3
$\begingroup$

Let $q$ be a natural number (the first cases of interest being $q = 10,12$ or $15$), and let $n = q^2+q+1$. Also, let $I_n$ be the $n\times n$ identity matrix, and let $A_n$ be the $n\times n$ diagonal matrix having coefficient $1,q,...,q$ on the diagonal.

Note that $I_n$ and $A_n$ are always congruent over the $p$-adic integers for all $p$ not dividing $q$. Further assume that $I_n$ and $A_n$ are congruent over $\mathbf{Q}_p$ for $p$ dividing $q$ (which is the case when $q=10,12$ or $15$). Does it follow that $I_n$ is congruent to $A_n$ over $\mathbf{Z}[1/q]$ ?

$\textbf{Motivation}$: A friend explained the paper "On the non-existence of certain finite projective planes", and a naïve hope is to push a bit further their strategy by answering the above question in the negative. This has for sure already been tried, and I would like to know some references about that.

$\textbf{Edit}$:In view of the question What's in the genus of the cubic lattice?, let me restate my question: the two quadratic forms over $\mathbf{Z}$ given by $\sum \limits_{i=1}^{111} x_i^2$ and $x_1^2+ \sum \limits_{i=2}^{111} 10x_i^2$ are in the same genus. But are they actually isomorphic over $\mathbf{Z}[1/10]$? (As Noam notes in the comments, they cannot be isomorphic over $\mathbf{Z}$ as determinant do not match). Can I hope for an answer from Magma, or would it be overwhelmed by the number of variables ?

Improve this question
$\endgroup$
8
  • $\begingroup$ They can't literally be isomorphic over $\bf Z$ because the determinants don't match. $\endgroup$
    – Noam D. Elkies
    Commented Dec 11, 2016 at 23:26
  • 2
    $\begingroup$ If we may invert $10$ then I don't need Magma because I can use $55$ copies of the identity $(x+3y)^2 + (3x-y)^2 = 10(x^2+y^2)$. $\endgroup$
    – Noam D. Elkies
    Commented Dec 11, 2016 at 23:47
  • 2
    $\begingroup$ @Noam How did you find that ? Because if there is a technique, there are other values of the parameters that are of great interest to me, as explained in the main question. Can you actually say something about the general question ? And could Magma help at this level of magnitude if necessary ? Anyway, thank you already for this wonderful answer. That's in some sense a bad news, but that was to be expected anyway... $\endgroup$
    – thierry stulemeijer
    Commented Dec 11, 2016 at 23:57
  • 3
    $\begingroup$ @thierrystulemeijer note that $10 = 1^2 + 3^2 = {\rm N}(1+3i)$ and $(1+3i)(x-iy) = (x+3y)+(3x-y)i$. Take the norm of both sides. $\endgroup$
    – KConrad
    Commented Dec 12, 2016 at 0:18
  • 1
    $\begingroup$ It's much the same thing, using $39$ copies of the formula for multiplying a Hamilton quaternion by say $2(1+i+j)$ or $3+i+j+k$ (and likewise for other $q \equiv 0,3 \bmod 4$, via the four-square theorem). $\endgroup$
    – Noam D. Elkies
    Commented Dec 12, 2016 at 1:05

1 Answer 1

Reset to default
3
$\begingroup$

The answer to the main question is yes ! (The question has been answered by the commentators).

As a consequence, it is not possible to push the Bruck-Ryser theorem by looking at congruences over rings of the form $\mathbf{Z}[1/q]$.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .