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Let $\bar{S}$ by the closure of S in $\mathbb{P}^n(\mathbb{R})$. If a polynomial with zero constant term is bounded on S, then its highest degree term vanishes on $S':=\bar{S} \setminus S \subset \mat …

I do not know about the local-to-global principle for testing rationality of a surface, but weakening "rational" to "unirational" it certainly fails. Indeed, del Pezzo surfaces violating the Hasse pr …

If you can check that the curve is geometrically irreducible, then you may try using the Hurwitz formula (you may use the formula in any case, but you would have to be more careful with the conclusion …

@Dror: I will address your last comment in the question about bounding the genus from below. Here I wanted to add a remark: the explicit bound you found is such that any prime larger than this bound …

This might be slightly off-topic, but I am surprised that no one mentioned Gauss' Quadratic Reciprocity: this is one of the very few fundamental results that allows you to deduce something about one p …

Here is one easy case of a curve of genus two whose Jacobian is isogenous to a product of two elliptic curves. Let $p,q,r$ be homogeneous separable polynomials of degree two in two variables $s,t$ th …

Just an expansion on my comment. I will assume that exact sequences need to have half the number of indices of the whole sequence. Then a sequence is exact for some choice of weights if and only if it …

Surely you can find the solutions to the equation if ab=0. Otherwise set $a^2=1/x$ and $b^2=1/y$ to turn your equation into the equation $1+x+y=0$. Is this enough?