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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?2012-05-21
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    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 though2012-05-21
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    But give me a concrete generating set!2012-05-22
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    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$2012-05-22
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    Indeed. So now we have a surjective map $k[x]/(x^2)\to K$ from a projective module to $k$. What is its kernel?2012-05-22
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    The ideal generated by $x$ in $k[x] / \langle x^2 \rangle$ ?2012-05-22
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    Yes. (Let us write $R=k[x]/(x^2)$ for simplicity, and let $F_0$ be the free module we already have) So: can you find a free module $F_1$ and a morphism $F_1\to F$ such that its image is precisely that ideal?2012-05-22
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    The map $R \twoheadrightarrow \operatorname{ker}(F_0 \twoheadrightarrow k)$ which takes a poly $f$ to $f$ in the quotient? Hmm maybe not, it sends constants in $k$ to constants in $k$ which shouldn't be in that ideal. What about sending $f$ to $f.x$ ?2012-05-22
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    You tell me. Deciding which of those two works is a matter of actually checking which of the two does what you want it to do.2012-05-22
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    well the 2nd one does the job and there's no reason for me to believe it's not $R$-linear. So I have $F_1 \to F_0 \to k$ now, the kernel of $F_1 \to F_0$ is again the ideal generated by $\langle x \rangle$. I guess this gives me my projective resolution!2012-05-22
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    There is never any reason to write «there's no reason for me to believe it's not R-linear»: just check if it is $R$-linear or not! :) In any case, indeed, you now have the (begining of) your projective resolution. Notice that what we have done is simply to follow the proof of the proposition that projective resolutions exist!2012-05-22
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    I know but I am a bit lazy. Thanks for the help I will attempt to work out the actual value of Ext now2012-05-22
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    Hello, if you are still around I haven't had much luck. It is clear that we have a projective resolution $ \cdots \to k[x] / \langle x^2 \rangle \to k[x] / \langle x^2 \rangle \to 0$, where the arrows mean the map $f \mapsto fx$. So this gives us the complex $0 \to \operatorname{Hom}( k[x] / \langle x^2 \rangle, k) \to \operatorname{Hom}( k[x] / \langle x^2 \rangle, k) \to \cdots$. So basically I am stuck trying to work out the kernel of the map on homsets. Can you give me any advice?2012-05-26
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    I think after observing that $\operatorname{Hom}(k[x] / \langle x^2 \rangle, k)$ is isomorphic to $k$ via the map $\phi : \theta \mapsto \theta(1)$, we can look at the homology of the complex $0 \to k \to k \to \cdots$ instead. And the corresponding maps in this sequence $h: k \to k$ are given by $h(a) = 0$, so the kernel is 0... which corresponds to $Ext_R^i (k,k) \cong k$ for all $i$. Is that analysis correct?2012-05-26
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    I mean kernel is $k$ in the above comment *, and the image is 02012-05-26
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    @MarianoSuárez-Alvarez Hello Mariano, I am pinging you on behalf of Paul who had posted 4 comments addressed to you as it seems, but forgot to include "@".2012-06-10
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    @MattN., thanks!2012-06-10
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    @PaulSlevin, indeed, that is correct! Congrats on your first Ext :)2012-06-10

2 Answers 2

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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 helps2012-06-10
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    Thank you! (for some reason your ping didn't ping me...)2012-06-10
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    I think you have to have commented for me to be able to do it!2012-06-10
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    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 : )2012-06-10
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    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 such2012-06-10
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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.2012-06-10
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    I think this has become very over-complicated - I will post my own answer below later!2012-06-10
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    @PaulSlevin Oooh, thank you! I'll have to have another look at my sums after dinner!2012-06-10