I'm trying to show that $\sum\_{i = 0}^{p-2} (i+1)^{-1} t^{i+n}$ where $0 \leq n \leq p-2$ spans the vector space $\mathbb{F}\_p[t]/(1-t)^{p-1}$ as a rank $p-1$ module over $\mathbb{F}\_p$.


In other words, I would like to show that the determinant of the following matrix is a unit in $\mathbb{F}\_p$. I've shown this for $p = 2, 3, 5, 7, 11, 13$. I have tried to use induction but failed. This is such a natural matrix I am hoping someone recognizes it!


$$\begin{pmatrix} 1 & 0 & -1 & (p-2)^{-1} & \cdots & 4^{-1} & 3^{-1} \\
2^{-1} & 1 & 0 & -1 & (p-2)^{-1} & \cdots & 4^{-1} \\
3^{-1} & 2^{-1} & 1 & 0 & -1 & \ddots & \vdots \\
\vdots & 3^{-1} & 2^{-1} & 1 & 0 & \ddots & (p-2)^{-1} \\
 (p-3)^{-1} & \ddots & \ddots & \ddots & \ddots & 0 & -1 \\
(p-2)^{-1} & (p-3)^{-1} & (p-4)^{-1} & \ddots & 2^{-1} & 1 & 0 \\
-1 & (p-2)^{-1} & (p-3)^{-1} & \cdots & 3^{-1} & 2^{-1} & 1
\end{pmatrix}$$




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Edit: At risk of overcrowding the above question, I am adding some context below the line. Feel free to ignore it.


Let us look at the action of $G := C\_p \rtimes C\_{p-1} \simeq (\mathbb{F}\_p, +) \rtimes (\mathbb{F}\_p^\*, \times)$. Then, $G$ acts on $x \in X := F\_p$ as follows, $(c, m)(x) = c + mx$. Let R be a $\mathbb{Z\_p}$-algebra, and $A=R[C\_p]$ be the permutation representation. If $x \in X$, we write $[x]$ as the corresponding element in the module. Let us fix $\sigma := (1,1)$ and $\tau := (0, a)$ to be the generators of $G$, where $a$ is a fixed primitive root of $\mathbb{Z}/p$.


Now, let $B$ be the kernel of the augmentation map $R[X] \to R$. This is a free representation of rank $p-1$ over $R$. Further, as a $C\_p$-module, $B$ is isomorphic to $R[X]/N$, where $N$ is generated by $(1+\sigma+\cdots+\sigma^{p-1})$. Further, $N$ is isomorphic to the trivial representation as a $C\_p$ module.


Let $V$ be a rank 1 $C\_{p-1}$-subrepresentation of $B$. We may extend the inclusion map $V \to B$ of $C\_{p-1}$ representations to a map of $G$ representations: $$f: R[G] \otimes\_{R[C\_{p-1}]} V \to B.$$


I wish to pick $V$ such that this map is surjective. By Nakayama's lemma, since we are working with local rings, it suffices to show this map is surjective mod $p$.


To get to the below phrasing, I rewrote $R[X]$ as $R[t]/(t^p-1)$. Here, $\sigma$ acts by taking $t^{i} \mapsto t^{i+1}$. Then, if we consider $B$ mod $p$, which we call $\overline{B}$, then $\overline{B} = \mathbb{F}\_p[t]/(t-1)^{p-1}$. There are $(p-1)$ 1-d $C\_{p-1}$ subrepresentations of $B$, where $t$ in $\mathbb{F}\_p^\times$ acts by multiplication with $t^r$, where $1 \leq r \leq p-1$. Further, from trial and error I noticed that picking $V$ to be the subrepresentation where $r = 1$, seems to make $f$ surjective. In other words, choosing $V$'s generator mod p to be $y := \sum\_{j = 0}^{p-2}a^{-j}t^{a^k} = \sum\_{i=0}^{p-2} i^{-1}t^i$ seems to make $f$ surjective. Note that the image of $f$ is $(y, \sigma(y), \cdots, \sigma^{p-2}(y))$.



How to see that the determinant of this matrix is nonzero for all primes?
As discussed in the comments, I don't see how to extract the desired matrix from the original question (about spanning the vector space). However, the matrix being nonzero IS equivalent to the following:


The polynomials $\sum\_{i = 0}^{p-2} (i+1)^{-1} t^{i+n}$ for $0\leq n\leq p-2$, along with the polynomial $t^{p-1}$, span the vector space $\mathbb{F}\_p[t]/(1-t)^p$.


To see this, the change of basis matrix from $1,t,\cdots, t^{p-1}$ to our set of polynomials is given by


$\begin{pmatrix} 1 & 2^{-1} & 3^{-1} & 4^{-1} & \cdots & (p-1)^{-1} & 0 \\
0 & 1 & 2^{-1} & 3^{-1} & 4^{-1} & \cdots & (p-1)^{-1} \\
(p-1)^{-1} & 0 & 1 & 2^{-1} & 3^{-1} & \ddots & \vdots \\
\vdots & (p-1)^{-1} & 0 & 1 & 2^{-1} & \ddots & 4^{-1} \\
 4^{-1} & \ddots & \ddots & \ddots & \ddots & 2^{-1} & 3^{-1} \\
3^{-1} & 4^{-1} & 5^{-1} & \ddots & 0 & 1 & 2^{-1} \\
0 & 0 & 0 & \cdots & 0 & 0 & 1
\end{pmatrix}$


(here we use that $(1-t)^p=1-t^p$.) Because of the form of the bottom row, the determinant of this matrix is equal to the determinant of the minor given by the first $(n-1)$ rows and columns, which is the transpose of the original matrix.


Now let us address this question about polynomials. Set $P(t)=\sum\_{i=0}^{p-2}(i+1)^{-1}t^i.$ Then the question is equivalent to showing that there is no choice of a polynomial $Q(t)$ of degree $\leq p-2$ and a constant $c\in\mathbb{F}\_p$ with $P(t)Q(t)=ct^{p-1}$ in $\mathbb{F}\_p[t]/(1-t)^p$. Assume otherwise.


First note that, if $p>2$, $t-1$ divides $P(t)$. Indeed, $P(1)=1^{-1}+2^{-1}+\cdots+(p-1)^{-1}$, which is zero modulo $p$. If $p=2$ then the matrix trivially has nonzero determinant.


Therefore, we have $0=P(1)Q(1)=c$, so we have $P(t)Q(t)=0$. This can only be possible if $P(t)$ is a multiple of $(t-1)^2$. This in turn would imply that $tP(t)=\sum\_{i=0}^{p-2}(i+1)^{-1}t^{i+1}$ is a multiple of $(t-1)^2$, and hence that the derivative $(tP(t))'=\sum\_{i=0}^{p-2}t^i$ is a multiple of $(t-1)$. But, by inspection, we see that its value at $1$ is $p-1$, a contradiction.


