Ideal norm in orders

Question

Let $\overline{T}$ be a Dedekind ring such that $\overline{T}/\overline{I}$ is finite for every nonzero ideal $\overline{I}$ of $\overline{T}$. Let $T$ be a subring of $\overline{T}$ with the same total ring of fractions (i.e. an order).

Let $I$ be an ideal of $T$ and let $\overline{I} = I\overline{T}$. The norm $N_T(I)$ of $I$ is defined to be the cardinality of $T/I$.

Question: Is there a formula relating $N_T(I)$ and $N_{\overline{T}}(\overline{I})$?

For example, it seems plausible that the discrepancy is measured by some "tor" group.

Remarks:

• If $I$ is projective then $N_T(I)$ and $N_{\overline{T}}(\overline{I})$ are equal.
• Localization reduces the problem to the case when $T$ is local (and $\overline{T}$ is semi-local), and both $I$ and the conductor of $T$ are proper ideals.
• (Thanks to Luc Guyot) If $T$ is a Bass ring ($\leftrightarrow$ every intermediate ring $T \subset R \subset \overline{T}$ is Gorenstein $\leftrightarrow$ every ideal is generated by two elements), and $T = \{a \in \overline{T} : a I \subset I \}$, then by [2, Proposition 5.8] $I$ is projective. It follows that $N_T(I)$ and $N_{\overline{T}}(\overline{I})$ are equal (by the first remark).
• (generalization of third remark) if $T$ is a Gorenstein integral domain and $T = \{a \in \overline{T} : a I \subset I \}$, then $I$ is projective. This follows from combining Theorem 6.2(4) with Proposition 7.2 of [1]. It follows that $N_T(I)$ and $N_{\overline{T}}(\overline{I})$ are equal (by the first remark).

[1] H. Bass, "On the ubiquity of Gorenstein rings", 1963.

[2] L. Levy and R. Wiegand, "Dedekind-like behavior of rings with 2-generated ideals", 1985.

Answer 1

I'll begin with a general remark which will be illustrated by a computation in an arbitrary order of quadratic number field.

If $\overline{I}$ contracts to $I$, i.e., if $\overline{I} \cap R = I$, then the inclusion $R \rightarrow \overline{R}$ induces an injective $R$-module homomorphism $R/I \rightarrow \overline{R}/\overline{I}$. As a result, $N_R(I)$ divides $N_{\overline{R}}(\overline{I})$ and in particular we have $N_R(I) \le N_{\overline{R}}(\overline{I})$. If for instance $I$ is a prime ideal, then $N_R(I)$ divides $N_{\overline{R}}(\overline{I})$.

The underlying question that I fail to answer is:

Question. Is it always true that $N_R(I)$ divides $N_{\overline{R}}(\overline{I})$, or at least that $N_R(I) \le N_{\overline{R}}(\overline{I})$?

Edit. The OP answer contains a proof that $N_R(I) \le N_{\overline{R}}(\overline{I})$ holds true for every non-zero ideal of $R$.

I will not address the above question. Instead, I'll introduce a condition on $R$ under which $N_R(I)$ divides $N_{\overline{R}}(\overline{I})$ for every non-zero ideal $I$ of $R$.

Proposition. If a non-zero ideal $I$ of $R$ is projective over its ring of multipliers $\varrho(I) \Doteq \{ r \in \overline{R} \, \vert \, rI \subseteq I\}$, then we have $$ N_{\overline{R}}(\overline{I}) = N_R(I) \vert \varrho(I)/R \vert. $$

Side note. that $\varrho(I) = \{ r \in K \, \vert \, rI \subseteq I\}$ where $K$ denotes the field of fractions of $R$, since $R$ is Noetherian.

Lemma 1 (OP's Claim). If $I$ is an invertible ideal of $R$ then $N_{\overline{R}}(\overline{I}) = N_R(I)$.

Proof. First, prove the statement for a non-zero principal ideal $I$. Then decompose the $R$-module of finite length $\overline{R}/\overline{I}$ as a direct sum of its localizations with respect to the maximal ideals of $R$ [4, Theorem 2.13]. Do the same for $R/I$ and compare the cardinalities of the summands.

Proof of the Proposition. By Lemma 1, we have $N_{\overline{R}}(\overline{I}) = N_{\varrho(I)}(I)$. Hence $N_{\overline{R}}(\overline{I}) = [\varrho(I) : R][R: I] = \vert \varrho(I)/R\vert N_R(I)$.

Note that if $R$ is an order whose ideals are two-generated (e.g., an order in a quadratic field or an order whose discriminant is fourth-power free [2, Theorem 3.6]), then every non-zero ideal of $R$ satisfies the hypothesis of the above proposition, see e.g., [1], [2] and Theorem 4.1, Corollaries 4.3 and 4.4 of Keith Conrad's notes. The OP discusses similar results in his remarks and his answer.

Let $m$ be a square-free rational integer. We set $K \Doteq \mathbb{Q}(\sqrt{m})$ and denote by $\mathcal{O}(K)$ the ring of integers of the quadratic field $K$.

Loose Claim. Given an order $R$ of $K$ and an ideal $I \subseteq R$, we shall compute $N_{\mathcal{O}(K)}(I\mathcal{O}(K))$ as a function of $N_R(I)$ and of a binary quadratic form associated to $I$.

To do so, we introduce some notation and definitions.

Setting $$\omega = \left\{ \begin{array}{cc} \sqrt{m} & \text{ if } m \not\equiv 1 \mod 4, \\ \frac{1 + \sqrt{m}}{2} & \text{ if } m \equiv 1 \mod 4, \\ \end{array}\right. $$ we have $$\mathcal{O}(K) = \mathbb{Z} + \mathbb{Z} \omega$$ and any order of $K$ is of the form $\mathcal{O}_f(K) \Doteq \mathbb{Z} + \mathbb{Z} f \omega$ for some rational integer $f > 0$ [2, Lemma 6.1]. Moreover, the inclusion $\mathcal{O}_f(K) \subseteq \mathcal{O}_{f'}(K)$ holds true if and only if $f'$ divides $f$. If $I$ is an ideal of $\mathcal{O}_f(K)$, then its ring of multipliers $\varrho(I) \Doteq \{ r \in \mathcal{O}(K) \, \vert \, rI \subseteq I\}$ is the smallest order $\mathcal{O}$ of $K$ such that $I$ is projective, equivalently invertible, as an ideal of $\mathcal{O}$ [2, Proposition 5.8]. Let us fix $f > 0$ and set $$R \Doteq \mathcal{O}_f(K), \quad \overline{R} \Doteq \mathcal{O}(K).$$

An ideal $I$ of $R$ is said to be primitive if it cannot be written as $I = eJ$ some rational integer $e$ and some ideal $J$ of $R$.

The main tool is the Standard Basis Lemma [5, Lemma 6.2 and its proof].

Lemma 2. Let $I$ be a non-zero ideal of $R$. Then there exist rational integers $a, e > 0$ and $d \ge 0$ such that $-a/2 \le d < a/2$, $e$ divides both $a$ and $d$ and we have $$ I = \mathbb{Z} a + \mathbb{Z}(d + e f \omega). $$ The integers $a, d$ and $e$ are uniquely determined by $I$. We have $\mathbb{Z}a = I \cap \mathbb{Z}$ and the integer $ae$ is equal to the norm $N_R(I) = \vert R /I \vert$ of $I$. The ideal $I$ is primitive if and only if $e = 1$.

Note that, since $\mathbb{Z}a = I \cap \mathbb{Z}$, the rational integer $a$ divides $N_{K/\mathbb{Q}}(d + e f \omega)$. We call the generating pairs $(a, d + ef \omega)$ the standard basis of $I$. Let us associate to $I$ the binary quadratic form $q_I$ defined by $$q_I(x, y) = \frac{N_{K/\mathbb{Q}}(xa + y(d + ef\omega))}{N_R(I)}.$$

Then we have $$eq_I(x, y) = ax^2 + bxy + cy^2$$ with $$b = Tr_{K/\mathbb{Q}}(d + ef \omega) \text { and } c = \frac{N_{K/\mathbb{Q}}(d + ef \omega)}{a}.$$ We define the content $c(q_I)$ of $q_I$ as the greatest common divisor of its coefficients, that is $$c(q_I) \Doteq \frac{\gcd(a, b, c)}{e}.$$

Remark. We have $c(q_I) = \frac{\gcd(a, d, ef)}{e} = \frac{f}{f'} = \vert \varrho(I) / R \vert$ where $f'$ is the divisor of $f$ such that $\varrho(I) = \mathcal{O}_{f'}$.

Claim. Let $I$ be a non-zero ideal of $R$. Then we have $$N_{\overline{R}}(\overline{I}) = N_R(I) \vert \varrho(I)/R \vert \text{ with } \vert \varrho(I)/R \vert = c(q_I).$$

Proof. Since $N_R(xI) = N_R(Rx) N_R(I)$ and $N_R(Rx) = N_{\overline{R}}(\overline{R}x) = \vert N_{K/\mathbb{Q}}(x) \vert$ for every $x \in R \setminus \{0\}$, we can assume, without loss of generality, that $I$ is primitive, i.e., $e = 1$. It follows immediately from the definitions that $$\overline{I} = \overline{R} I = \mathbb{Z}a + \mathbb{Z}a \omega + \mathbb{Z}(d + f \omega) + \mathbb{Z}v$$ where
$$v = \left\{ \begin{array}{cc} f \omega^2 + d \omega & \text{ if } m \not\equiv 1 \mod 4, \\ f \frac{m - 1}{4} + (d + f) \omega & \text{ if } m \equiv 1 \mod 4. \\ \end{array}\right.$$ Now it suffices to compute the Smith Normal Form $\begin{pmatrix} d_1 & 0 \\ 0 & d_2 \\ 0 & 0 \\ 0 & 0 \end{pmatrix}$ of the matrix $A \Doteq \begin{pmatrix} a & 0 \\ 0 & a \\ d & f \\ v_1 & v_2 \end{pmatrix}$ where $(v_1, v_2)$ is the matrix of $v$ with respect to the $\mathbb{Z}$-basis $(1, \omega)$ of $\overline{R}$. The coefficient $d_1$ is the greatest common divisor of the coefficients of $A$ and is easily seen to be $\gcd(a, d, f) = \gcd(a, b, c)$. The coefficient $d_2$ is the greatest common divisor of the $2 \times 2$ minors of $A$ divided by $d_1$ and is easily seen to be $\frac{a \gcd(c(q_I), q_I(0, 1))}{d_1} = \frac{a c(q_I)}{d_1}$. Thus $N_{\overline{R}}(\overline{I}) = d_1 d_2$ has the desired form.

---

[1] J. Sally and W. Vasconcelos, "Stable rings", 1974.
[2] C. Greither, "On the two generator problem for the ideals of one-dimensional ring", 1982.
[3] L. Levy and R. Wiegand, "Dedekind-like behavior of rings with $2$-generated ideals", 1985.
[4] D. Eisenbud, "Commutative algreba with a view toward algebraic geometry", 1995.
[5] T. Ibukiyama and M. Kaneko, "Quadratic Forms and Ideal Theory of Quadratic Fields", 2014.

Answer 2

I am recording for the benefit of others what is to my knowledge the full extent of what is known about the general problem. Luc Guyot has provided a nice and explicit answer for the case of quadratic orders.

I do not mark this post as "the answer" as the original question has not yet been answered.

Let the discrepancy of a $T$-ideal $I$ be defined as $ds(I) = N_{\overline{T}}(\overline{I})/N_T(I)$ (non-standard definition).

When does $ds(I) = 1$?

The following theorem is the main tool of the paper [1]. The statement uses the module index notation of [2].

Theorem [1; Theorem 1]:

1. $[\overline{T}:\overline{I}] \subset [T:I]$.
2. $[\overline{T}:\overline{I^{-1}}] \subset [I:T]$.
3. $[{T}:{I^{-1}}] \subset [\overline{I}:\overline{T}]$.

Moreover, the following are equivalent:

• Any subset relation among (1), (2), (3) is an equality.
• All subset relations among (1), (2), (3) is an equality.
• $I$ is invertible.

This theorem has the following corollaries for the "discrepancy". Recall that the different of $T$ is defined to be $\mathfrak D_{T} = (T^\vee)^{-1}$ where $T^\vee$ is the dual of $T$ for the trace form.

Corollary: $ds(I) \geq 1$ with equality if and only if $I$ is invertible.

Corollary: The following are equivalent:

• The discrepancy of $\mathfrak D_{T}$ is $1$.
• For every ideal $I$ of $T$, $ds(I) = 1$ if and only if $T = (I:I)$.
• $T$ is Gorenstein.

Everything in these corollaries follows immediately from the theorem except the second point of the second corollary which follows from the well-known equivalence $T=(I:I) \iff I \text{ invertible}$ when $T$ is Gorenstein (cf. e.g. [3; Proposition 5.8] or [4; Proposition 2.7]).

Quadratic case

[Following the notation in Luc Guyot's answer]

Using the above corollaries we revisit the quadratic case. The discrepancy is invariant under homotheties and so we may assume the ideal $I$ is primitive ($e = 1$). By [5; Lemma 6.5], the ideal $I$ satisfies $R = (I:I)$ if and only if $\gcd(a,b,c) = 1$. Indeed, the formula for the discrepancy in Luc Guyot's answer is precisely $\gcd(a,b,c)$. (By the remark in Luc Guyot's answer, we even have $ds(I) = f/f'$ where $f$ is the conductor of $T$ and $f'$ is the conductor of $(I:I)$.) Thus the formula $ds(I) = c(q_I)$ is consistent with the second corollary.

Upper bound

We will derive an upper bound for $ds(I)$ which is independent of $I$. I assume that $T$ is a domain for simplicity. We may suppose that $T \neq \overline{T}$ and set $S = \overline{T}$. Let $\mathfrak f$ denote the conductor of $T$.

Upper bound: For any T-fractional ideal $I$, $ds(I) \leq |S/T||S/\mathfrak f|.$

Two $T$-fractional ideals are in the same genus if they are locally isomorphic; equivalently, there exists an invertible T-ideal which multiplies one ideal into the other.

Claim: Any $T$-fractional ideal $I$ is in the same genus as a $T$-fractional ideal $J$ such that $\mathfrak f \subset J \subset S.$

Proof: Let $P$ be a prime ideal of $T$ and let $S_P$ denote the integral closure of $T$ (integral closure commutes with localization). It suffices to construct a $T_P$-fractional ideal which is isomorphic to $I_P$ such that $\mathfrak f_P \subset J_P \subset T_P$ where subscript denotes tensoring with $T_P$. $S_P$ is a finite product of local Dedekind rings so it is a PID. Hence $I_PS_P = \alpha S_P$ for some $\alpha$ in $Quot(T)$. Let $J_P = \alpha^{-1}I_P$. Then $J_P \subset S_P$, but also $$J_P \supset J_P \mathfrak f_P = J_P S_P \mathfrak f_P = \mathfrak f_P.$$

Claim: The discrepancy $ds(I)$ is constant on genera.

Proof: This is proven by localizing and using that an invertible ideal of $T$ is locally principal (this latter fact follows from [5; Proposition 2.3]).

Putting these claims together, we have that for $I$ any $T$-fractional ideal, $ds(I) = ds(J)$ for some $T$-fractional ideal $J$ such that $\mathfrak f \subset J \subset S$. From [1; Theorem 1], $|T/J| \leq |S/SJ|$. We also have $S\mathfrak f = \mathfrak f \subset SJ \subset S$, and so $|S/SJ| \leq |S/\mathfrak f|$. Write $M' = M/\mathfrak f$ for any module containing $\mathfrak f$. Putting the inequalities together we have

$$ds(I) = |S/SJ|/|T/J| \leq |S/\mathfrak f|/ |T/J| = |S'|/(|T'|/|J'|) = |S/T| |J/\mathfrak f| .$$

The last term is bounded from above by $|S/T| |S/\mathfrak f|$.

Conclusion

The discrepancy function satisfies the inequality, $1 \leq ds(I) \leq |\overline{T}/T||\overline{T}/\mathfrak f|$, for any $T$-fractional ideal $I$, and admits an explicit and natural formula in terms of conductors in the quadratic case. However it appears to be unknown whether the discrepancy function can be given a "closed form" in general (e.g., an expression in terms of the conductor of $T$, the differents or discriminants of $T$ and $\overline{T}$, Ext or Tor groups over $T$ or $\overline{T}$).

References:

[1] I. Del Corso, R. Dvornicich, Relations among Discriminant, Different, and Conductor of an Order, 2000.

[2] A. Fröhlich, Local fields, from J. W. S. Cassels and A. Fröhlich, Algebraic number theory, 1967.

[3] L. Levy and R. Wiegand, Dedekind-like behavior of rings with 2-generated ideals, 1985.

[4] J. Buchmann and H. W. Lenstra, Jr., Approximating rings of integers in number fields, 1994.

[5] V. M. Galkin, $\zeta$-functions of some one-dimensional rings, 1973.