Orders in number fields and their Picard groups

Written by GPT-6.1 Sol (OpenAI) in Codex, Ultra setting, October 2026. Self-checked by the writing AI. Independent full-lesson AI review is pending. Public domain (CC0).

The full ring of integers is not the only useful integral ring inside a number field. A chosen generator may give a smaller ring, and norm equations naturally involve these smaller rings. Their ideals need not factor uniquely. Invertible ideals retain a group law, and the conductor measures precisely the discrepancy between their class group and that of the full ring.

Orders and their conductors

Let \(K\) be a number field of degree \(n\), and put \(B=O_K\). An order \(A\) is a subring of \(B\) containing \(1\), of rank \(n\) over \(\mathbf Z\). Since it is a subgroup of the free group \(B\), it is itself free, and its full rank means \[ m=[B:A]<\infty,\qquad mB\subseteq A. \tag{1} \] Its fraction field is \(K\): the rational span of \(A\) is \(K\), and rational linear combinations are fractions with integral denominators.

An order is Noetherian, because every ideal is a finitely generated subgroup of its finite free abelian group. Every nonzero ideal contains a nonzero rational integer. Indeed, for \(0\ne\alpha\) in it, a monic integral polynomial for \(\alpha\) with nonzero constant term writes that constant term as \(\alpha\) times an element of \(A\). Hence every nonzero ideal has finite quotient. Every nonzero prime is therefore maximal, since its quotient is a finite domain. The order has dimension one: it has such maximal ideals, and the only possible prime chains are \((0)\subsetneq\mathfrak p\).

The integral closure of \(A\) in \(K\) is \(B\). Elements of \(B\) are integral over \(\mathbf Z\subseteq A\); conversely, integrality over \(A\), followed by integrality of \(A\) over \(\mathbf Z\), implies integrality over \(\mathbf Z\). We use this transitivity from Algebraic integers and rings of integers. Thus \(A\ne B\) is not integrally closed.

Define the conductor ideal \[ \mathfrak f=\{x\in B\mid xB\subseteq A\}. \tag{2} \] It is nonzero by (1).

Proposition 11.1. The conductor is the largest \(B\)-ideal contained in \(A\).

Proof. If \(x,y\) satisfy (2), so does \(x+y\). If \(b\in B\), then \((bx)B\subseteq xB\subseteq A\). Hence \(\mathfrak f\) is a \(B\)-ideal; taking \(1\in B\) shows \(\mathfrak f\subseteq A\). If a \(B\)-ideal \(J\) lies inside \(A\), then every \(x\in J\) has \(xB\subseteq J\subseteq A\), so \(J\subseteq\mathfrak f\). \(\square\)

Invertibility and local generators

A nonzero fractional \(A\)-ideal is a finitely generated \(A\)-submodule \(I\subset K\). It is a full abelian lattice: it contains \(\alpha A\) for some \(\alpha\ne0\), and lies in \(d^{-1}A\) after clearing the denominators of finitely many generators. It is invertible if \(IJ=A\) for another fractional ideal. Then \[ I^{-1}=\{x\in K\mid xI\subseteq A\}. \tag{3} \] Products, inverses and nonzero principal ideals are invertible. Their group modulo principal ideals is the Picard group \[ \operatorname{Pic}(A)= \{\text{invertible fractional }A\text{-ideals}\}/\{\alpha A\mid\alpha\in K^\times\}. \tag{4} \] For \(A=B\), this is \(\operatorname{Cl}_K\).

We need three elementary facts about local generators. Write \(A_{\mathfrak p}\) for localization at a maximal ideal.

Local generator lemma. A fractional ideal \(I\) is invertible if and only if \(I_{\mathfrak p}\) is principal for every maximal ideal \(\mathfrak p\).

Proof. If \(IJ=A\), write \(1=\sum x_i y_i\), with \(x_i\in I,y_i\in J\). In a local ring at least one product is a unit. If \(x_1y_1\) is a unit, every \(x\in I_{\mathfrak p}\) satisfies \[ x=x_1(xy_1)(x_1y_1)^{-1}, \] so \(I_{\mathfrak p}=x_1A_{\mathfrak p}\).

Conversely suppose this local principality holds. A local generator can be taken to be \(a_{\mathfrak p}\in I\) by clearing its denominator outside \(\mathfrak p\). Clear the local denominators of a finite generating set of \(I\) to find \(s\in A\setminus\mathfrak p\) with \(sI\subseteq a_{\mathfrak p}A\). Then \(s/a_{\mathfrak p}\in I^{-1}\), and \(s\in II^{-1}\). Thus the integral ideal \(II^{-1}\) is contained in no maximal ideal and equals \(A\). The module (3) is a fractional ideal: for any fixed nonzero \(a\in I\) it lies in \(a^{-1}A\), and it is finitely generated by Noetherianity. \(\square\)

Membership in a submodule is detected at all maximal ideals. To see this, the annihilator of a nonzero element of a quotient module is a proper ideal; choose a maximal ideal containing it, and that element remains nonzero after localization. This also proves that a module map which is an isomorphism at every maximal ideal is an isomorphism.

Finite quotient lemma. A module locally free of rank one over a finite commutative ring \(R\) is free of rank one.

Proof. There are finitely many maximal ideals \(\mathfrak m_i\). Choose an element whose residue generates the one-dimensional space \(P/\mathfrak m_iP\) for each \(i\). The Chinese remainder theorem, applied to these pairwise comaximal ideals and to the module \(P\), combines them into one element \(x\). Locally \(P\) is free of rank one, and the coefficient of \(x\) in a local basis has nonzero residue, hence is a unit. Thus \(R\to P,\ r\mapsto rx\), is a local isomorphism everywhere and therefore an isomorphism. \(\square\)

For an integral ideal put \(N_A I=|A/I|\); for a fractional ideal use its positive lattice covolume divided by that of \(A\). The determinant description of lattice index gives \(N_A(\alpha A)=|N_{K/\mathbf Q}\alpha|\) and the corresponding positive scaling law for fractional ideals. If \(I\) is invertible and \(J\) integral, the finite quotient lemma applies to \(I/JI\) over \(A/J\), because its localizations have rank one. Hence \[ |I/JI|=|A/J|,\qquad N_A(IJ)=N_A I\,N_A J \tag{5} \] when both ideals are integral; scaling extends this to fractional invertible ideals. The index and field-norm facts used here were proved in Norms, class groups, and modules over Dedekind domains; the finite quotient argument supplies the extra step for orders.

Ideals away from the conductor

A nonzero integral \(A\)-ideal is prime to \(\mathfrak f\) if \(I+\mathfrak f=A\); a nonzero integral \(B\)-ideal is prime to it if \(J+\mathfrak f=B\).

Theorem 11.3. Extension and contraction are mutually inverse multiplicative bijections \[ I\longmapsto IB,\qquad J\longmapsto J\cap A \tag{6} \] between these two sets of ideals. All the \(A\)-ideals in (6) are invertible, and \[ A/(J\cap A)\simeq B/J,\qquad N_A I=N_B(IB). \tag{7} \] Every class of \(\operatorname{Pic}(A)\) contains an integral ideal prime to any prescribed nonzero rational integer.

Proof of the bijection. If \(I+\mathfrak f=A\), then \(IB+\mathfrak f=B\). Write \(1=u+t\) with \(u\in I,t\in\mathfrak f\). For \(x\in IB\cap A\), \[ x=xu+xt\in I, \] because \(x\in A\) gives \(xu\in I\), and \(x\in IB,tB\subseteq A\) gives \(xt\in I\). Thus \(IB\cap A=I\).

If \(J+\mathfrak f=B\), choose \(1=v+t\), with \(v\in J,t\in\mathfrak f\). Then \(v=1-t\in A\), so \(v\in J\cap A\), proving \(J\cap A+\mathfrak f=A\). For \(z\in J\), \[ z=zv+zt\in(J\cap A)B, \] since \(zt\in J\cap\mathfrak f\subseteq J\cap A\). Hence \((J\cap A)B=J\). The map in (7) is injective. It is surjective because every \(z\in B\) has \(z\equiv zt\pmod J\), with \(zt\in A\). This proves equal norms. Extension preserves products, so its inverse does also.

To prove invertibility, let \(\mathfrak p\) contain \(\mathfrak f\). Then \(I_{\mathfrak p}=A_{\mathfrak p}\). If \(\mathfrak p\) does not contain \(\mathfrak f\), take \(t\in\mathfrak f\setminus\mathfrak p\). Every \(b\in B\) equals \(tb/t\) in the localization, with \(tb\in A\); consequently \(B_{A\setminus\mathfrak p}=A_{\mathfrak p}\). This ring is integrally closed and one-dimensional Noetherian, hence a DVR by the criterion in Discrete valuation rings and Dedekind domains. Its ideals are principal. The local generator lemma proves \(I\) invertible.

For the final assertion, take an integral invertible ideal \(L\) representing the inverse of the desired class, and let \(q>0\) be the prescribed integer. There are finitely many maximal ideals containing \(q\), since \(A/qA\) is finite. At each, choose the residue of an element of \(L\) which locally generates it. Module CRT gives one \(x\in L\) with all these residues. A nonzero such \(x\) gives \[ I=xL^{-1}\subseteq A,\qquad I_{\mathfrak p}=A_{\mathfrak p}\quad(\mathfrak p\mid q). \] Thus \(I+qA=A\), and \([I]=[L]^{-1}\) is the required class. For \(q=1\), any integral representative suffices. Negative prescribed integers give the same condition as their absolute values. In particular, choosing \(q\in\mathfrak f\cap\mathbf Z\setminus\{0\}\) makes \(I\) prime to \(\mathfrak f\). \(\square\)

The conductor exact sequence

Write \(\bar A=A/\mathfrak f\), \(\bar B=B/\mathfrak f\). The inclusion \(\bar A\subseteq\bar B\) identifies their unit groups with a subgroup inclusion.

Theorem 11.4. There is an exact sequence \[ 1\longrightarrow A^\times\longrightarrow B^\times \longrightarrow\bar B^\times/\bar A^\times \xrightarrow{\delta}\operatorname{Pic}(A) \longrightarrow\operatorname{Cl}_K\longrightarrow1. \tag{8} \] Consequently the unit index and \(\operatorname{Pic}(A)\) are finite, and \[ |\operatorname{Pic}(A)| =h_K\,\frac{[\bar B^\times:\bar A^\times]}{[B^\times:A^\times]}. \tag{9} \]

Proof. The case \(A=B\) reduces to identity maps and a trivial residue quotient. Assume \(A\ne B\). For \(u\in\bar B^\times\), define the lattice \[ L_u=\{b\in B\mid b\bmod\mathfrak f\in u\bar A\}. \tag{10} \] It is a fractional \(A\)-ideal and contains \(\mathfrak f\). To examine it at \(\mathfrak p\supseteq\mathfrak f\), lift \(u\) to \(\alpha\in B\). The ring \(B_{A\setminus\mathfrak p}\) is finite integral over the local ring \(A_{\mathfrak p}\); its maximal ideals lie over \(\mathfrak pA_{\mathfrak p}\), and there are finitely many since the residue algebra is finite. All therefore contain \(\mathfrak f\), so the unit residue makes \(\alpha\) a unit. As \(\mathfrak f\) is a \(B\)-ideal, \(\alpha\mathfrak f_{\mathfrak p}=\mathfrak f_{\mathfrak p}\); reduction of (10) gives \[ (L_u)_{\mathfrak p}=\alpha A_{\mathfrak p}. \] Outside the conductor, \(A_{\mathfrak p}=B_{A\setminus\mathfrak p}\), and \(L_u\) equals that ring locally. Thus \(L_u\) is invertible, \(L_uB=B\), and local multiplication gives \(L_uL_v=L_{uv}\). Changing \(u\) by a unit of \(\bar A\) leaves \(L_u\) unchanged. Set \(\delta(u)=[L_u]\); this defines the indicated homomorphism.

At \(B^\times\), a unit whose residue lies in \(\bar A^\times\) belongs to \(A^\times\): both it and its inverse differ from elements of \(A\) by conductor elements. Thus the kernel is exactly \(A^\times\).

At the residue quotient, if \(u\) is the residue of \(\gamma\in B^\times\), then \(L_u=\gamma A\), since \(\gamma\mathfrak f=\mathfrak f\). Conversely, if \(L_u=\gamma A\), extension gives \(\gamma B=B\), hence \(\gamma\in B^\times\). Reduction gives \(u\bar A=\bar\gamma\bar A\); because both residues are units of \(\bar B\), their quotient is a unit of \(\bar A\). This proves exactness there.

Extension of invertible ideals defines the map to \(\operatorname{Cl}_K\). It kills every \(L_u\). Suppose an invertible ideal \(I\) lies in its kernel, so \(IB=\gamma B\). Replace \(I\) by \(\gamma^{-1}I\); now \(IB=B\), so \(I\subseteq B\), and \[ \mathfrak f I=\mathfrak f(IB)=\mathfrak f. \] The invertible module \(I/\mathfrak f I\) is locally free of rank one over the finite ring \(\bar A\). By the finite quotient lemma choose a generator \(u\). Extending to \(\bar B\), the equality \(IB=B\) says that \(u\bar B=\bar B\), so \(u\in\bar B^\times\). Thus \(I/\mathfrak f=u\bar A\), or \(I=L_u\). This proves exactness at \(\operatorname{Pic}(A)\).

Finally every class of \(B\) has an integral representative prime to \(\mathfrak f\), by the last part of Theorem 11.3 applied to \(B\) and an integer in \(\mathfrak f\). Contracting it by (6) proves surjectivity. The residue quotient is finite and \(h_K\) is finite by Finiteness of the class number. Exactness gives finite \(B^\times/A^\times\), finite \(\operatorname{Pic}(A)\), and (9). \(\square\)

This also shows that units of an order have the same rank as units of its maximal order: their index is finite, so Dirichlet's unit theorem gives rank \(r_1+r_2-1\).

Quadratic orders and proper ideals

Now let \(K\) be quadratic, of field discriminant \(d_K\), and choose \(B=\mathbf Z+\mathbf Z\omega\), where \(\omega=(d_K+\sqrt{d_K})/2\). Since \(B/\mathbf Z\) is infinite cyclic, every order is uniquely \[ A_f=\mathbf Z+fB=\mathbf Z+\mathbf Z f\omega,\qquad f\geq1. \tag{11} \] Here \(f=[B:A_f]\). If \(\omega^2=t\omega+s\) with \(t,s\in\mathbf Z\), an element \(x=a+fb\omega\in A_f\) has \(x\omega\in A_f\) exactly when \(f\mid a\). Hence the conductor ideal and discriminant are \[ \mathfrak f=fB,\qquad \Delta=f^2d_K. \tag{12} \] The discriminant formula follows directly by squaring the determinant of the embedding basis \((1,f\omega)\). The conductor ideal \(\mathfrak f\) and its integer \(f\) are different objects.

Conversely every nonsquare \(\Delta\equiv0,1\pmod4\) occurs uniquely this way. Write \(\Delta=k^2d\) with \(d\) squarefree. If \(d\equiv1\pmod4\), take \(d_K=d,f=k\). Otherwise \(d\equiv2,3\pmod4\), and the discriminant congruence forces \(k\) even; take \(d_K=4d,f=k/2\). This proves existence and uniqueness.

The multiplier ring of \(I\) is \((I:I)=\{x\in K\mid xI\subseteq I\}\). The ideal is proper if \((I:I)=A_f\).

Proposition 11.2. For a quadratic order, a fractional ideal is proper if and only if it is invertible.

Proof. If \(IJ=A_f\) and \(xI\subseteq I\), then \(xA_f=xIJ\subseteq IJ=A_f\), so \(x\in A_f\).

For the converse write \(I=\alpha(\mathbf Z+\mathbf Z\tau)\), with \(\tau\) quadratic, and let \[ a\tau^2+b\tau+c=0,\qquad \gcd(a,b,c)=1,\quad a>0 \tag{13} \] be its primitive integral minimal polynomial. An element preserving \(\mathbf Z+\mathbf Z\tau\) must be \(r+s\tau\) with \(r,s\in\mathbf Z\). Its product with \(\tau\) lies in that lattice exactly when \(a\mid bs\) and \(a\mid cs\). Bézout for \(a,b,c\) makes these equivalent to \(a\mid s\). Thus \[ (I:I)=\mathbf Z+\mathbf Z a\tau. \tag{14} \] If \(I\) is proper, this is \(A_f\). Using \(\tau+\bar\tau=-b/a\), \(\tau\bar\tau=c/a\), we obtain \[ aI\bar I =N\alpha\,(\mathbf Za+\mathbf Za\tau+\mathbf Z(-b)+\mathbf Zc) =N\alpha\,A_f. \] The last equality uses \(\gcd(a,b,c)=1\). Hence \(I\) has inverse \((a/N\alpha)\bar I\). \(\square\)

For example, in \(A=\mathbf Z[\sqrt{-3}]\), the ideal \[ J=(2,1+\sqrt{-3})=2B,\qquad B=\mathbf Z[(1+\sqrt{-3})/2], \] has multiplier ring \(B\), so it is not invertible over \(A\). Its norm is \(2\), while \(J^2=4B\) has norm \(8\). This explicitly shows why arbitrary ideal norms need not multiply in an order. It does not contradict (5).

Forms of a nonfundamental discriminant

The correspondence of Quadratic fields: ideal classes and binary quadratic forms extends to \[ C(\Delta)\simeq \begin{cases} \operatorname{Pic}(A_f),&\Delta<0,\\ \operatorname{Pic}^+(A_f),&\Delta>0. \end{cases} \tag{15} \] The narrow Picard group in the second line uses totally positive principal generators. Ordinary \(\operatorname{Pic}(A_f)\) is the group in (8)–(9).

Here is a proof of the extension, including the primitivity that no longer follows from a field discriminant. The determinant-and-index formula makes it possible to orient an ideal basis by \[ \alpha\bar\beta-\bar\alpha\beta=-\sqrt\Delta\,N_{A_f}I \] and use \(N(\alpha x-\beta y)/N_{A_f}I\). Choose the first basis element with positive norm; for real fields the positive-cone argument in lesson 10 permits this. Put \(\tau=\beta/\alpha\) and use (13). Since \(I\) is invertible, it is proper, and (14) identifies \(A_f=\mathbf Z+\mathbf Za\tau\). Its discriminant is \(a^2(\tau-\bar\tau)^2=\Delta\). Orientation and \(N\alpha>0\) give \(a(\tau-\bar\tau)=\sqrt\Delta\) and \[ N_{A_f}I=N\alpha/a. \] The norm form is therefore exactly the primitive form \((a,b,c)\) in (13).

Conversely a primitive form \((a,b,c)\) of discriminant \(\Delta\), with \(a>0\), gives \[ I=\mathbf Za+\mathbf Z\frac{-b+\sqrt\Delta}{2}. \] The multiplier calculation (14) identifies its order as \(A_f\), and Proposition 11.2 makes it invertible. Its index in \(A_f\) is \(a\), so its norm form is the original form. Proper basis changes and positive-norm scalings act exactly as in Theorem 10.1. For injectivity, the oriented ratio is \(\beta/\alpha=(-b+\sqrt\Delta)/(2a)\); equal forms thus give equal ratios and ideal scalings of positive norm. This proves (15), with composition given by ideal multiplication.

The reduction proof of Theorem 10.2 did not require a fundamental discriminant, so for \(\Delta<0\) reduced primitive forms count \(|\operatorname{Pic}(A_f)|\). The prime-representation proof also applies whenever \(p\nmid\Delta\): its explicit form \((p,b,c)\) remains primitive, and at \(2\) the same Kronecker condition applies.

The quadratic order class-number formula

Corollary 11.5. For a quadratic order \(A_f\), real or imaginary, \[ h(A_f)=h_K\, \frac{f}{[B^\times:A_f^\times]} \prod_{p\mid f}\left(1-\frac{(d_K/p)}p\right), \qquad h(A_f)=|\operatorname{Pic}(A_f)|. \tag{16} \] For real orders this is the ordinary Picard class number. Proper form classes instead count the narrow group.

Proof. Here \(A_f/\mathfrak f\simeq\mathbf Z/f\mathbf Z\), so its unit-group order is \(f\prod_{p\mid f}(1-1/p)\). By CRT it remains to count units in \(B/p^eB\). Reduction onto \(B/pB\) is surjective on units: an inverse modulo \(p\) lifts by a finite geometric series, since \(pB/p^eB\) is nilpotent. Its kernel consists of \(1+pz\) and has size \(p^{2(e-1)}\).

The full-ring prime decomposition proved in Decomposition of primes in extensions gives, including \(p=2\), \[ B/pB\simeq \begin{cases} \mathbf F_p\times\mathbf F_p,&(d_K/p)=1,\\ \mathbf F_{p^2},&(d_K/p)=-1,\\ \mathbf F_p[T]/(T^2),&(d_K/p)=0. \end{cases} \] In the ramified case the last description also follows by reducing the quadratic integral-basis polynomial, which has one repeated linear root. The unit counts are respectively \((p-1)^2,p^2-1,p(p-1)\). Thus \[ |(B/p^eB)^\times| =p^{2e}(1-1/p)(1-(d_K/p)/p). \] Dividing the product by \(|(\mathbf Z/f\mathbf Z)^\times|\) gives \([\bar B^\times:\bar A_f^\times]=f\prod_{p\mid f}(1-(d_K/p)/p)\). Substitute this into (9). For \(f=1\), the empty product and unit index are one and the formula is \(h(B)=h_K\). \(\square\)

In imaginary quadratic fields other than \(\mathbf Q(i)\) and \(\mathbf Q(\sqrt{-3})\), all units are \(\pm1\), so the index is one. In these two fields a proper suborder has only \(\pm1\), while the full rings have four and six units; the indices are two and three. In real fields it is the least positive power of a full-ring fundamental unit that belongs to the order, by the finite index proved above.

Examples

The maximal orders of discriminants \(-3,-4,-7\) have class number one by lesson 10.

The factor at \(3\) for \(d_K=-3\), and at \(2\) for \(d_K=-4\), is one because those primes ramify. For \(d_K=-3\), \(2\) is inert; for \(d_K=-7\), \(2\) splits. The unit indices are essential, even when the conductor is small.

For a nonquadratic example take \(B=\mathbf Z[\alpha]\), \(\alpha^3=2\), and \(A=\mathbf Z+2B\). Lessons 2, 8 and 9 prove that \(B\) is maximal, \(h_K=1\), and \(B^\times=\{\pm(\alpha-1)^j\}\). Modulo \(2B\), the ring is \(\mathbf F_2[T]/(T^3)\); its units are the four elements with constant term one. The image of \(A\) is the constants. The largest ideal inside those constants is zero, since multiplication of \(1\) by \(T\) leaves them, so the conductor is \(2B\). The residue of \(\alpha-1\) is \(1+T\), of order four: its square is \(1+T^2\), and its fourth power is one. Hence the residue unit quotient and the full-ring unit index both have order four. Formula (9) gives \(\operatorname{Pic}(A)=1\). This uses the general conductor theorem, rather than the quadratic formula (16).

The three order discriminants \(-12,-16,-28\) each have one reduced form, \[ x^2+3y^2,\qquad x^2+4y^2,\qquad x^2+7y^2. \tag{17} \] These forms belong to orders, not to the maximal-order correspondence for field discriminants. Class number one of an order also does not assert unique factorization of all its ideals or elements. The noninvertible ideal \(J\) above is a counterexample to that inference.

Exercises

1. Compute the conductor ideals and conductor integers of \(\mathbf Z[\sqrt{-3}]\) and \(\mathbf Z[3i]\).
2. Prove that proper and invertible fractional ideals coincide for quadratic orders.
3. Compute \(h(\mathbf Z[\sqrt{-27}])\) and list all reduced forms of discriminant \(-108\).
4. Use the three class numbers in (17) to characterize primes \(x^2+3y^2\), \(x^2+4y^2\), and, for odd primes, \(x^2+7y^2\).
5. Prove the conductor exact sequence (8), including the map from residue units and its kernel.

Solutions

1. In \(B=\mathbf Z[(1+\sqrt{-3})/2]\), the first order is \(\mathbf Z+2B\); its conductor integer is \(2\), and its conductor ideal is \(2B=(2,1+\sqrt{-3})\). In \(B=\mathbf Z[i]\), the second order is \(\mathbf Z+3B\), with conductor integer \(3\) and ideal \(3B=(3,3i)\). These are the largest \(B\)-ideals inside the respective orders by (12), not merely ideals generated by the integer inside the smaller ring.

2. Invertibility implies properness by multiplying \(xI\subseteq I\) by an inverse. Conversely express \(I=\alpha[1,\tau]\), with primitive minimal polynomial \(aT^2+bT+c\). Preservation of this lattice by \(r+s\tau\) forces \(a\mid bs,cs\), hence \(a\mid s\); so its multiplier ring is \([1,a\tau]\). Properness identifies this with the given order. Then \[ aI\bar I=N\alpha[a,a\tau,-b,c]=N\alpha[1,a\tau] \] by the coefficient gcd. Thus \((a/N\alpha)\bar I\) is an inverse. This proves the equivalence without assuming unique ideal factorization.

3. The conductor is \(6\), since \(\sqrt{-27}=3\sqrt{-3}=6\omega-3\). The unit index is \(3\); at \(2,3\), the symbols are \(-1,0\). Formula (16) gives \(h=3\).

For the reduced list, lesson 10's bound gives \(a\leq6\), and \(b\) is even. At \(a=1\), \((1,0,27)\) survives. At \(a=2\), neither \(b=0\) nor \(b=2\) gives a primitive integral triple. At \(a=3\), \(b=0\) gives \((3,0,9)\), not primitive, and \(b=\pm2\) gives nonintegral \(c\). At \(a=4\), only \(b=\pm2,c=7\) survives. At \(a=5\), \(b=0,\pm2,\pm4\) gives no integral \(c\). At \(a=6\), the only integral possibilities have \(b=\pm6,c=6\), again not primitive. Thus the complete list is \[ (1,0,27),\qquad(4,2,7),\qquad(4,-2,7), \] whose three proper classes agree with (16).

4. For a prime not dividing an order discriminant, the extended prime criterion and class number one force representation by the principal form (17). The reciprocity computations in Corollary 10.5 therefore give \[ \begin{aligned} p=x^2+3y^2 &\iff p=3\text{ or }p\equiv1\pmod3,\\ p=x^2+4y^2 &\iff p\equiv1\pmod4,\\ p=x^2+7y^2\quad(p\text{ odd}) &\iff p=7\text{ or }p\equiv1,2,4\pmod7. \end{aligned} \tag{18} \] For the first form \(p=3\) is represented by \((0,1)\), and \(2\) is not represented: \(y\ne0\) gives at least \(3\), while \(y=0\) gives a square. For the second form \(2\) is likewise excluded. For the third, \(7\) is represented by \((0,1)\) and \(2\) is excluded. These are exactly the primes omitted by \(p\nmid\Delta\). In particular the last congruence rule cannot be applied to the even prime \(2\). For odd \(p\ne7\) it is precisely the stated congruence equivalence.

5. For a unit residue \(u\) of \(B/\mathfrak f\), take the inverse image \(L_u\) of \(u(A/\mathfrak f)\) in \(B\). At a conductor prime it is generated by a lift of \(u\), which is a unit in the normalization localized there; outside the conductor the two rings agree. The local generator lemma makes \(L_u\) invertible, and local products give \(L_uL_v=L_{uv}\), defining \(\delta\).

If its class is principal, \(L_u=\gamma A\) and \(L_uB=B\) imply \(\gamma\in B^\times\). Reduction says \(u/\bar\gamma\in(A/\mathfrak f)^\times\). Conversely every such global unit gives a principal \(L_u\). This proves the residue kernel; a global \(B\)-unit has trivial residue class exactly when it and its inverse lie in \(A\), proving the preceding kernel.

For an invertible class killed by extension, normalize its representative to \(IB=B\). Then \(\mathfrak f I=\mathfrak f\), and \(I/\mathfrak f\) is a rank-one locally free module over the finite ring \(A/\mathfrak f\). Its global generator is a unit after extending to \(B/\mathfrak f\), so \(I=L_u\) for some unit residue \(u\). This proves the Picard kernel. Finally a \(B\)-class has a representative prime to \(\mathfrak f\), whose contraction is invertible and extends back to it. Surjectivity finishes every term of (8).

What this lesson does not prove

We use the integral closure and integrality transitivity of lesson 1; lattice indices, field norms and full-ring ideal norms of lesson 4; the normal one-dimensional local-domain/DVR criterion of lesson 3; the full quadratic prime decomposition, including \(2\), of lesson 5; class finiteness of lesson 8; and the unit group of lesson 9. Reduction and the representation arguments are proved in lesson 10. The congruences in (18) also use quadratic reciprocity and both supplements, fully proved in Cyclotomic fields, Theorem 12.5; this later-course provider does not use the order class-number formula.

The local generator, finite quotient, conductor comparison, exact sequence, quadratic class-number formula, and order–form extension are proved here. The general discriminant theory for orders associated with Noether's work belongs to Orders and the discriminant theorem in the Noether course; only the direct quadratic basis calculation (12) is needed here. Ring class fields and higher residue conditions for primes \(x^2+ny^2\) belong to class field theory.

References

• Keith Conrad, The Conductor Ideal of an Order, Section 5, equation (5.7) and Theorem 5.3, pp.15-16. The conductor exact sequence and size formula are source context for the fully supplied proof; no expression adaptation grant inferred.