# Calculate Hasse-Minkowski invariant

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 $$c(g)$$

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, $$(a,-2)=(\pi u,\pi \bar{\pi})$$ )