I'm trying to show (in-)equivalence of two quadratic forms and got stuck calculating the Hasse-Minkowski invariant. After diagonalisation of the quadratic form I get that $c(f)=(1,1)$ and $c(g)=(2,\frac{1}{2}(7-\sqrt{17})(2,\frac{1}{2}(7+\sqrt{17})(\frac{1}{2}(7-\sqrt{17},\frac{1}{2}(7+\sqrt{17})$.

My question: How can I calculate $c(f)$ and $c(g)$? Especially $c(g)$


1 Answer 1


It would have been nicer to tell us what is your base field $K$. Is $K=\mathbb{Q}(\sqrt{17})$ ?

We have $(u,v)=1$ (or $0$, depending on your convention) if and only if $\langle 1,-u,-v\rangle$ is isotropic, so $c(f)=1$.

For $c(g)$: the symbol is bimultiplicative and only depends on square classes of $a$ and $b$. Setting $a=\frac{7-\sqrt{17}}{2}$ and $b=\frac{7+\sqrt{17}}{2}$, we get $c(g)=(2,a)(2,b)(a,b)=(2,ab)(a,b)$. Now $ab=8=2^2 2$ so, $c(g)=(2,2)(a,b)$.

The form $\langle 1,-2,-2\rangle$ is isotropic, since the vector $(2,1,1)$ is isotropic, so $(2,2)=1$.

Thus $c(g)=(a,b)$. Since $b=a(ab)$ mod squares, $(a,b)=(a,8a)=(a,2a)$.

Now you can check easily that for any $a$, $(a,a)=(a,-1)$ since $\langle 1,-a,-a\rangle$ is isotropic iff $\langle -a,a^2,a^2\rangle\simeq \langle 1,-a,-(-1)\rangle$ is.

Finally, $c(g)=(a,2)(a,a)=(a,2)(a,-1)=(a,-2)$.

It remains to check if $(a,-2)$ is trivial or not. The best way to do it is to work over a completion wrt to some prime ideal. If you want something potentially non trivial, you will have to try a prime lying above $a$ or $2$. For nondyadic primes, you have known formulas that you may find in the literature, but you can also proceed by hand.

You need to check if $x^2-ay^2+2z^2=0$ has a non trivial solution. You can always assume that $x,y,z$ lies in the ring of integers $O_K$, and mod out by a prime ideal dividing $a$ or $2$. Then you will be reduced to check if something is a square in a finite field.

Since you did not give enough information on your base field, I cannot do much more, so I let you perform this very last step.

Note that it is known that $(a,b)=1$ if and only if the local invariants (that is, the invariants obtained after completion to every prime ideal and $\infty$) are all equal to 1 except maybe for a single exception.

Since $(a,-2)_\mathfrak{p}=1$ for all $\mathfrak{p}$ not dividing $2a$, you are reduced to a finite amount of computations: you will have to try the prime ideals dividing $2a$ and $\infty$.

(If $K=\mathbb{Q}(\sqrt{17})$, note that $a=\pi^3 u$, where $\pi=\frac{-3+\sqrt{17}}{2}$ is irreducible in $O_K$ and $u=4+\sqrt{17}$ is a unit, and that $2=\pi \bar{\pi}$. Hence, $(a,-2)=(\pi u,\pi \bar{\pi})$ )


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