$\DeclareMathOperator\SL{SL}\DeclareMathOperator\Aut{Aut}$Fix $d < 0$, a fundamental quadratic discriminant and $n$ a positive integer. Suppose $Q$ is a primitive binary quadratic form of discriminant $d$. Let us define the following,

  1. Representations of $n$ by $Q$: $R(Q, n) = \{ (x, y) \in \mathbb{Z}^2 | Q(x, y) = n\}$
  2. $\Aut(Q)$ as the subgroup of $\SL_2(\mathbb{Z})$ that fixes $Q$ under the usual action of $\SL_2(\mathbb{Z})$ on binary quadratic forms.
  3. $R(d, n) = \coprod_QR(Q, n)/\Aut(Q)$ as $Q$ runs through a complete set of representatives for each class of properly equivalent forms. Typically, we may select the reduced forms.
  4. $I(d, n) = \{I \textrm{ proper ideals of the ring of integers of } K \textrm{ of norm } n \}$ where $K = \mathbb{Q}[\sqrt{(d)}]$ the quadratic numberfield of discriminant $d$.
  5. The Epstein zeta function for $Q$: $\zeta(s, Q) = \frac{1}{2} \sum_{(x, y) \neq (0, 0)}{(Q(x, y))}^{-s}$ for $\mathfrak{R}(s) > 1$ for $(x, y) \in \mathbb{Z}^2$
  6. The Dirichlet zeta function for $K$: $\zeta(s, K) = \sum_{I}{Nm(I)}^{-s}$ for $\mathfrak{R}(s) > 1$ as $I$ runs through all the proper integer ideals of $K$.

As a fact I know that when $d < -4$, $\sum_{[Q]}\zeta(s, Q) = \zeta(s, K)$. Also in I read that in general $|R(d, n)| = \sum_{m|n}{\chi_{d}(m)} = |I(d, n)|$ where $\chi_{d}$ is the Kronecker symbol mod $|d|$. All these facts tend to point out that there is a one-one correspondence between $R(d,n)$ and $I(d, n)$. My question is that whether such a correspondence exists? If so could you explain this correspondence? Does this hold for $d > 0$ also?

  • $\begingroup$ Yes, this is indeed the case and is well known. A nice exposition is written up in chapter 7 of "Algebraic Theory of Quadratic Numbers" by Mak Trifkovic. $\endgroup$ Jan 28, 2021 at 17:42
  • $\begingroup$ @PeterHumphries Sorry if I have missed it. I went through the book, especially chapter 7. But I didn't find what I was looking for. The exposition is about the correspondence between the narrow class group and the equivalence classes of forms - which is well known. Not quite what I was looking for. I am looking for the correspondence between ideals of a fixed norm $n > 0$ and the integral representations $(x, y)$ of $n$ by the forms. $\endgroup$
    – Melanka
    Jan 28, 2021 at 19:10
  • $\begingroup$ There is a recent book that follows binary quadratic forms and quadratic number fields all through the book, Lehman bookstore.ams.org/dol-52 There is a more concentrated discussion in Cohen, A Course in Computational Algebraic Number Theory, especially section 5.2 in pages 225-230. In brief, a simple mapping from forms to ideals is one-to-one for positive forms, also indefinite forms when the principal form integrally represents $-1,$ and two-to-one otherwise (so that forms $f$ and $-f$ are distinct). $\endgroup$
    – Will Jagy
    Jan 28, 2021 at 19:57
  • $\begingroup$ By definition, two sets have the same cardinality if and only if there is a one-to-one correspondence between them. So it is not clear what your question is. $\endgroup$
    – GH from MO
    Jan 28, 2021 at 20:35
  • 1
    $\begingroup$ By Siegel's mass formula, $|R(d,n)|$ is a product of local densities given explicitly in Siegel's original paper (Annals of Mathematics, 1935). It equals $\sum_{m\mid n}\chi_d(m)=|I(d,n)|$. About an explicit correspondence: define an ordering on both $R(d,n)$ and $I(d,n)$, and map the $k$-th element of $R(d,n)$ to the $k$-th element of $I(d,n)$. $\endgroup$
    – GH from MO
    Jan 28, 2021 at 22:36

2 Answers 2


As explained below, there is a correspondence between non-zero locally principal ideals $I$ in the order of disc. $D$ and quadratic forms $Q$ of disc. $D$. The form $Q$ is naturally defined on $I$ via $x \in I \mapsto N(x)/N(I).$ So $Q(x) = n$ iff $N(x) = n N(I)$ if $N(x I^{-1}) = n,$ so representations of $n$ by $Q$ correspond to ideals in the class of $I^{-1}$ having norm $n$.

If we sum over all ideal classes, or equivalently over all $Q$, (and divide by the number of automorphisms of $Q$, which is the group of units in the order, to count ideals rather than elements that generate them) we will get the number of ideals of norm $n$.


The discovery of this correspondence must go back to the 19th century, maybe to Dirichlet or Dedekind? And Will Jagy's answer gives a very concrete description of it.

One can also give a more conceptual description of it. Indeed, it is fairly easy, and classical, to describe the quadratic form attached to an ideal class in conceptual terms; I explain this at the end of my answer. There is also a conceptual approach to going from quadratic forms to ideal classes, but as far as I know it is more recent, and due to Melanie Wood (a version of it is mentioned in this answer).

Namely, if $Q(x,y)$ is a quadratic form over $\mathbb Z$ which is primitive, then one can consider $\mathrm{Proj} \mathbb Z[x,y]/Q(x,y)$, which turns out to be finite flat of degree $2$ over $\mathbb Z$, therefore affine, of the form $\mathrm{Spec} A$ for some quadratic order $A$. And in fact $A$ is isomorphic to $R[(D +\sqrt{D})/2]$, where $D$ is the discriminant of $Q$ (so $A$ is the quadratic order of discriminant $D$). (There is no canonical identification of $A$ with this order though; we can compose with the automorphism $\sqrt{D} \mapsto -\sqrt{D}$ to get another isomorphism.)

Now $\mathrm{Proj}$ comes with a canonical line bundle $\mathcal O(1)$, and so this is an element of $\mathrm{Pic} A$, which can be thought of as the class group of non-zero locally principal ideals in $A$. (The usual class group when $D$ is a fundamental discriminant.)

Using the isomorphism $A \cong R[(D +\sqrt{D})/2],$ we get an element of the class group of $R[(D +\sqrt{D})/2],$ or really a pair $I,I^{-1}$ of elements, because applying the automorphism $\sqrt{D} \mapsto -\sqrt{D}$ switches $I$ and it's inverse.

This gives a map
$$\text{ primitive quadratic forms } Q \text{ of discriminant } D \longrightarrow \text{ pairs } I, I^{-1} \text{ in the class group of } R[(D +\sqrt{D})/2].$$

To see it is a bijection, one can give an explicit inverse (as I mentioned above, this is more classical):

If $I$ is a locally principal ideal in $R[(D +\sqrt{D})/2],$ then the formula $$x \in I \mapsto N(x)/ N(I) \in \mathbb{Z}$$ defines a quadratic form $Q$ on the free rank two $\mathbb{Z}$-module $I$. (Note that $I^{-1}$ will give the same quadratic form, since formation of norms is invariant under conjugation.)

  • $\begingroup$ You misunderstood the question. The OP wants a correspondence betwen representations of $n$ by quadratic forms of discriminant $d$ (modulo equivalences and automorphs) and ideals of norm $n$ in $\mathbb{Q}(\sqrt{d})$. $\endgroup$
    – GH from MO
    Jan 28, 2021 at 22:48
  • 1
    $\begingroup$ Ah yes, thanks for pointing that out. $\endgroup$
    – Matt E
    Jan 29, 2021 at 2:11
  • $\begingroup$ @GHfromMO: Added something which hopefully addresses the actual question. $\endgroup$
    – Matt E
    Jan 29, 2021 at 2:21
  • $\begingroup$ Good. My answer is similar to yours, but in more concrete (or more classical) terms. $\endgroup$
    – GH from MO
    Jan 29, 2021 at 2:26

We shall give an explicit correspondence between $R(d,n)$ and $I(d,n)$ based on the explicit correspondence between representatives $Q$ of proper equivalence classes of quadratic forms of discriminant $d$ and representatives $I$ of the narrow ideal class group of $\mathbb{Q}(\sqrt{d})$.

We recall that to each representative quadratic form $Q$, there exists a unique representative ideal $I$ and an oriented basis $(\omega_1,\omega_2)$ of $I$ such that $$Q(r,s)=N(r\omega_1+\omega_2 s)/N(I).$$ Oriented basis means that $I=\mathbb{Z}\omega_1+\mathbb{Z}\omega_2$ and $\bar\omega_1\omega_2-\omega_1\bar\omega_2=N(I)\sqrt{d}$, where $\sqrt{d}$ is a fixed square-root of $d$ (independent of $I$). Now to each integral representation $Q(x,y)=n$, we associate the ideal $(x\omega_1+y\omega_2)/I$ of norm $n$. The range of this map is clearly $I(d,n)$. Moreover, two integral representations $Q(x,y)=n$ and $Q'(x',y')=n$ give rise to the same ideal if and only if $Q=Q'$ and $(x,y)$ only differs from $(x',y')$ by an automorph of $Q$. That is, our map induces a bijection between $R(d,n)$ and $I(d,n)$.

  • $\begingroup$ Thank you for the answer. But I still do have one question. Why is it that the ideal representative $I$ of $Q$ unique? For example, $cI$ and the basis $(c \omega_1, c \omega_2)$ would also be a valid representative for any positive integer $c$. I think for $I$ to be unique we need to add an extra condition. Given $n > 0$, represented by $Q$ there is a unique such ideal $I$ with norm $n$. Now in this case $I$ is unique if such a $I$ exists. $\endgroup$
    – Melanka
    Jan 29, 2021 at 15:06
  • $\begingroup$ To show existence, suppose $n = m e^2$ where $m > 0$ is properly represented by $Q$ and $e > 0$. Then for some $\sigma \in SL_2(\mathbb{Z})$, $\sigma Q(x, y) = mx^2 + b'xy + c'y^2 = \frac{(mx + \gamma y)(mx + \bar{\gamma y)}}{m}$ by factorising $\sigma Q$ for some $\gamma$ in the ring of integers. So $\sigma Q(x, y) = frac{(emx + e \gamma y)(emx + e \bar{\gamma y)}}{e^2 m} = frac{(emx + e \gamma y)(emx + e \bar{\gamma y)}}{(e \mathfrak{m})(e \bar{\mathfrak{m}})}$ and we set $I = e \mathfrak{m}$ where $\mathfrak{m} \bar{\mathfrak{m}} = (m)$ as ideals in the ring of integers. $\endgroup$
    – Melanka
    Jan 29, 2021 at 15:08
  • $\begingroup$ Now we do a base change to get back $Q$.But this relies on proving that $(m, \gamma)$ is a basis of $\mathfrak{m}$. Now I need to prove it. $\endgroup$
    – Melanka
    Jan 29, 2021 at 15:09
  • $\begingroup$ @Melanka: $cI$ and $I$ represent the same ideal class, so both cannot be representatives. I fixed a set of (pairwise inequivalent) representatives $\{Q\}$ for the forms and a set of (pairwise inequivalent) representatives $\{I\}$ for the ideals. I assumed the knowledge of the correspondence $Q\leftrightarrow I$, my proof started from there. You can find the details of this correspondence in the Appendix of gofile.io/d/iqnq4A $\endgroup$
    – GH from MO
    Jan 29, 2021 at 18:13
  • $\begingroup$ Sorry, I missed the point about fixing a representative $I$ for each class. Thanks for pointing it out. $\endgroup$
    – Melanka
    Jan 29, 2021 at 18:34

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