# How to see that the determinant of this matrix is nonzero for all primes?

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

• 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 $3,5,7,13$. – Noam D. Elkies Feb 15 at 5:41
• 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 – Francesco Polizzi Feb 15 at 6:33
• If there were one more row / column it would be a circulant matrix. – Tim Campion Feb 15 at 6:35
• But we have only $p-2$ polynomials, how can they span a space of dimension $p-1$? – Fedor Petrov Feb 15 at 13:11
• 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 $1,t,\cdots, t^{p-2}$. – Venkataramana Feb 15 at 14:22

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.