# Thursday, August 19 :::{.remark} Some useful prerequisites: - Number theory (e.g. places) - Class field theory - See Cassels-Frolich (up through ch. 5 and 6) - AG (although we'll avoid the language of schemes) - Galois and group cohomology - Bjorn Poonen's book ::: :::{.remark} On notation: - $k\powers{n}$ will denote $n$th powers in $k$, and similarly for $k\units$. - $k^\unram$ denotes an unramified extension. ::: :::{.remark} Setup: let $k = \QQ$ or more generally a number field or a function field over $\FF_q$. Consider a system of polynomial equations over $\kx{m}$: \[ \begin{cases} f_1(x_1, \cdots, x_m) & = 0 \\ \quad \vdots & \vdots \\ f_n(x_1, \cdots, x_m)& =0 . \end{cases} \] Some natural questions: ::: :::{.remark title="Topic 1: Are there any common solutions?"} More generally, does $X \da V(f_1, \cdots, f_n)$ have any rational points? How many rational points are there? Finitely many, or infinitely many? ::: :::{.remark title="Topic 2: what is the distribution of points?"} - How many points are there of height at most $N$, where $\height(a/b) = \max(\abs a, \abs b)$? - Are they Zariski dense? I.e. are there solutions outside of the ideal $\gens{f_i}$? - Are they *potentially* dense, i.e. dense after some finite extension $k\embeds k'$? - Choosing $k\injects \CC$ or $\QQpadic$, are the solutions dense in the analytic topology on $X(\CC), X(\QQpadic)$? If not, what is the closure? There are many conjectures around these questions, but few general results! ::: :::{.remark title="Topic 3: Local to Global Principles"} Topic 3: local to global principles. Given $X\slice{\QQ}$, if $X(\QQpadic)\neq \emptyset$ for all $p$ and $X(\RR) \neq \emptyset$, does this imply that $X(\QQ) \neq \emptyset$? More generally, for $X\slice{k}$ with $X(K_v) \neq \emptyset$ for all places $v$ of $K$, is this enough to imply $X(k)\neq \emptyset$ If so, we say $X$ satisfies the **Hasse principle**. If not, are there obstructions? ::: :::{.remark title="Topic 3': Weak and Strong Approximation"} As an example, \[ X(k) \injects \prod_{v\in P(k)} X(k_v) \] where $p(k)$ are the places of $k$. Is this map dense? Note the topology is the product topology, so a basis for opens are sets with finitely factors with opens, and the remaining are the entire space. Strong approximation is an adelic version of this. Obstructions to this principle: if this is not dense, what is the closure $X(k)$ in $\prod X(k_v)$ or $X(\AA)$ for $\AA$ the adeles? One example we'll consider is the Brauer-Manin obstruction. ::: :::{.remark title="Topic 4: effectiveness and decidability questions."} Given a variety $X\slice{\QQ}$, is there an actual algorithm that decides if $X(\QQ)= \emptyset$? This is known over $\ZZ$, but open for $\QQ$ and most (not all) number fields. Are there special classes of varieties where the answer of yes? For curves, this is only known contingent on open problems (the abc conjecture, the section conjecture, Birch-Swinnerton-Dyer, etc). Given a special $X\slice{k}$ can you find $X(k)$? ::: :::{.remark} Other possible topics: - The Mordell-Weil theorem for $X$ an abelian variety, and a generalization, the Néron-Lang theorem which works over other fields. - Falting's theorem, that curves of genus 2 have finitely many rational points. ::: ## Examples of Hasse Principles :::{.example title="?"} Let $a\in \QQ$, does $x^2 = a$ satisfy a local to global principle? This is related to Chebotarev density. Claim: any positive number $a$ such that $v_p(a)$ is even for all $p$ is necessarily a square. This follows from writing $a = \pm \prod p_i^{n_i}$ where $n_i\in \ZZ$ and is equal to zero for all but finitely many $i$, then its square root is obtained by halving all of the $n_i$. Note that $a \in (\RR\units)^2$ implies $a$ is positive, and $a\in (\QQpadic\units)^2$ implies that $n_p$ is even. ::: :::{.example title="?"} Let $a\in \QQ$ and take $x^n=a$, or more generally $f(x) = a$ for $f\in \QQ[x]$, where $f(x) - a$ is irreducible. Corollary of Chebotarev density: the set of primes where $f-a\mod p$ has no linear factors has positive density. This means that an even stronger theorem is true: there exists a $c<1$ such that if $f-a$ has no roots $\mod p$ for a set of primes of density $d>c$, then $f-a$ has no roots. So this satisfies the Hasse principle. ::: :::{.example title="Conics"} Take $X \da V(ax^2+by^2 +cz^2) \subseteq \PP^2$ for $a,b,c\in \QQ$. This also satisfies the Hasse principle, but the proof is harder. Note that $x^2+y^2+z^2 = 0$ has no rational points (excluding zero since we're in $\PP^2$) since it has no solutions over $\RR$. It is potentially dense, noting that one can take $\QQ[i]$ over $\QQ$ and get rational points $0, 1, \infty$. Given one point, one can stereographically project to yield infinite many points by just taking lines through the fixed point and letting slopes vary. \todo[inline]{Something about using $\OO(1)$ to give an embedding into $\PP^1$. Start with $\OO(-1)$, dualize, project?} ::: :::{.example title="Severi-Brauer varieties"} Taking $X\slice{k}$ such that $X\slice{\bar k} \cong \PP^n\slice{\bar k}$ satisfy the Hasse principle. ::: :::{.example title="Quadrics"} A theorem by Hasse-Minkowski shows that these also satisfy the Hasse principle. ::: :::{.example title="Genus 1 curves"} The **Selmer curve** $3x^3 + 4y^3 + 5z^3=0$ does *not* satisfy the Hasse principle, which can be understood in terms of the Tate-Shafarevich group or Brauer-Manin obstructions. ::: :::{.remark} Note that it doesn't make sense to say a single variety satisfies the Hasse principle, but rather a class. But it makes sense to say a single variety *doesn't*. ::: :::{.remark} A common generalization is that these are all torsors for an algebraic group, i.e. a homogeneous space, for which there are cohomological methods to understand the Hasse principle. ::: :::{.remark} A variety $X\slice{k}$ is *geometrically integral* in the affine case if when $X = V(f_1,\cdots, f_n)$, the ring $\bar{k}[x_1,\cdots, x_n]$ is an integral domain. ::: :::{.theorem title="?"} Suppose $K$ is a number field and $X\slice{K}$ is geometrically integral. Then $X(K_v) \neq \emptyset$ for all but finitely many $v$. ::: :::{.proof title="Sketch/idea"} \envlist 1. Write $X = V(f_1, \cdots, f_n)$ with a nonempty smooth locus $X^{\smooth} \subseteq X$ which is a variety (just adjoin inverses of partial derivatives appearing in minors of Jacobian matrices). So $X^{\smooth}/\OO_{K, S} = \OO_K\left[ {1\over N} \right]$ which is smooth over $\OO_{K, S}$ 2. Use Lang-Weil to show that $X^\smooth(\OO_{K, S} / \mfp) \neq \emptyset$ for almost all $\mfp$. 3. Use smoothness and Hensel's lemma to get $X^{\smooth }(\OO_{K, S}\complete{\mfp})$. :::