Local to global principle for a pair of bilinear equations?

Let $$A_{i, j}, B_{i, j}, C, D \in \mathbb{Q}$$, and consider the following pair of equations $$A_{1, 1} x_1 y_1 + A_{1, 2} x_1 y_2 + A_{2, 1} x_2 y_1 + A_{2, 2} x_2 y_2 = C$$ $$B_{1, 1} x_1 y_1 + B_{1, 2} x_1 y_2 + B_{2, 1} x_2 y_1 + B_{2, 2} x_2 y_2 = D.$$ I was interested in figuring out if this system of equations satisfies the local to global principle. In other words, suppose I can find a solution $$(\mathbf{x}, \mathbf{y})$$ of the system in $$\mathbb{R}^4$$ and in $$\mathbb{Q}^4_p$$ for all primes $$p$$, then there exists a solution $$(\mathbf{x}, \mathbf{y}) \in \mathbb{Q}^4$$.

I would appreciate any comments or suggestions or counterexamples. Thank you very much!

• Intersections of two quadrics in $\mathbb{P}^4$ are del Pezzo surfaces of degree 4. For those there are counter-examples to the Hasse principle going back to Birch and Swinnerton-Dyer. I don't know if any known example has this particular form. There are more knowledgable people active here who will give a better answer. Commented May 1, 2020 at 7:28
• Actually, what the OP is looking at is the intersection of 2 planes with a smooth quadric $Q\subset \mathbb{P}^3$ -- that is, the intersection of a line with $Q$. This is an ordinary second degree equation, which certainly satisfies the local to global principle.
– abx
Commented May 1, 2020 at 8:06
• @abx, I had the same thought until I realised that C and D are not necessarily zero... the projective closure is a dP4 with four singular points at infinity. Commented May 1, 2020 at 8:15
• May be I'm extremely stupid and missing something obvious, but jf we just pick $y$'s at random we get two linear equations in $x$. If the determinant is zero for all $y$'s then either equations are proportional in which case we have only one equation which always have solutions or both of our equations factor as $(ax_1+bx_2)(cy_1+dy_2)$ and moreover $a$ and $b$ are the same for them. But doing the same thing switching $x$ and $y$ shows that $c$ and $d$ are the same as well and we are back to the proportionality case. All in all solution either exists or not in all fields of characteristics $0$ Commented May 1, 2020 at 9:58
• I think that Aleksei may have a point; The blow-up of $X$ in a rational point not on a line is a cubic surface with $2$ skew lines; these are all well-known to be rational, so it seems that $X$ is actually rational, rather than just unirational. This means that there should actually be a complete parametrisation of all the solutions, which I expect is what Aleksei's approach is leading to (writing down an explicit complete parametrisation). Commented May 1, 2020 at 11:13

Let $$X$$ denote the closure of your variety in $$\mathbb{P}^4_{\mathbb{Q}}$$. Explicitly: \begin{align*} &A_{1, 1} x_1 y_1 + A_{1, 2} x_1 y_2 + A_{2, 1} x_2 y_1 + A_{2, 2} x_2 y_2 = C z^2 \\ &B_{1, 1} x_1 y_1 + B_{1, 2} x_1 y_2 + B_{2, 1} x_2 y_1 + B_{2, 2} x_2 y_2 = D z^2. \end{align*} As remarked in comments, this defines a del Pezzo surface of degree $$4$$ with $$4$$ singular points at infinity (for general choices of coefficients).
This contains the $$2$$ lines at infinity $$x_1=x_2=z=0, \quad y_1=y_2=z=0.$$ In particular $$X$$ contains a smooth rational point (choose a point on a line which is not one of the $$4$$ singular points). But it is well-known that a singular del Pezzo surface of degree $$4$$ with a smooth rational point is unirational (see e.g. [1, Theorem B]). Thus the rational points are Zariski dense, so there is a rational point with $$z \neq 0$$, which gives a solution to the original equations.
Here by "general", to apply [1, Theorem B] one needs to know that $$X$$ is irreducible with only finitely many singular points and is not a cone.