First of all, since every other book somehow mentions that this is trivial, I apologize if it turns out that I am just misunderstanding something in the definitions. So here goes:
The motivation for this is the proof that $Ext_{\Gamma}(k, k) = P(y_1, y_2, ...)$ where $\Gamma$ is a commutative, graded connected Hopf algebra of finite type over a field $k$ found in Ravenel's green book Chapter 3, page 62. With this, let me set up my question.
Let $P[x]$ and $E[x]$ be the polynomial and exterior algebra on one generator over a field $k$ respectively. With $x$ as the primitive element (that is to say, $\Delta(x) = 1 \otimes x + x \otimes 1$), they are both coalgebras over $k$.
I would like to understand the following computation: $Ext_{E(x)}(k, E(x)) = P(y)$.
(The earlier question was wrong oops)
The way one does this is to take an injective resolution $0 \rightarrow k \rightarrow E(x) \rightarrow E(x) \rightarrow ...$ where the maps are $d_i(x) =1, d_i(1) = 0$ then applying $Hom_{E(x)}(k, )$ to get a complex with zero differentials. It then suffices to compute each term $Hom_{E(x)}(k, E(x))$. These $Hom$-sets are what I am interested in knowing an explicit computation of.
Thanks!
Maybe it would help to work with algebras rather than coalgebras: since everything is finite-dimensional, we can dualize. The dual to the coalgebra $E(x)$ is again the exterior algebra $E(x)$, and we can work with $E(x)$-modules rather than comodules.
In this case, the goal is to show that $\mathrm{Ext}^{\bullet}_{E(x)}(k, k)$ is a polynomial algebra, generated by a nontrivial map $k \to k[1]$ in the derived category of $E(x)$-modules (which corresponds to the extension $0 \to k \to E(x) \to k \to 0$).
One can write down a minimal resolution of $k$ as an $E(x)$-module that looks like
$$ \dots \to E(x) \to E(x) \to E(x) \to 0 ,$$ where each of the successive maps is given by multiplication by $x$. This computes $\mathrm{Ext}^\bullet_{E(x)}(k, k)$ as a graded vector space (there's one element in each dimension as desired). The main point is that all the products (compositions) of the element in degree zero are nontrivial. One way to do this computation is to write down compositions of maps of chain complexes: it suffices to show that the maps of chain complexes of the form
$$ \dots \to E(x) \to E(x) \to E(x) \to 0 \\ \dots \to E(x) \ \to 0 \ \ \to 0 \ \ \to 0 $$ which are the identity or zero degreewise, are not nullhomotopic. In fact, one can't construct a homotopy even at the first level.
But if you work with coalgebras, one way to see this is to use the cobar complex which is discussed in Ravenel's book: in this case, the cobar complex has entirely zero differentials and gives the polynomial algebra as desired.
(Note also that the analogous construction applied to a polynomial algebra yields an exterior algebra. It's probably worth pointing out that this connection between polynomial and exterior algebras is a special case of Koszul duality for associative algebras.)