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Let $f(X,Y) \in \mathbb{Q}[X,Y]$ be a quadratic polynomial. The Hasse-Minkowski theorem says that $f(X,Y) = 0$ has a solution $(x,y) \in \mathbb{Q}^2$ iff it has a solution in $\mathbb{R}^2$ and $\mathbb{Q}^2 _p$ for every prime $p$, and this famously fails for higher-order forms like cubics, etc., with the failure measured by the Tate-Shafarevich group.

Is there a nice, simple geometric way to think about this?

Edit: More specifically, what does it "look like" when you piece together local solutions into a global solution? Is there any good way to visualize this?

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I mean, in some sense you should think about the primes of $\mathbb{Z}$ as being points on a curve, and the localization of $\mathbb{Z}$ at $p$ as "focusing in at $p$", or investigating the curve "at $p$". From this, the local to global principle says precisely that one has a property on this curve everywhere (globally) if you have the property at every point (locally). I don't know if this is what you wanted though? It seems more like you're asking literally what the solution looks like. I mean, would a good example of the type of thing you're looking for be the following analogy: a poly. –  Alex Youcis Oct 29 '13 at 0:24
    
$f(x_1,\ldots,x_n)\in\mathbb{Z}[x_1,\ldots,x_n]$ has a solution in $\mathbb{Z}_p$ if and only if has a solution in every $\mathbb{Z}/p^n\mathbb{Z}$? Here we can actually "see" the solution given the solutions for $\mathbb{Z}/p^n\mathbb{Z}$. We just form a tuple by weaving consistent solutions together. Do you mean literally that explicitly? –  Alex Youcis Oct 29 '13 at 0:25
    
Yes, that's exactly the sort of thing I'm looking for. Actually, my difficulty might come from something much more basic: I just don't have much intuition for the p-adic numbers, and so while it's clear what a real or rational point on a curve (like a conic section) is, I can't really picture a p-adic point. And yet, the procedure for constructing a rational solution given real and p-adic solutions seems pretty straight-forward, at least in method. –  AndrewG Oct 29 '13 at 9:08
    
The above theorem really tells you how to picture a $p$-adic point. If you wanted to check if something had a $\mathbb{Z}$-point, the first thing to check is whether or not it had a $\mathbb{Z}/p^n\mathbb{Z}$-point for all $n$. This is the obvious obstruction to having a $\mathbb{Z}$-point, at least as far as $p$ is concerned. Of course, these solutions should be consistent with each other (so that reduction maps the chosen solution to the chosen solution in a lower power of $p$). –  Alex Youcis Oct 29 '13 at 9:19
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It would be great if there was a single geometric object that could capture such a consistent $\mathbb{Z}/p^n\mathbb{Z}$ solution for all $n$--this object is precisely $\mathbb{Z}_p$ (as the stated theorem I told you showed). This is literally how I think about $\mathbb{Z}_p$. It's the geometric space where consistent solutions to a system of equations in every $\mathbb{Z}/p^n\mathbb{Z}$ naturally live. –  Alex Youcis Oct 29 '13 at 9:20
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1 Answer 1

Here is one way in which you can interpret the Hasse-Minkowski result "geometrically."

Proposition. Let $X$ be a smooth conic over a number field $K$. The following are equivalent:

  1. The set $X(K)$ of $K$-rational points of $X$ is nonempty.
  2. The curve $X$ is isomorphic over $K$ to $\textbf{P}^1$.

The equivalence follows from an application of the Riemann-Roch theorem. Pick a rational point $x \in X(K)$. There is a rational function with only a simple pole at $x$. Such a function gives an isomorphism between $X$ and $\textbf{P}^1$ defined over $K$.

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