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Let $k$ be a field. I want to compute $\operatorname{Ext}_{k[x] / \langle x^2 \rangle}(k,k)$.

However I have no idea how to do this? I cannot even think how to construct a projective resolution that would give me a useful answer. Any help would be appreciated.

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Can you think of a surjection $F\to k$ with $F$ a free module? – Mariano Suárez-Alvarez May 21 '12 at 17:16
Free module over $k[x] / \langle x^2 \rangle $? Let $S$ be a generating set for $k$ over this ring, and then consider $\left( k[x] / \langle x^2 \rangle \right)^{\oplus S} \twoheadrightarrow k$... I am not sure how to use this though – Paul Slevin May 21 '12 at 20:42
But give me a concrete generating set! – Mariano Suárez-Alvarez May 22 '12 at 1:44
I'm a bit confused about how $k$ is a module over $k[x] / \langle x^2 \rangle$, is it via the map $f \mapsto f(0)$? If so, then what about the set $S = {1}$? That is every element of $k$ can be written $a1$ where $a$ is the image of the constant poly $a \in k[x] / \langle x^2 \rangle$ – Paul Slevin May 22 '12 at 9:28
@PaulSlevin, indeed, that is correct! Congrats on your first Ext :) – Mariano Suárez-Alvarez Jun 10 '12 at 19:35
up vote 6 down vote accepted

Full credit goes to Mariano for this answer.

Define $R:= k[x] / \langle x^2 \rangle$. We think of $k$ as an $R$-algebra via the homomorphism that sends a poly $f \mapsto f(0)$. It is very clear that $k$ is generated by one element, namely the element $1 \in R$. So there is a surjection $R \twoheadrightarrow k$. Clearly $\ker (R \twoheadrightarrow k)$ is the ideal generated by the polynomial $x$ in $R$, again one element. So using the standard construction I now have an exact sequence $R \to R \twoheadrightarrow k$, where the left map is the homomorphism generated by $1 \mapsto x$. Again, after some thought it's clear that $\ker(R \to R)$ is again the ideal generated by the poly $x$ in $R$. We can keep repeating this construction to get a projective (free) resolution $$\cdots \to R \to R \to R \to k \to 0$$ which gives the chain complex (chopping off $k$) $$ \cdots \to R \to R \to R \to 0$$ which yields the complex of hom-sets $$0 \to \operatorname{Hom}_R(R,k) \to \operatorname{Hom}_R(R,k) \to \operatorname{Hom}_R(R,k) \to \cdots $$ You can stop here by noticing each of these maps are zero, which tells us the homology at each point is $\operatorname{Hom}_R(R,k) \cong k$. We have this isomorphism because of the fact that $\operatorname{Hom}_R\left(\coprod_{\alpha \in \mathcal{A}}R, M\right) \cong \prod_{\alpha \in \mathcal{A}}M$ in general.

Hence for all $i \ge 0$, $$\operatorname{Ext}_R^i (k,k) \cong k.$$

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@MattN hope this helps – Paul Slevin Jun 10 '12 at 20:33
Thank you! (for some reason your ping didn't ping me...) – Rudy the Reindeer Jun 10 '12 at 20:52
I think you have to have commented for me to be able to do it! – Paul Slevin Jun 10 '12 at 20:53
Heh. I see. : ) I think I'll re-read your answer more thoroughly in a few days -- you use a few things that I haven't revised (like e.g. $R$-algebras and the thing about mapping $f \mapsto f(0)$). +1 for now : ) – Rudy the Reindeer Jun 10 '12 at 20:56
Basically all $R$-algebra means is that $k$ is a module over $R$ via some map $\phi:R \to k$. It's not explicit when we say $k$ is an $R$-module but as far as I can see it's the only way to sensibly define it as such – Paul Slevin Jun 10 '12 at 21:02

As suggested by you I am posting my (as of now tentative!) answer:

To construct a free resolution for $k$ over $k[x]/\langle x^2 \rangle$ we proceed as in the proof of the theorem that every $R$-module has a free (and hence projective) resolution:

We observe that $k$ is generated by $S = \{1\}$. Hence, to get a surjective map from a free module to $k$ we take the free module over $\{1\}$, $F(S) = F(\{1\}) = k[x]/\langle x^2 \rangle$, and define a map as follows: $$ \pi: F(\{1\}) \to k$$ $$ a_0 + a_1 x + \langle x^2 \rangle \mapsto a_0$$

The kernel of $\pi$ is $\langle x \rangle$. Next we produce the free module $F(\operatorname{Ker}{\pi}) = F(\langle x \rangle)$ and define a map $$ \pi_1: F(\langle x \rangle) \to F(\{1\})$$ $$ e_{ax} \mapsto ax$$ $\operatorname{Ker}{\pi_1} = \{0\}$ and $\operatorname{Im}{\pi_1} = \operatorname{Ker}{\pi} = \langle x \rangle$.

Hence we have an exact sequence $$ 0 \to F(\operatorname{Ker}{\pi}) \xrightarrow{d_1 = \pi_1} F(\{1\}) \xrightarrow{d_0 = \pi} k \to 0$$

We chop off $k$ and apply $\operatorname{Hom}{(-,k)}$ to get $$ 0 \xrightarrow{\overline{d_0}=0} \operatorname{Hom}{(F(\{1\}),k)} \xrightarrow{\overline{d_1}} \operatorname{Hom}{(F(\operatorname{Ker}{\pi}),k)} \xrightarrow{\overline{d_2}=0} 0$$

Now we see that for $i \geq 3$, $\operatorname{Ext^i}{(k,k)} = 0$ since the modules in the chain are all $0$. For $i = 2$, the sequence is exact and we also get $\operatorname{Ext^2}{(k,k)} = 0$. For $k=0$ we know that $\operatorname{Ext^0}{(k,k)} = \operatorname{Hom_{k[x]/\langle x^2 \rangle}}{(k,k)}$.

To compute $\operatorname{Ext^1}{(k,k)} = \operatorname{Ker}{\overline{d_1}}$ we have to compute $\overline{d_1}$.

For this we want to know when given $\varphi \in \operatorname{Hom}{(k[x]/\langle x^2 \rangle, k)}$ we have $\varphi \circ d_1 = 0$. This is true when $\varphi$ is zero on the image of $d_1$ and since the image of $d_1 = \langle x \rangle$, this is true for all $\varphi$ that are zero on $\langle x \rangle$.

I'm not entirely sure how to write this but perhaps we can write this set as $\operatorname{Hom}{(R/\langle x \rangle, R/\langle x \rangle)} \subset \operatorname{Hom}{(R, R/\langle x \rangle)}$ where $R = k[x]/ \langle x^2 \rangle$?

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My post started with "Hi Paul." but apparently, SE doesn't like friendly greetings and automatically removes them. – Rudy the Reindeer Jun 10 '12 at 15:17
I think this has become very over-complicated - I will post my own answer below later! – Paul Slevin Jun 10 '12 at 19:05
@PaulSlevin Oooh, thank you! I'll have to have another look at my sums after dinner! – Rudy the Reindeer Jun 10 '12 at 19:05

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