3 ms·
Trouble starts when your field doesn't always have a solution for f(x_1,...,x_n) = y^2, like rational numbers (but I'm sure you can find more examples). But may
by enedil 3y ago
Trouble starts when your field doesn't always have a solution for f(x_1,...,x_n) = y^2, like rational numbers (but I'm sure you can find more examples).
But maybe that can be mitigated as well?
- JadeNB 3y agoI'm not an expert in this area, but, precisely for the reason you mention, I'd expect it's easier to solve a polynomial inequality over ℚ by solving it over ℝ and intersecting down to ℚ, rather than by working directly over ℚ.
- macrolocal 3y agoAs long as you don’t bump into Gal(ℝ/ℚ) somewhere along the way.