+t_2} P $.   
The point is that one needs (a finite bit of) the chain map lifting $y$, but only the cocycle $x$.   
Saves effort.
(The $M_s$ and $N_s$ are the terms in projective resolutions of $M$ and $N$, of course.)
A: In this case, you're dealing with a Koszul algebra ---this is precisely your observation that the spectral sequence is concentrated in the diagonal. Its Koszul dual is a polynomial algebra, and this gives you the multiplicative structure on $\mathsf{Ext}$: each diagonal entry is one dimensional with generator $x_s$, and you have that $x_sx_t=x_{t+s}$, so $\mathsf{Ext}$ is a polynonimal algebra generated by $x_1$. 
This can be seen at the level of the (reduced) bar construction $B\Gamma$, whose dual computes your $E_2$-page. It has zero differential, and it is generated in each homological degree $s$ by $x_s = [x\vert\cdots\vert x]$ with $x$ appearing $s$ times. The product is dual to the coproduct of $B\Gamma$, which is simply $\Delta(x_s) = \sum x_i\otimes x_j$. 
