Let $k$ be some field (say of characteristic zero, if it matters) and define $$R=k[x]/(x^2).$$ I want to compute $$\mathrm{Ext}^\bullet_R(k,k)$$ and, in particular, the ring structure on it (though I think I can do this part if I can compute the Ext modules).

I know that we can think of elements of $\mathrm{Ext}^m_R(k,k)$ as length $m+2$ exact sequences of the form $$0\to k\to X_m\to\ldots\to X_1\to k\to0$$ modulo some sensible equivalence relation, and that we can also think of it as $$\mathrm{Ext}^m_R(k,k) = H^m(\mathrm{Hom}_{R\hbox{-}\mathsf{mod}}(P_\bullet,k)) = H^m(\mathrm{Hom}_{R\hbox{-}\mathsf{mod}}(k,I^\bullet))$$ for some projective (or injective) resolution $P_\bullet$ (or $I^\bullet$, respectively) of $k$.

However, when it comes to the hands-on part of actually computing this, I hit a mental block. Since $R$ is a local ring (with maximal ideal $(x)$) we know that projective modules are exactly the free modules, and so computing a projective resolution will probably be easiest...?

I would appreciate hints and partial answers over explicit answers (though it's likely I might have to ask for more hints if I still struggle...). At this stage I'll take whatever I can get.

Edit: Here is a partial answer, all that remains is the question of the ring structure.

Note that $k\cong R/(x)$ and so we have an epimorphism $\pi\colon R\twoheadrightarrow k$ given by $x\mapsto0$ (the quotient map). If we write $R=k[\varepsilon]$ where $\varepsilon$ is such that $\varepsilon^2=0$ then we obtain the following free resolution of $k$: $$\ldots\xrightarrow{\cdot\varepsilon}k[\varepsilon]\xrightarrow{\cdot\varepsilon}k[\varepsilon]\twoheadrightarrow k\to0.$$

Now any morphism $k[\varepsilon]\to k$ must send $\varepsilon$ to some element $\eta\in k$ such that $\eta^2=0$. But $k$ is a field, and so we are forced to choose $\eta=0$. This means that any such morphism is determined entirely by where it send $1\in k$, and it can send it to any $x\in k$. Thus $$\mathrm{Hom}_{R\hbox{-}\mathsf{mod}}(k[\varepsilon],k)\cong k.$$ So taking $\mathrm{Hom}_{R\hbox{-}\mathsf{mod}}(-,k)$ of the free resolution gives us the sequence $$0\to k\xrightarrow{\cdot0}k\xrightarrow{\cdot0}\ldots$$ which has homology $H_n=\ker d_n/\mathrm{im}\,d_{n+1}=k/0\cong k$ for all $n\geqslant0$. Thus $$\mathrm{Ext}^\bullet_R(k,k)\cong\bigoplus_{n\geqslant0}k$$

So my question now is about the ring structure of $\mathrm{Ext}^\bullet_R(k,k)$, and also about thinking of $\mathrm{Ext}$ as being extensions of $k$ by $k$. Unless I'm wrong, this means that we should be able to construct, taking $n=1$, short exact sequences $$0\to k\hookrightarrow X\twoheadrightarrow k\to0$$ and the collection of all such sequences should be isomorphic to $k$. The first thing that sprang to mind was to take $X=R$ and the epimorphism multiplication by $x\varepsilon$ for $x\in k$, but then I struggle to find a monomorphism into $R$ with the right kernel, and also taking $x=0$ means that the map fails to be an epimorphism.

  1. What is the correct choice of $\,\,\to X\to\,\,$?
  2. How can we compute explicitly the ring structure on $\mathrm{Ext}^\bullet_R(k,k)$?

Edit 2: Following the ideas in the comments, I'm trying to explicitly spell out the following isomorphism, but I'm struggling to understand how the quotients are realised on both sides (i.e. the equivalence relations): workings I feel like the right-hand side should just be chain maps modulo homotopy equivalence, even though the $\mathrm{Hom}$ complex is just of maps of chains. I'm pretty certain that the lifts $\hat{f}_\bullet$ that we construct are in fact chain maps.

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    $\begingroup$ You can write down a very explicit and very small projective resolution. $\endgroup$ – Qiaochu Yuan Apr 17 '16 at 21:03
  • $\begingroup$ @QiaochuYuan my first thought was that $k=\langle1\rangle$ is free and thus projective, but that didn't seem right to me somehow, since $k\not\cong R^n$ for any $n$...? $\endgroup$ – Tim Apr 17 '16 at 21:06
  • $\begingroup$ For ring structure, use the fact that $\mathrm{Ext}_R^\bullet(k,k) \cong H^\bullet( \mathrm{End}_{Ch(R)}(P_k^\bullet))$, where $P_k^\bullet$ is the projective resolution of $k$; the multiplication on the left hand side is the same as the (homology of) the composition of the right. Alternatively, use Koszul theory (but I guess the point you started this computation is to understand Koszul theory with more hands-on experience...?). $\endgroup$ – Aaron Apr 27 '16 at 5:46
  • $\begingroup$ @Aaron I haven't actually heard of Koszul theory! How did you obtain that isomorphism in your comment? Is the endomorphism ring simply the ring of chain maps $P^\bullet_k\to P^\bullet_k$? $\endgroup$ – Tim Apr 27 '16 at 10:50
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    $\begingroup$ Oh, your example is the smallest non-trivial example of Koszul algebra; but may be you should read up on that after you are finished with this. The isomorphism is evident as groups/graded vector space; you just need to write everything out explicitly to see why it is true.... The elements of the endomorphism complex are maps between chain complexes but not necessarily chain maps, but then when you take cohomology, you will be effectively considering the chain maps. The $k$-th degree component is given by maps of degree $k$; the differential is the obvious choice... $\endgroup$ – Aaron Apr 27 '16 at 11:10

To compute the ring structure in the Yoneda ring of $A=\mathbb k[x]/(x^2)$ you can take several paths. First, note that $A$ is Koszul, so its Yoneda ring is its Koszul dual. the Koszul dual to $A$ is simply $\mathbb k[x]$, a polynomial ring in one variable. This can be read off quite directly from the bar construction $BA$ of $A$, since it has basis elements $t^n = [t\vert\cdots \vert t]$ ($n$ copies of $t$) which are all cycles, and the coproduct in $BA$ is simply given by $\Delta(t^n) = \sum_{i+j=n} t^i\otimes t^j$. Then $(BA)^\vee$ computes $\operatorname{Ext}_A(\mathbb k,\mathbb k)$ and $\Delta^\vee = \,\smile$, so you're done.

Put in another way, you can consider a model of this algebra in the category of dg algebras. One such model, which is the minimal model, is the free algebra $TX$ on generators $x_i$ of degree $i$ and differential given on generators by $dx_{n+1} = \sum_{i+j=n} (-1)^i x_ix_j$. The map $TX\to A$ sends $x_0\to x$ and kills all other generators. It is known then that (since the model is minimal) $X^\vee$ identifies canonically with $\operatorname{Ext}_A(\mathbb k,\mathbb k)$ and that the cup product is dual to the quadratic part of the differential $d_2 : X\to X\otimes X$, which in this case is just $d$ (since $A$ is Koszul). So again you recover that $x^i\smile x^j = x^{i+j}$ where the Koszul sign rule gives you the abscence of signs.


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