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Next day: apparently my original question is harder, by far, than the other bits. So: it is a finite check, I was able to confirm by computer that, if the polynomial below satisfies $$ f(a,b,c,d) \equiv 0 \pmod {27}, \;\; \mbox{THEN} \; \; a,b,c,d \equiv 0 \pmod 3, $$ and if $$ f(a,b,c,d) \equiv 0 \pmod {125}, \;\; \mbox{THEN} \; \; a,b,c,d \equiv 0 \pmod 5, $$

ORIGINAL: $f$ is a polynomial in four variables. Take matrices $$ 1 = \left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right), $$

$$ i = \left( \begin{array}{rrrr} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{array} \right), $$

$$ j = \left( \begin{array}{rrrr} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{array} \right), $$

$$ k = \left( \begin{array}{rrrr} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{array} \right), $$

Then take $$ f(a,b,c,d) = \det (a \cdot 1 + b \sqrt 3 i + c \sqrt 5 j + d \sqrt{15} k), $$ $$ =a^4-6 a^2 b^2+9 b^4-10 a^2 c^2-30 b^2 c^2+25 c^4+120 a b c d-30 a^2 d^2-90 b^2 d^2-150 c^2 d^2+225 d^4$$. Note that everything is commutative; $$ i^2 = 1, j^2 = 1, k^2 = 1; \; ij=ji=k, ki=ik=j,jk=kj=i. $$ It is also possible to re-write this with the square roots absorbed into the definitions of $i,j,k.$

So, questions include: does it make sense to anyone that, as I checked by brute force, that if $$ f(a,b,c,d) \equiv 0 \pmod {81} $$ then $a,b,c,d \equiv 0 \pmod 3?$ Same for $625$ and $5.$ Need to think about how to check $5$ completely.

Finally, is it true that this thing represents the same numbers as $x^2 - 15 y^2,$ and what is such a thing called anyway? It might be a field norm, I dunno.

Oh, from a closed question at https://math.stackexchange.com/questions/931769/integer-solution-to-diophantine-equations which I found interesting.

http://en.wikipedia.org/wiki/Go_I_Know_Not_Whither_and_Fetch_I_Know_Not_What

EDIT: It turns out we may use $27$ in place of $81.$ Evidently explaining this is the hard part. Confirmed, anyway. See what I can do with $125$ instead of $625.$

EDIT 2: Figured out how to program it; if the polynomial is divisible by $125,$ each variable is indeed divisible by $5.$

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    $\begingroup$ This is the norm of $a+b\sqrt{3}+c\sqrt{5}+d\sqrt{15}$ from the quartic field to the rationals, so it's a norm form. I don't think $1,\sqrt{3},\sqrt{5},\sqrt{15}$ form a basis of the ring of integers over the usual integers, so the congruences don't automatically follow from facts about norms, but it should not be hard to prove them that way. You can also use that to get information on numbers (integers?, rationals?) represented by it. I don't know how hard it would be to answer your last question. $\endgroup$ Sep 16, 2014 at 2:30
  • $\begingroup$ @FelipeVoloch, thanks. What about the first question, if the polynomial is divisble by $81,$ then all variables are divisible by $3?$ David suspects that is lots more work. $\endgroup$
    – Will Jagy
    Sep 16, 2014 at 2:39
  • $\begingroup$ That's what I meant by congruences and it's not automatic. By pure thought, if the value of polynomial is divisible by a suitably high power of three, then the variables are divisible by three. To figure out what suitably high means, requires a calculation which I won't do. $\endgroup$ Sep 16, 2014 at 2:45
  • $\begingroup$ @FelipeVoloch, thanks. I will finish it with computer, I need to fiddle a bit for the prime $5$ because the 4th powers times coefficients get a bit too large for ordinary C++ integers. I see what you mean about your use of the word congruences, I think I noticed that in your comment at first, then i got distracted by David's answer. $\endgroup$
    – Will Jagy
    Sep 16, 2014 at 2:48

1 Answer 1

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Yes, this is a field norm; it is the norm of $a + b \sqrt{3} + c \sqrt{5} + d \sqrt{15}$, from $K = \mathbb{Q}(\sqrt{3}, \sqrt{5})$ down to $\mathbb{Q}$. Note that $a+b \sqrt{3} + c \sqrt{5} + d \sqrt{15}$ acts on the basis $(1, \sqrt{3}, \sqrt{5}, \sqrt{15})$ by $$a \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} + b \begin{pmatrix} 0 & 3 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 5 \\ 0 & 0 & 1 & 0 \end{pmatrix} + c \begin{pmatrix} 0 & 0 & 5 & 0 \\ 0 & 0 & 0 & 5 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{pmatrix} + d \begin{pmatrix} 0 & 0 & 0 & 15 \\ 0 & 0 & 5 & 0 \\ 0 & 3 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{pmatrix}.$$ Now conjugate by the matrix whose diagonal entries are $(1, \sqrt{3}, \sqrt{5}, \sqrt{15})$ to get your matrix. The entries are no longer rational, so I can't think of the result as describing the action on $K$, but the determinant is the same.

$\mathbb{Q}(\sqrt{15})$ has class number $2$ and $K$ is the class field. So, for a prime $p$ other than $2$, $3$, $5$, we have that $\pm p$ is a value of $x^2-15 y^2$ if and only if $p$ splits principally in $\mathbb{Q}(\sqrt{15})$ if and only if $p$ splits in $K$ if and only if $\pm p$ is a value of $f$. Also, neither $x^2-15 y^2$ nor $f$ can be $3 \bmod 4$, so the sign is the same in the two cases.

However, they don't take the same set of composite values. Look at $-119 = 7 \times 17$. We have $61^2 - 15 \cdot 16^2 = -119$, but, if $7 | f(a,b,c,d)$ then $7^2 | f(a,b,c,d)$.

I found this by hunting for two primes which are non-principally split in $\mathbb{Q}(\sqrt{15})$. In terms of quadratic forms, which I know you love, I needed primes of the form $3 x^2 - 5 y^2$, and I found $7=3 \cdot 2^2 - 5$ and $-17 = 3 \cdot 6^2 - 5 \cdot 5^2$. Then their product was of the form $x^2-15 y^2$.

Since these primes split non principally in $\mathbb{Q}(\sqrt{15})$, they don't split further in the class field. (We can also directly compute $\left( \frac{3}{7} \right) = \left( \frac{3}{17} \right) = -1$.) So things divisible by one power of $7$ or $17$ are not norms from $K$.

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  • $\begingroup$ oh, very nice, David. I cannot tell whether you have explained the first thing, that when $f(a,b,c,d)$ is divisible by $81,$ all the variables are divisible by $3.$ $\endgroup$
    – Will Jagy
    Sep 16, 2014 at 2:34
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    $\begingroup$ I am honored to be confused with GH. $\endgroup$ Sep 16, 2014 at 2:46
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    $\begingroup$ It is good to be in good company, folks. $\endgroup$
    – GH from MO
    Sep 16, 2014 at 3:18
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    $\begingroup$ @FelipeVoloch Nonetheless, sorry about that. $\endgroup$ Sep 16, 2014 at 3:28
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    $\begingroup$ While ${\bf Z}(\sqrt{3},\sqrt{5})$ is not the full ring of integers $O_K$, we can recover $O_K$ by dividing some elements of ${\bf Z}(\sqrt{3},\sqrt{5})$ by $2$, so the distinction betwen ${\bf Z}(\sqrt{3},\sqrt{5})$ and $O_K$ shouldn't affect questions of divisibility by powers of $3$ and $5$. $\endgroup$ Sep 16, 2014 at 4:03

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