I am having trouble with Silverman's exercise 1.10(b). The converse of (a) is easy because there is no integer solution to the equation when $p \equiv 3$ mod $4$. However, this method does not work for (b) because we always get empty set when intersecting whatever $V_p (p \equiv 3 $ mod $4)$ with $\mathbb{Q}$. I tried to intersect two varieties $V_p, V_q$ with something else, for example $\mathbb{Q}({\sqrt{p}})$ and try to tell the difference, but this is hard since it is very possible for $V_q$ to have nonempty intersection with $\mathbb{Q}({\sqrt{p}})$ and I think it is true that if so then $V_p, V_q$ are isomorphic over $\mathbb{Q}({\sqrt{p}})$. Are there some invariants I can use in this circumstance to show that two varieties are not isomorphic over $\mathbb{Q}$?
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6$\begingroup$ The chance that someone has Silverman open in front of them (or even on hand) when reading this is pretty low; you would be more likely to get an answer if you actually wrote out the question. $\endgroup$– tracingJan 20, 2015 at 1:14
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$\begingroup$ Thanks a lot for the hint! I got a hint from my instructor and have solved the problem. $\endgroup$– T.C YangJan 20, 2015 at 2:20
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1$\begingroup$ actually, posting that hint for further people searching for that problem could have been helpful $\endgroup$– Stijn CambieJun 27, 2016 at 18:24
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2 Answers
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Using the Quadratic Reprocity Law, $\Big(\frac{p}{q}\Big)\Big(\frac{q}{p}\Big)=(-1)^\frac{(q-1)(p-1)}{4}=-1$, which implies that one of them (let's say p) is a square mod the other one, so $x^2+y^2=pz^2$ has solution in $\mathbf{F}_{q}$ because the equation $x^2+y^2=z^2$ has non trivial solution in $\mathbf{Z}$ but $x^2+y^2=qz^2$ has no solution in $\mathbf{F}_{q}$ because $q\equiv3\mod 4$ so they cannot be isomorphic over $\mathbf{Q}$.
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$\begingroup$ There is a question: actually all elements in finite fields can be written as a sum of two squares, so your discussion about Quadratic reciprocity law and so on is redundant—— you can directly get that they have different number of solutions without assuming $p,q\equiv 3 \pmod 4$. Besides, I don't think you can deduce they are not isomorphic from their difference in number of solutions. $\endgroup$– RichardMay 24, 2022 at 13:58
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You could calculate the Hilbert symbols $(\frac{p,-1}{q})_2$ for varying primes $q$.
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$\begingroup$ This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review $\endgroup$ Jul 10, 2016 at 10:47
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$\begingroup$ Why does this not provide an answer to the question? $\endgroup$– user3267Jul 10, 2016 at 10:52