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. 



multiplicative structure of Ext
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*


