Heegner Points and Binary Quadratic Forms - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T19:37:23Z http://mathoverflow.net/feeds/question/110351 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/110351/heegner-points-and-binary-quadratic-forms Heegner Points and Binary Quadratic Forms unknown (google) 2012-10-22T17:49:48Z 2012-10-23T00:47:09Z <p>I've been trying to read Gross' paper on Heegner points on $X_0(N)$ and I am stuck on a few details. The definition he is working with is that a heegner points is a pair $y=(E,E')$, where $E$ and $E'$ are elliptic curves admitting an isogeny that has cyclic kernel of order $N$ and where $E$ and $E'$ both have complex multiplication by the order $\mathcal{O}$ of discriminant $D$ in a quadratic imaginary field $K$. Gross goes on to explain that we may assume the lattice for $E$ is a fractional ideal $\mathfrak{a}$ and the lattice for $E'$ is $\mathfrak{b}$ such that the ideal $\mathfrak{n}=\mathfrak{a}\mathfrak{b}^{-1}$ is proper ideal of $\mathcal{O}$ such that the quotient $\mathcal{O}/\mathfrak{n}$ is cyclic of order $N$. It is the next line that I don't understand:</p> <p>"Such an ideal will exist if and only if there is a primitive binary quadratic form of discriminant $D$ which properly represents $N$...". The line goes on, but this is one of the things I'm stuck on. I've tried googling some notes/papers on binary quadratic forms, but I can't find anything that helps me understand what a binary quadratic form representing $N$ has to say about an order admitting a cyclic quotient. An explanation or a good reference would be much appreciated. </p> <p>The second and, I think, more important part of my confusion is a bit later on in the same section: Gross goes on to explain that if we have such an $\mathfrak{n}$, we can construct a heegner point as follows. Let $\mathfrak{a}$ be an invertible $\mathcal{O}$-submodule of $K$ and let $[\mathfrak{a}]$ denotes its class in $Pic(\mathcal{O})$. Let $\mathfrak{n}$ be a proper $\mathcal{O}$-ideal with cyclic quotient of order $N$, put $E=\mathbf{C}/\mathfrak{a}$, $E'=\mathbf{C}/\mathfrak{a}\mathfrak{n}^{-1}$. They are related by an obvious isogeny and thus determine a Heegner point, denoted $(\mathcal{O},\mathfrak{n},[\mathfrak{a}])$. </p> <p>Next, given $y=(\mathcal{O},\mathfrak{n},[\mathfrak{a}])$, we can find the image of it in the upper-half plane by picking an oriented basis $\langle\omega_1,\omega_2\rangle$ of $\mathfrak{a}$ such that $\mathfrak{a}\mathfrak{n}^{-1}=\langle\omega_1,\omega_2/N\rangle$. Then $y$ corresponds to the orbit of $\omega_1/\omega_2$ under $\Gamma_0(N)$. Lastly, since $\tau\in K$ it follows that it satisfies $A\tau^2+b\tau+C=0$ for some integers $A,B,C$ such that $gcd(A,B,C)=1$. </p> <p>Finally, what I don't understand is that Gross claims that $D=B^2-4AC, A=NA'$ from some $A'$ and $gcd(A',B,NC)$. I don't see what the $\tau$ we cooked up has to do with the discriminant of our order. I have read a paper that defined a Heegner point to be a quadratic imaginary point in the half-plane such that $\Delta(\tau)=\Delta(N\tau)$. I have seen how this would help with part of the claim above, but I don't see why in this situation, $\Delta(\tau)=\Delta(N\tau)$. In fact, it seems that everything I'm confused about here is the fact that it seems to be the case that $$D=\Delta(\tau)=\Delta(NT),$$ where $\Delta$ denotes discriminant.</p> <p>Any insight into these two questions would very appreciated.</p> http://mathoverflow.net/questions/110351/heegner-points-and-binary-quadratic-forms/110358#110358 Answer by Faisal for Heegner Points and Binary Quadratic Forms Faisal 2012-10-22T19:38:26Z 2012-10-22T23:42:16Z <p>I think the following facts, which you can find in Cox's book <em>Primes of the Form $x^2+ny^2$</em>, will alleviate your confusion. First off, if ${\mathfrak a}=[\alpha,\beta]$ is a proper ideal of <code>${\mathcal O}$</code> then one can show that $$f(x,y) := \frac{N(\alpha x-\beta y)}{N{\mathfrak a}}$$ is a primitive binary quadratic form of discriminant <code>$D = {\rm disc}(\mathcal O)$</code>. Moreover, the map that associates such an ${\mathfrak a}$ to such an <code>$f(x,y)$</code> induces an isomorphism from the class group <code>${\rm Pic } ({\mathcal O})$</code> onto the form class group <code>$C(D)$</code>. The inverse of this map is given by $$f(x,y) := ax^2+bxy+cy^2 \mapsto [a,(-b+\sqrt{D})/2] = [a,a\tau],$$ where <code>$\tau$</code> is the unique point in the upper-half plane such that <code>$f(\tau,1)=0$</code>. It's not hard to show that we'll have <code>${\mathcal O} = [1,a\tau]$</code> for all such <code>$\tau$</code> (see Addendum below). In particular, we see that <code>${\mathcal O}/[a,a\tau] \cong {\mathbb Z}/a{\mathbb Z}$</code> is cyclic.</p> <p>The last piece of the puzzle is this: a positive integer <code>$N$</code> is represented by a form <code>$f(x,y)$</code> in <code>$C(D)$</code> if and only if <code>$N$</code> is the norm of some ideal in the corresponding ideal class in <code>${\rm Pic}({\mathcal O})$</code> (loc. cit., Theorem 7.7(iii)). On the other hand, <code>$N$</code> is properly represented by such an <code>$f(x,y)$</code> if and only if <code>$f(x,y)$</code> is properly equivalent to <code>$Nx^2+bxy+cy^2$</code> for some <code>$b,c \in {\mathbb Z}$</code>. Now the results mentioned in the preceding paragraph will take you home.</p> <p><strong>Addendum:</strong> Given a proper ideal $\mathfrak a$ of <code>$\mathcal O$</code>, we can recover <code>$\mathcal O$</code> as the set <code>${\mathfrak a}^\vee = \{x \in K \mid x\mathfrak a \subset \mathfrak a \}$</code>. This last set is easy to compute in the following special case. Let <code>$K=\mathbb Q(\tau)$</code> be quadratic and suppose that <code>$ax^2+bx+c$</code> is the minimal polynomial of <code>$\tau$</code>, where <code>$a,b,c$</code> are coprime integers. Then <code>$[1,\tau]^\vee = [1,a\tau]$</code> (loc. cit., Lemma 7.5).</p> <p>By applying this to Gross's $\mathfrak a = [\omega_1, \omega_2] = \omega_2 [\tau, 1]$, which is a proper ideal in some order <code>$\mathcal O$</code>, we find that <code>$\mathcal O = [A\tau, 1]$</code>. Consequently,<code>$$D = {\rm disc}({\mathcal O}) = \det \begin{pmatrix}1 &amp; A\tau \\ 1 &amp; A\bar{\tau} \end{pmatrix}^2 = B^2 - 4AC.$$</code> The assertion about <code>$A$</code> can be gotten in a similar manner.</p>