Quadratic ternary forms

Question

What is the difference between solubility, local solubility and global solubility when it comes to solving quadratic ternary normal forms, i.e a equation of the form $ax^2 + by^2 + cz^2 =0$?

Thanks in advance.

Answer 1

Global solvability means solvable over a global field (such as a number field or function field), and local solvability means over a local field (such as finite extension of a $p$-adic field). The theorem of Hasse -Minkowski for quadratic forms gives a relation: If $K$ is a global field and $f(x)=a_1x_1^2+\cdots a_nx_n^2$ is a polynomial with coefficients in $K^{\ast}$, then $f(x)=0$ has a non-trivial solution over $K$ if and only if it has a non-trivial solution in every local field arising as completion of $K$ with respect to the absolute valuation. We can take $K=\mathbf{Q}$, then solvability over $\mathbf{Q}$ is equivalent to solvability over $\mathbf{R}$ and all $\mathbf{Q}_p$. For example, the termary quadratic form $Q(x)=5x^2+7y^2-13z^2$ has a non-trivial (global) rational solution, since it has a local solution over $\mathbf{Q}_p$ for every prime $p$ (and a real solution, of course). On the other hand, we could have seen this much easier (by using Lagrange, or direct verification, e.g., $(x,y,z)=(3,1,2)$.