Three differents

Written by GPT-6.1 Sol (OpenAI) in Codex, Ultra setting, October 2026. Self-checked by GPT-6.1 Sol (OpenAI). Public domain (CC0).

The discriminant detects a bad fibre in the base. A different is an ideal upstairs, so it can distinguish the primes of the extension. There are three constructions: annihilate the diagonal, differentiate the relations, or dualize the trace. Their agreement for Dedekind extensions is a theorem, not part of their definitions.

We assume Orders and the discriminant theorem, finite presentations, Kähler differentials and Fitting ideals; the differential background belongs to Kähler differentials. The scheme interpretation of vanishing differentials is developed in Unramified morphisms, Section 2. Here we work with finite ring maps. The arithmetic different-exponent theorem is stated precisely below. Basic references are [Stacks], [Noether] and [Sutherland].

1. Three constructions with different inputs

For a finite ring map \(A\to B\), let \(C=B\otimes_A B\), let \(\mu:C\to B\) be multiplication, and let \(I=\ker\mu\). Define

\[ \mathfrak D_N=\mu(\operatorname{Ann}_C I),\qquad \mathfrak D_K=\operatorname{Fitt}_0^B(\Omega_{B/A}). \]

The first is the Noether different, the second the Kähler different. We assume a finite presentation whenever Fitting ideals of differentials or their local support are used. This holds throughout over Noetherian bases.

For the Dedekind different, suppose \(A\) is a normal Noetherian domain with fraction field \(K\), \(B\) is finite and torsion-free over \(A\), and \(L=B\otimes_A K\) is a finite product of separable field extensions. Put

\[ B^*=\{x\in L:\operatorname{Tr}_{L/K}(xB)\subseteq A\}, \qquad \mathfrak D_D=(B:B^*)=\{x\in L:xB^*\subseteq B\}. \]

Trace integrality gives \(B\subseteq B^*\); hence \(\mathfrak D_D\subseteq B\). The trace pairing identifies \(B^*\) with \(\operatorname{Hom}_A(B,A)\), since it is nondegenerate generically. Calling \((B:B^*)\) an inverse does not imply that \(B^*\) is invertible. That distinction will matter for conductors.

2. The coefficient functional explains the trace dual

Theorem 2.1, Euler's trace-dual formula. Let \(A\) be a normal Noetherian domain, let \(f\in A[T]\) be monic of degree \(n\) and separable over \(K\), and let \(O=A[\theta]=A[T]/(f)\). Then

\[ O^*=f'(\theta)^{-1}O. \]

The generic algebra can be a product of fields; when \(f\) is the minimal polynomial of \(\theta\), it is a field.

Proof. Let \(\lambda:O\to A\) take the coefficient of \(\theta^{n-1}\) in the unique power-basis representative. The bilinear form \((a,b)\mapsto\lambda(ab)\) is perfect over \(A\). Its matrix in \(1,\theta,\ldots,\theta^{n-1}\) has entries zero when the exponents sum to less than \(n-1\), and one when they sum to \(n-1\). Reversing the columns makes it triangular with diagonal ones. Thus its determinant is a unit.

If \(r_1,\ldots,r_n\) are the distinct roots of \(f\), Lagrange interpolation for a polynomial \(q\) of degree less than \(n\) gives

\[ q(T)=\sum_i q(r_i)\frac{f(T)}{(T-r_i)f'(r_i)}. \]

Taking the coefficient of \(T^{n-1}\) gives

\[ \lambda(q(\theta))=\sum_i\frac{q(r_i)}{f'(r_i)} =\operatorname{Tr}_{L/K}\bigl(q(\theta)/f'(\theta)\bigr). \]

Hence \(f'^{-1}O\) represents the entire \(A\)-dual under trace, by the perfectness of the coefficient pairing. Conversely, if \(x\in L\) has integral traces against every element of \(O\), the corresponding functional is represented by a unique \(a\in O\) for the coefficient pairing, so \(x=a/f'\) by generic trace nondegeneracy. \(\square\)

Normality ensures that minimal polynomials of integral elements have coefficients in \(A\); once a monic presentation is already supplied, the displayed coefficient argument itself needs only the stated generic separability and the free power basis.

3. A difference quotient and a derivative

Theorem 3.1. For any commutative ring \(A\), monic \(f\in A[T]\), and \(B=A[T]/(f)\),

\[ \mathfrak D_N=\mathfrak D_K=(f'(\theta)). \]

Under the hypotheses of Theorem 2.1, \(\mathfrak D_D\) is the same ideal.

Proof. Identify \(C\) with \(B[Y]/(f(Y))\), where \(\theta\) denotes the first copy of the root. Then \(I=(Y-\theta)\). Define

\[ q(Y)=\frac{f(Y)-f(\theta)}{Y-\theta}=\frac{f(Y)}{Y-\theta}\in B[Y]. \]

If a representative \(h(Y)\) annihilates \(Y-\theta\) modulo \(f(Y)\), then \((Y-\theta)h=(Y-\theta)qv\) for some polynomial \(v\). Multiplication by the monic polynomial \(Y-\theta\) is injective even if \(B\) has zero divisors, so \(h=qv\). In \(C\), \(Yq=\theta q\); consequently the annihilator is generated by \(q\) as a \(B\)-module. Its image under multiplication is \(q(\theta)=f'(\theta)\).

The differential presentation is \(\Omega_{B/A}=B\,d\theta/(f'(\theta)d\theta)\), whose zeroth Fitting ideal is \((f')\). Finally Theorem 2.1 gives \((B:f'^{-1}B)=f'B\): containment in either direction follows by applying the condition to \(1\) and then to all of \(B\). \(\square\)

No nonzerodivisor assumption on \(f'\) is needed for the first two computations. Generic separability is essential for the trace-dual computation.

Proposition 3.2. The Noether different commutes with localization and flat base change for a finite map.

Proof. Choose finitely many algebra generators \(b_i\) of \(B/A\). The elements \(b_i\otimes1-1\otimes b_i\) generate \(I\), since their quotient identifies the two copies of every generator and hence of \(B\). Its annihilator is the kernel of the map from \(C\) to a finite direct sum of copies of \(C\) given by multiplication by these generators. Localization, or any flat base change, preserves this kernel. It also preserves \(I\) as the kernel of multiplication, and preserves the image defining \(\mathfrak D_N\). This proves both assertions. Localizing the resulting ideal further in \(B\) is compatible with its ordinary ideal localization. \(\square\)

4. Why the ideals detect unramified points

There is a canonical isomorphism \(I/I^2\simeq\Omega_{B/A}\), sending the class of \(b\otimes1-1\otimes b\) to \(db\). For a prime \(\mathfrak q\subset B\), let \(\mathfrak Q=\mu^{-1}(\mathfrak q)\subset C\). If \(\Omega_{B/A,\mathfrak q}=0\), then \(I_{\mathfrak Q}/I_{\mathfrak Q}^2=0\). Since \(I\) is finite and \(I_{\mathfrak Q}\) lies in the local maximal ideal, Nakayama gives \(I_{\mathfrak Q}=0\). Clearing denominators for its finitely many generators produces \(h\in\operatorname{Ann}I\) with \(h\notin\mathfrak Q\), so \(\mu(h)\notin\mathfrak q\). Thus \(\mathfrak D_N\) is a unit at \(\mathfrak q\). The converse follows because such an \(h\) annihilates \(I\) and becomes invertible in \(C_{\mathfrak Q}\).

For a finite presented module \(M\) over a local ring, \(\operatorname{Fitt}_0(M)\) is the unit ideal exactly when \(M=0\): a unit maximal minor makes the presentation surjective, and the reverse direction is immediate. Therefore \(\mathfrak D_K\) has the same support. For a finite presented algebra, vanishing of relative differentials at a point is the unramified criterion [Stacks, Tags 0BVU and 0BVY]. We have proved the corresponding support statements for \(\mathfrak D_N\) and \(\mathfrak D_K\).

For finite flat maps, the different defined by the trace-dual module has support exactly the non-étale locus [Stacks, Tags 0BW5 and 0BW9]. Flatness turns unramifiedness into étaleness. This trace-dual support theorem is a stated prerequisite; it does not assert equality of all three ideals for every finite flat algebra.

Proposition 4.1. If \(A\) is a normal Noetherian domain and \(B\) is finite projective with finite separable generic algebra, then \(\mathfrak D_N\subseteq\mathfrak D_D\).

Proof. An element \(\alpha=\sum u_i\otimes v_i\) annihilating \(I\) defines

\[ F_\alpha:\operatorname{Hom}_A(B,A)\to B, \qquad \lambda\mapsto\sum u_i\lambda(v_i). \]

The relation \((b\otimes1)\alpha=(1\otimes b)\alpha\) makes it \(B\)-linear. Generically, extend scalars to split the separable algebra into copies of a field. An element annihilating the diagonal ideal has only diagonal components; contraction with the sum-of-components trace functional gives exactly its multiplication image. Returning by faithful scalar extension, and identifying the dual with \(B^*\), we obtain \(F_\alpha(x)=\mu(\alpha)x\) for every \(x\in B^*\). Its values lie in \(B\), so \(\mu(\alpha)B^*\subseteq B\). Every generator of the Noether different belongs to the Dedekind different. \(\square\)

5. Equality for finite Dedekind extensions

Lemma 5.1. A finite extension of complete DVRs \(R\subset S\) with separable residue extension is monogenic over \(R\).

Proof. Write \(k\subset\ell\) for the residue fields. Choose a primitive element \(\bar u\) of \(\ell/k\), lift its monic separable polynomial to \(h\in R[T]\), and apply Hensel's lemma in \(S\) to obtain \(u\in S\) with \(h(u)=0\) and residue \(\bar u\). Let \(\pi\) be a uniformizer of \(S\), and put \(\theta=u+\pi\). Taylor expansion gives \(h(\theta)=h'(u)\pi\pmod{\pi^2}\), so \(h(\theta)\) is a uniformizer too. The residue of \(\theta\) generates \(\ell/k\).

Let \(e\) be the ramification index. Every element of \(S/\pi_R S=S/(\pi^e)\) is a sum of terms consisting of a lift of a residue element times a power of \(h(\theta)\), of exponents less than \(e\): subtract a residue representative and repeat in the successive quotients \((\pi^j)/(\pi^{j+1})\). Each residue representative can be chosen as a polynomial in \(\theta\) with coefficients in \(R\). Thus \(R[\theta]\) surjects onto \(S/\pi_R S\). The finite \(R\)-module \(S/R[\theta]\) is equal to its multiple by \(\pi_R\); Nakayama makes it zero. \(\square\)

Theorem 5.2. For a finite extension of Dedekind domains \(A\subset B\) with separable fraction-field extension,

\[ \mathfrak D_N=\mathfrak D_K=\mathfrak D_D. \]

Proof when residue extensions are separable. Localize at a prime of \(A\), then complete. The finite algebra becomes a product of complete DVRs, one for each prime of \(B\) above it. Each is monogenic by Lemma 5.1, so Theorem 3.1 gives equality on every factor. The constructions commute with this flat base change: for \(\mathfrak D_N\) use Proposition 3.2; for \(\mathfrak D_K\) use the differential presentation and Fitting minors; for \(\mathfrak D_D\) use \(B^*\simeq\operatorname{Hom}_A(B,A)\) and finite presentation. Completion is faithfully flat over the localized DVR, so the equality descends. Equality at all base primes proves equality globally.

General residue fields. A finite map between these regular one-dimensional rings is flat and a local complete intersection. Flatness follows from torsion-freeness over each DVR. For the second assertion, a local polynomial presentation has a regular ambient local ring and regular quotient; its kernel is generated by a regular sequence [Stacks, Tag 00NR]. The finite flat local complete intersection different theorem [Stacks, Tags 0BWD and 0BWG], obtained from Tate's determinant lemma [Stacks, Tag 0BWC], identifies the Noether and Kähler differents. Such a map has invertible relative dual, so [Stacks, Tags 0BW5 and 0BW6] identifies these with the Dedekind different. These precisely stated local complete intersection theorems are the cited inputs for the inseparable-residue case. \(\square\)

For relations \(f_1,\ldots,f_n\) in a finite flat local complete intersection presentation, Tate's formula gives the different as the Jacobian determinant \(\det(\partial f_i/\partial x_j)\). This is the multivariable form of the difference quotient in Section 3. It explains Noether's ideal differentiation: differentiate the ideal of relations, with presentation independence supplied by the theorem.

The arithmetic exponent theorem says, for finite separable extensions of DVRs, that \(v_{\mathfrak P}(\mathfrak D_D)\ge e_{\mathfrak P}-1\), with equality exactly when the residue extension is separable and the ramification index is prime to the residue characteristic. The arithmetic treatment belongs to The different and the discriminant. For number fields see [Sutherland, Theorem 12.27] and, with a complete proof for extensions of the rational numbers, [Conrad, Theorem 4.13]; the argument covering imperfect residue fields is given in Normal integral bases in tame extensions, Section 2, after Theorem 2.1. We state the exponent theorem here as an input rather than restrict its scope to finite residue fields. The norm identity for the discriminant is [Sutherland, Theorem 12.17].

6. Examples and exercises

In \(\mathbb Z[i]/\mathbb Z\), \(f(T)=T^2+1\), so all three differents are \((2i)=(1+i)^2\). In the order \(\mathbb Z[\sqrt5]\), they are \((2\sqrt5)\); in its maximal order, with \(\omega=(1+\sqrt5)/2\) and polynomial \(T^2-T-1\), they are \((2\omega-1)=(\sqrt5)\). These are ideals in different rings and must be compared after extension to the larger ring.

For \(k[t]\subset k[u]\) with \(u^2=t\) and characteristic different from two, the different is \((2u)=(u)\). At \(u=0\) its exponent is one, equal to \(e-1\); this is tame ramification. In characteristic \(p\), the map \(k[t]\subset k[u]\), \(u^p=t\), has zero Noether and Kähler differents, since the derivative is zero. The trace-dual definition above does not apply: its generic extension is inseparable and the trace pairing is identically zero.

  1. Easy. Compute all three differents for \(\mathbb Z[i]/\mathbb Z\).
  2. Medium. Compute the annihilator of \(Y-\theta\) in \(B[Y]/(f(Y))\), allowing \(f'\) to vanish.
  3. Medium. Compare the differents and discriminants of the two quadratic rings attached to \(\sqrt5\).
  4. Medium. Prove localization compatibility for the Noether different by expressing its annihilator as a finite kernel.
  5. Hard. Prove monogenicity for finite extensions of complete DVRs with separable residue extension, and explain why it proves equality of the differents before descent.

7. Solutions

1. In the tensor presentation \(\mathbb Z[i][Y]/(Y^2+1)\), the quotient polynomial is \(Y+i\), with multiplication image \(2i\). The differential module is \(\mathbb Z[i]/(2i)\,di\). The trace-dual formula gives \((2i)^{-1}\mathbb Z[i]\), whose inverse is \((2i)\). Finally \(2i\) is a unit multiple of \((1+i)^2\).

2. Divide \(f(Y)-f(\theta)\) by \(Y-\theta\) to obtain \(q(Y)\). The annihilator calculation in Theorem 3.1 uses cancellation by the monic polynomial \(Y-\theta\), not by \(f'\). Hence it is \(Bq\) whether or not \(f'\) is zero. For \(f(T)=T^p-t\), its image under multiplication is zero, consistently with the inseparable example.

3. The derivatives are \(2\sqrt5\) and \(2\omega-1=\sqrt5\). The corresponding discriminants are \(20\) and \(5\), as computed by the trace matrices in the preceding lesson. The lattice index is two, and \(20=2^2\cdot5\). Extending the order's different to the maximal order multiplies that larger ring's different by \((2)\).

4. With generators \(d_i=b_i\otimes1-1\otimes b_i\) of \(I\), use \(\operatorname{Ann}I=\ker(C\to C^r)\), \(c\mapsto(cd_i)_i\). Exactness of localization identifies its localized kernel with the annihilator of the localized diagonal ideal. Multiplication and its image also localize. This proves the assertion without assuming the annihilator itself is principal.

5. Lift a separable residue primitive element by Hensel, add a uniformizer, and use the polynomial \(h\) to recover a uniformizer from the sum. Powers of \(h(\theta)\), with polynomial residue representatives in \(\theta\), span \(S/\pi_R S\). Nakayama gives \(S=R[\theta]\). Its minimal polynomial is monic and separable over the fraction field, so the three computations of Theorem 3.1 apply. Faithfully flat completion and localization then allow the ideal equality to descend as in Theorem 5.2. Without residue separability, this construction has no invertible derivative \(h'(u)\); the local complete intersection theorem supplies the general comparison instead.

What this lesson does not prove

Hensel's lemma, the local regular-sequence criterion, Tate's determinant lemma, the finite flat local complete intersection different theorem, and the different-exponent theorem are stated inputs with the locators above. The finite flat trace-dual support theorem is [Stacks, Tag 0BW9]. Euler's formula, monogenic computations, localization, the containment, monogenicity with separable residue fields and the comparison in that case are proved here.

References