3

I have annoyed various faculty members enough with this, so I will turn to you to check that my proof of Singer's Theorem is satisfactory. I have been politely told before on Math.SE that I give way too much background in questions, so I will just dive right in:

Theorem. Let $q$ be a prime power and $m\geq 1$ an integer. Then the symmetric design defined by the points and hyperplanes of $PG(m,q)$ has an automorphism $\varphi$ of order $(q^{m+1}-1)/(q-1)$, and $G=\langle \varphi \rangle$ is a cyclic group acting regularly on points of $PG(m,q)$.

Proof. The group $GF^*(q^{m+1})$ is cyclic and $GF(q^{m+1})$ is an $(m+1)$-dimensional vector space over $GF(q)$. Let $V=GF^*(q^{m+1})/GF^*(q)$. Suppose $\omega$ is a generator for $GF(q^{m+1})$. Then $\langle \omega\rangle$ is a cyclic group under multiplication, and $GF^*(q)\leq \langle \omega\rangle$. Furthermore, $|GF^*(q)|=q-1$. In other words, if we let $v=(q^{m+1}-1)/(q-1)$, then $\langle \omega^v \rangle=\{1,\omega^v,\omega^{2v},...,\omega^{(q-2)v}\}$, and as such, $GF(q)=\{0,1,\omega^v,\omega^{2v},...,\omega^{(q-2)v}\}$. Treat $GF(q^{m+1})$ as a vector space over $GF(q)$. The elements $\omega^i$ and $\omega^j$ for $i,j\in\mathbb{N}$ span the same $1$-dimensional subspace of $GF^*(q^{m+1})$ if and only if there exists $\alpha\in GF^\star(q)$ such that $\omega^i=\alpha\omega^j$. In this way, we establish a bijection between the $1$-dimensional subspaces of $GF(q^{m+1})$ and the cosets $x_i$ of $GF(q)$, where $x_i:=\{0,\omega^i,\omega^{v+i},\omega^{2v+i},...,\omega^{(q-2)v+i}\}$.

Suppose $\varphi$ is a map such that $x_i\mapsto x_{i+1}$, where the subscripts are modulo $v$ (i.e., $x_{v-1}\mapsto x_0$). Then $\varphi$ is an automorphism of $V$ and $\langle\varphi\rangle$ is obviously regular on $V$.

----NOTE: The following part is what I am not sure of. It seems way too cheap after all that setup.

To show that $\varphi$ is an automorphism of the symmetric design defined by the points and hyperplanes of $PG(m,q)$, consider the obvious fact that $\varphi(x_i)=\omega x_i$. This clearly maps blocks to blocks (that is, hyperplanes to hyperplanes) and preserves incidence of the points and hyperplanes, and is hence an automorphism of the design. $\square$

So have I done it?

Community
  • 1
The Count
  • 3,668
  • 1
    @MorganRodgers Yes, I am trying to write a more explicit version of the construction given in elementary texts like van Lint and Wilson. This stuff is too often assumed trivial or easy and I want my thesis to be highly readable for anyone like me that stumbles upon it in the future. Thanks very much! – The Count Jun 17 '17 at 14:13
  • @MorganRodgers I thought you might like to know that I got compliments on this proof, specifically. I said a fellow on Stack Exchange helped me check it. Thanks again! – The Count Jul 19 '17 at 20:26

1 Answers1

2

Singer only shows it for planes and is quite explicit but the idea is the same (and your modern proof is much cleaner looking and more understandable): viewing the points of the projective space $\mathrm{PG}(n-1,q)$ as nonzero elements of $\mathbb{F}_{q^{n}}$, and using the fact that the multiplicative group of the field is cyclic. We actually have that if $\omega$ is a generator of the field, then the first $(q^{n}+1)/(q+1)$ powers of $\omega$ are pairwise linearly independent over $\mathbb{F}_{q}$. Also, multiplication by any element of $\mathbb{F}_{q^{n}}$ is a linear operation (in particular, is $\mathbb{F}_{q}$-linear). Very nice proof.

(https://pdfs.semanticscholar.org/ef43/358b151dfef7af486a4e40e4056f1e0a8bde.pdf)

xxxxxxxxx
  • 13,688