# Ideal norm in orders

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.

• You might get something by looking at minimal ring extensions. There are some restrictions one what can happen on the "residue side" see Section 5.1 of arxiv.org/pdf/1909.10860.pdf and the references there.
– hans
Aug 18, 2020 at 15:29
• Regarding your first remark, can you prove it without further assumptions? It is clear in the number field setting and works as well in the more general context of Del Corso and Dvornicich paper that you cite below. But is it the same context as yours?. Nov 12, 2020 at 18:11

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.

• Thank you for a very nice and helpful answer. For the time being I will leave the question open in the hopes that someone may attempt to say something about the general case. I have used your reference to [1] to add a useful remark to the original question. For the application I have in mind, one may assume that $T$ is the full ring of multipliers of the ideal $I$, in which case the problem is even solved by [1, Proposition 5.8]. Do you know whether Proposition 5.8 might extend in some fashion to more general classes of orders?
– AWO
Aug 18, 2020 at 2:53
• Thanks for the edit, I appreciate! And thanks for the feedback. Regarding [1, Proposition 5.8], I don't know of any generalisation at the moment ("Stable rings" by Sally and Vasconcelos, 1973, might be worthwhile reading), but I'll keep you posted if anything dawns on me (I like the question). I also hope that someone will come with something more general (bookmark on; I feel that my computation fails to capture some significant and general features of the problem at hand). Aug 18, 2020 at 7:17
• It looks like $e$ divides $a$, $b$, and $c$, and $q_I(0,1) = c/e$, so it looks like we have $\gcd(c(q_I), q_I(0,1)) = c(q_I)$? It also looks like $(a,b,c) \mapsto (ea,eb,ec)$ under $I \mapsto eI$, so it seems like the formula simplifies to $N_{\overline T}(\overline I) = N_T(I) \gcd(a,b,c)$ even if $I$ is not primitive, or even $N_{\overline T}(\overline I) = N_T(I) f/f'$ by your remark.
– AWO
Aug 20, 2020 at 22:40
• Thanks, indeed, it can be simplified. I was actually editing my answer while you where writing this comment. It seems that you made also some progress, very interesting. Aug 20, 2020 at 23:26

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]).

[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.

• In "Upper Bound" and "Conclusion", what are the assumptions made on $T$? Is $T$ any domain whose proper quotients are finite? Aug 20, 2020 at 23:43
• Throughout the post, $T$ is assumed to be an order of a Dedekind domain $\overline T$ whose proper quotients are finite. In "Upper Bound" and "Conclusion", it is further supposed that $T$ is not integrally closed.
– AWO
Aug 21, 2020 at 2:22
• Thanks for making this clear. I was confused by the sentence "I assume that $T$ is a domain for simplicity." since $T$ is a domain throughout. Aug 21, 2020 at 7:02
• In the original post I intended for a "Dedekind ring" to mean a finite product of Dedekind domains. Otherwise I would have been bothered that the problem as stated did not strictly make sense in the local setting since the localization of an order may have zero divisors.
– AWO
Aug 21, 2020 at 15:29
• Is the context of "Relations among Discriminant, Different, and Conductor of an Order" the same as your question, i.e., is it possible to identify R, K, F, S and A of the cited paper in your setting? Nov 12, 2020 at 17:56