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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!

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


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 $(y, \sigma(y), \cdots, \sigma^{p-2}(y))$.

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    $\begingroup$ The $i=0$ term of $\sum_{i=0}^{p-2} i^{-1} t^{i+n}$ is $t^n/0$. Do you mean $\sum_{i=0}^{p-2} (i+1)^{-1} t^{i+n}$? $$ $$ The matrix you exhibit has determinant $1 \bmod p$ for each prime $p < 120$. I guess you checked $\det \neq 0$ also for $p=2$ and $p=11$, not just $3,5,7,13$. $\endgroup$ Commented Feb 15, 2021 at 5:41
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    $\begingroup$ This is a Toeplitz matrix over $\mathbb{F}_p$. There are some general results about the invertibility of such matrices, see for instance dl.acm.org/doi/10.1145/236869.237081 $\endgroup$ Commented Feb 15, 2021 at 6:33
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    $\begingroup$ If there were one more row / column it would be a circulant matrix. $\endgroup$
    – Tim Campion
    Commented Feb 15, 2021 at 6:35
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    $\begingroup$ But we have only $p-2$ polynomials, how can they span a space of dimension $p-1$? $\endgroup$ Commented Feb 15, 2021 at 13:11
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    $\begingroup$ Then in fact no matrix computation is needed. The polynomial $\sum _{i=1}^{p-2} \frac{t^i}{i+1}$ is invertible in the ring $\frac{F_p[t]}{(t-1)^{p-1}}$ and hence the linear independence follows from the linear independence of the polynomials $1,t,\cdots, t^{p-2}$. $\endgroup$ Commented Feb 15, 2021 at 14:22

1 Answer 1

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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.

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