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This issue began as an attempt to understand the projective resolution of $R = k[x]/(x^{n+1})$ that appears in Weibel (Exercise 9.1.4)

$$\cdots\to R^e \xrightarrow{u} R^e \xrightarrow{v} R^e \xrightarrow{u} R^e \xrightarrow{\mu} R \to 0$$

where $u = x\otimes 1 - 1\otimes x$, $v = \sum_{i=0}^n x^{n-i}\otimes x^i$ and $\mu$ is multiplication. I found this paper paper that constructs a projective resolution for the more general ring $A = k[x]/(f)$ where $f$ is a degree $n$ monic polynomial

I was hoping to make use of the contracting homotopies that appear in this paper and apply them to the resolution for $R$ to show that it is in fact a resolution, but I'm struggling to make sense of them, so I'll just define them and go through what I understand so far.

There are three maps that make up the contracting homotopies, $s_0:A\to A^e$, $s_1:A^e\to A^e$ and $s_2:A^e\to A^e$, defined as

$$s_0(P)= 1\otimes P,\quad s_1(P\otimes 1) = \frac{T(P)}{T(x)}, \quad s_2(P\otimes 1) = \bar{Px}\otimes 1$$

where $P$ is a polynomial in $A$ and $T(P) = 1\otimes P - P\otimes 1$. The resolution, and its contracting homotopies is

$$ \cdots \substack{\xrightarrow{v}\\[-1em] \xleftarrow[s_2]{}\\[-1em]} A^e \substack{\xrightarrow{u}\\[-1em] \xleftarrow[s_1]{}\\[-1em] } A^e \substack{\xrightarrow{v}\\[-1em] \xleftarrow[s_2]{}\\[-1em] } A^e \substack{\xrightarrow{u}\\[-1em] \xleftarrow[s_1]{}\\[-1em] } A^e \substack{\xrightarrow{\mu}\\[-1em] \xleftarrow[s_0]{}\\[-1em] } A \to 0 $$

where $u(a\otimes b) = (a\otimes b)T(x)$ and $v(a\otimes b) = (a\otimes b)\frac{T(f)}{T(x)}$. To be clear, the arrows on the bottom read $s_2$, $s_1$, $s_2$, $s_1$, $s_0$ and the arrows on top read $v$, $u$, $v$, $u$, $\mu$, from left to right. I had issues formatting the arrows so apologies for that.

If I've understood things correctly, I think that $\frac{T(P)}{T(x)}$ can be written

$$\frac{T(P)}{T(x)} = \sum_{i=0}^np_i\sum_{j=0}^{i-1}x^j\otimes x^{i-j-1}$$

and then $s_1(P\otimes 1) = \sum_{i=0}^np_i\sum_{j=0}^{i-1}x^j\otimes x^{i-j-1}$ and $v(a\otimes b)$ can be written in a similar way. Then of course the aim is to show that

$$s_1u + vs_2 = \mbox{id}, \quad s_2v + us_1 = \mbox{id}, \quad s_0\mu + us_1 = \mbox{id}, \quad \mu s_0 = \mbox{id} $$

Demonstrating $\mu s_0 = \mbox{id}$ is easy enough, so I've had no issues there.

That's all I understand so far, so on to my questions. There are a few things that I don't understand about this:

  1. Why are $s_1$ and $s_2$ only defined on $P\otimes 1$? It's not obvious to me that $P\otimes 1$ generates $A^e$, so I'm not sure why it's only necessary to know how $s_1$ and $s_2$ act on these elements (Edit: This question has now been answered in the comments).
  2. What exactly does $\bar{Px}$ mean? From the notation in the paper my thinking is that $P\in A$ is considered as an element of $k[x]$ and then $\bar{Px}$ is $(Px)\mbox{ mod}f$ as it were? However, I'm not sure, so any clarification would be appreciated.
  3. (Edit: Re a point made in the comments: All the calculations done in this question are done using the definitions from the paper. "My $u$" that the comment refers to I think is the $u$ I define at the beginning of this post, but that $u$ is never actually used, it's only present because it's the definition given in Weibel, which was the motivating problem that lead to me asking this question) Ignoring for the moment my issue with the definition of $s_1$, when I calculate $(s_0\mu + us_1)(P\otimes 1)$ I get \begin{align} (s_0\mu + us_1)(P\otimes 1) & = s_0(\mu(P\otimes 1)) + u(s_1(P\otimes 1))\\ & = 1\otimes P + \frac{T(P)}{T(x)}T(x) \\ & = 1\otimes P + 1\otimes P - P\otimes 1\\ & = 2(1\otimes P) - P\otimes 1 \end{align}which, unless I'm missing some obvious relation in $A^e$, is not equal to $P\otimes 1$. I've gone over this countless times looking for a mistake but I can't find one. Assuming I've not made a mistake, this is solved by putting a minus sign in front of $\frac{T(P)}{T(x)}$ in the definition of $s_1$. But I'm wary of declaring that there is a typo in the paper, so I come to you for a second opinion. Have I made a mistake somewhere? Or is this a typo in the paper?

I've made a tentative attempt at calculating the other two relations but they've proven to be a bit more complicated and not fully understanding the definition of $s_2$ doesn't help either, so I've not made much progress there. The authors of the paper assure the reader that the relations can be shown by direct calculation, so I can only assume that it's possible. In any case, any help with this would be greatly appreciated!

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    $\begingroup$ they are defined on $P\otimes 1$ as they are defined to be "right A-maps" i.e. homomorphisms of right $A$-modules. The sign error is because their $T(x)$ has the opposite sign to your $u$. $\endgroup$ Jun 6 '20 at 16:21
  • $\begingroup$ @MatthewTowers So if I understand you, if the maps are homomorphisms of right $A$-modules, then we can write $s_1(P_1\otimes P_2) = s_1((P_1\otimes 1)P_2) =s_1(P_1\otimes 1)P_2$? With the sign error, I never actually use that first $u$ in any calculations, I included it as context. All calculations were done using the definitions that appear in the paper. So in question 3 $u$ is multiplication by $T(x)=1\otimes x - x\otimes 1$, and not multiplication by $x\otimes 1 - 1\otimes x$, so I'm fairly certain that the sign is the right way around. Or have I misunderstood your comment? $\endgroup$
    – SeraPhim
    Jun 7 '20 at 10:37

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