An integral degree-three Gersten counterexample

Proof. The polynomial

is Eisenstein at five. Therefore \pi=1-\zeta is a uniformizer of V, its residue field is \F_5, and 5=u\pi^4 for some u\in V^\times. The homogeneous equation P defining \X reduces to

If \alpha is a root of s^5-s+1, then Frobenius sends \alpha to \alpha-1. Its orbit has length exactly five: the rth iterate is \alpha-r, which first returns at r=5. Thus s^5-s+1 is irreducible over \F_5, and so is its homogenization \overline P. This remains irreducible on adjoining the independent variable Z. A factorization of the primitive homogeneous polynomial P over L would, by Gauss’ lemma, give primitive positive-degree homogeneous factors over V and hence a factorization of \overline P. Consequently C is integral. It is a projective plane curve, so it has dimension one.

Put

In the regular local ring V[x,y]_{(\pi,x,y)} of dimension three, F\equiv-\pi modulo the square of the maximal ideal. Thus quotienting by F gives the regular local ring A=R_{\mathfrak m} of dimension two. A regular local ring is a domain. Since F is monic in x, the ring R is free over V[y] and hence flat over V. Its localization A is also V-flat, so V embeds into A. Together with A/\mathfrak m_A=\F_5, this proves mixed characteristic (0,5). Expanding F=0 gives

In particular, \pi\in(x,y)A and \mathfrak m_A=(x,y)A. Since 5 is divisible by \pi, the term 5x is in (x,y)^2A, as are all remaining terms on the right. Hence \pi\in\mathfrak m_A^2. Since 5=u\pi^4, also 5\in\mathfrak m_A^2, so A is ramified in the usual mixed-characteristic sense. The reduction of the equation has no linear term at m, so the one-dimensional special fiber is singular there; A is not smooth over V.

Modulo y, the equation becomes

This is Eisenstein at \pi. The finite free V-algebra B=V[t]/(f) is a domain. Every maximal ideal lies over the maximal ideal of V, and B/\pi B=\F_5[t]/(t^5), so B is local. Since f(t)=0 implies \pi\in(t)B, its maximal ideal is (t). Its dimension is one, making it a DVR. The localization defining A/yA leaves this local ring unchanged, proving (1). Its fraction field has degree five over L and residue degree one, hence total ramification. The constant term gives

The relation (1-t)^5=\zeta also follows from f(t)=0. As S[1/\pi]=D, the equation y=0 on the generic affine chart gives a single closed point. There is no additional point with Y=Z=0 on C, since substitution in P would force X=0. This proves the assertion about q_0 on all of C.

We have constructed the distinguished point with residue degree one. It remains to establish the residue-degree constraint on points that specialize away from m. A closed point of C has residue field finite over L. Properness of \X/V extends its E-point to \Spec\OO_E. On the special fiber Y=0 forces X=0, so m is the only point with Y=0. If the closed image is elsewhere, X/Y in k_E is a root of s^5-s+1. Thus k_E contains \F_{5^5}, and 5\mid[k_E:\F_5].□

Proof. The coefficient cofiber sequence is a sequence of K(D)-modules. Its connecting map is therefore compatible with multiplication by an integral K(D) class, up to the suspension sign. By (2), \zeta=\eta^5 and t=1-\eta. Both \eta and 1-\eta are nonzero: the first has fifth power \zeta\ne0, and the second is the uniformizer t of S. Matsumoto’s Steinberg relation [7, §5, Lemma 5.6, Theorem 5.10, and Corollary 5.11], with the identification of symbols and products in Quillen K_2(D) [16, III.6.1, IV.1.7.1, IV.1.10.1, and IV.7.2], gives

The sign does not affect the vanishing. Exactness of (3) yields an integral lift \widetilde a_D\in K_3(D) of a_D. The boundary [\zeta]\in K_1(D)[5] of \beta_D itself is still nonzero, even though \zeta=\eta^5. Being a fifth power makes the class of \zeta vanish in D^\times/D^{\times5}, not in the integral group K_1(D)=D^\times. The Steinberg relation is what kills the boundary of the product a_D.

Localization and dévissage for the DVR S give [11, §7, Proposition 3.2]

The last group is zero by Quillen’s finite-field computation [10]. We may therefore choose b\in K_3(S) restricting to \widetilde a_D. This is a second lift, distinct from the coefficient lift to K_3(D).

Let c=j_*b\in G_3(A)=K_3(A). Its restriction to \Spec A[1/y] is zero because the closed immersion j has empty pullback there. The element y is nonzero in the domain A: its quotient S has dimension one, whereas A has dimension two. Hence y becomes invertible in \Frac A, and c|_{\Frac A}=0 integrally.

Base change of j along T\to\Spec A is i:\Spec D\hookrightarrow T, since S[1/\pi]=D. Flat base change for coherent pushforward [16, V.3.7.2] here is the natural isomorphism

for a finite S-module M, with both sides viewed as A[1/\pi]-modules. Thus restriction of scalars commutes with localization. Naturality of coefficient reduction gives the commutative diagram

Both vertical arrows mean restriction followed by coefficient reduction. The left arrow sends b to a_D, which proves the stated identity.□

Proof. We first identify the degree-three group, then locate the image of \beta_E in degree two, and finally compute multiplication by \beta_E on integral K_1. The strongly convergent motivic-to-K spectral sequence for a characteristic-zero field is

It has a pairing with the integral spectral sequence that induces the integral/coefficient K-module product on the abutment [3, Theorem 16.2 and Appendix C.2–C.3]. The comparison with motivic complexes preserves products [14, Lemma 3.4 and Theorem 6.1]. Write s=a-b and j=-b, so a term H^s(E,\Z/5(j)) has total K-degree 2j-s. Negative weights vanish. For j\geq0, the field motivic groups vanish for s>j, and the norm-residue and motivic–étale comparison theorems identify them, for every s\leq j, with

see [15, Theorems 6.16–6.17] and [16, VI.4.1–4.2]. In particular, the groups vanish for s<0. These comparisons apply because 5 is invertible in E.

The 5-cohomological dimension of a finite extension of \Q_5 is two [16, VI.§7, before Proposition 7.3]. Thus the only potentially nonzero terms satisfy 0\leq s\leq\min(j,2). For r\geq2 a differential changes (s,j) to (s+2r-1,j+r-1), so none can join two nonzero terms. Total degree three has the single term (s,j)=(1,2), yielding the natural identification

In particular, this K-group is killed by five. In total degree two the weight-one quotient of the filtration is H^0(E,\mu_5); the only other possible piece is H^2(E,\mu_5^{\otimes2}) in weight two.

Let \alpha_E be the weight-one image of \beta_E. To see that it is nonzero, restrict to an algebraic closure \overline E. The same comparison there has only s=0 terms, so its total degree-two group has just the weight-one piece H^0(\overline E,\mu_5). Matsumoto’s presentation of field K_2 ([7, §5]; see also [16, III.6.1]) gives K_2(\overline E)/5=0: every symbol \{u,v\} equals 5\{r,v\} after choosing r^5=u. The coefficient sequence therefore identifies K_2(\overline E;\Z/5) with K_1(\overline E)[5] through \partial_{\mathrm{coef}}. The restriction of \beta_E is nonzero, since its boundary is the primitive root [\zeta]. Naturality of the filtration shows that \alpha_E has nonzero image in H^0(\overline E,\mu_5). It is consequently a generator of H^0(E,\mu_5); only this nonvanishing is needed.

It remains to determine the actual product from these filtration pieces. Integral K_1(E)=E^\times has only the weight-one piece H^1(E,\Z(1)): on the diagonal 2j-s=1, weights j\geq2 give s>j and vanish, weight zero gives H^{-1}(E,\Z(0))=0, and negative weights vanish. The identification \Z(1)\simeq\mathbb G_m[-1] identifies reduction of [u] with its Kummer class in H^1(E,\mu_5)=E^\times/E^{\times5}.

Write F^j for the decreasing weight filtration. The preceding calculations give

The class \beta_E lies in F^1K_2(E;\Z/5) and has image \alpha_E in its quotient by F^2. The mixed filtered pairing satisfies

Thus the possible weight-two part of \beta_E contributes nothing: the product \beta_E\cdot[u] is determined by \alpha_E and the weight-one class of [u]. Product compatibility identifies it, under (7), with \alpha_E\cup[u]. Since \alpha_E generates the constant one-dimensional \F_5-module \mu_5, this cup product is an isomorphism. The product factors through fifth powers because its target is killed by five and [u^5]=5[u]. This proves (5) with its asserted multiplicative identification.□

Proof. Restriction of scalars obeys the tensor projection formula at the level of K-theory spectra [16, V.3.3.2]. Smashing its base-field factor with M_5 gives the mixed pairing identity

The second equality uses the norm description of field K_1 transfer [16, III.1.7.1 and the following paragraph]. This proves (8).

For normalized valuations, the norm of the prime ideal of E is the [k_E:k_L]th power of the prime ideal of L. Hence

Every a\in K_3(E;\Z/5) has the form \theta_E(u) by Lemma 4.1. Applying the formula to that u proves the assertion for all a. The factor is the residue degree, not the total degree [E:L].□

Proof. The multiplicative set \Sigma is directed by taking products: hk gives a common refinement of h and k. Transitivity of localization gives

Each refinement is a principal localization, so the corresponding scheme map is affine and flat. The stages and their limit are affine Noetherian schemes, hence separated. Quillen’s continuity theorem, in its coherent K' conclusion [11, §7, Proposition 2.2], therefore applies to their integral G-groups.

To pass to coefficients, use the natural short exact sequence

for each stage Y=U_h and for Y=T. Filtered colimits of abelian groups are exact, and thus commute with the quotient by five and the kernel of multiplication by five. Integral continuity in degrees three and two identifies the outer terms in the resulting comparison of short exact sequences. It consequently identifies the middle terms, proving the coefficient assertion in (9).

Since h\notin\mathfrak m, its image in the local divisor ring S is a unit. Thus q_0=\Spec D belongs to U_h. More generally, consider a closed point q\in C with residue field E whose extension \Spec\OO_E\to\X specializes to m. The inverse image of the chart Z\ne0 contains the closed point of \Spec\OO_E, and hence is all of \Spec\OO_E. On that chart the reduction of h at m is nonzero. Its pullback to \OO_E is therefore a unit, so q lies in U_h. It follows that each point omitted from U_h specializes away from m. Lemma 2.1 then gives 5\mid[k_E:\F_5]. Finally, C is an integral projective curve and U_h is nonempty, so its closed complement consists of finitely many closed points.□

Proof of Theorem 1.1. Take the integral class c\in K_3(A) constructed in Proposition 3.1, and put

Coherent G-theory provides proper pushforward even when C is singular. The point immersion is finite, and the structural map p:C\to\Spec L is proper with C carrying the ample line bundle \mathcal O_C(1). Coherent pushforward and its composition law therefore apply to both maps [11, §7, paragraph 2.7 and the discussion following (2.8)].

Flat base change for the point immersion on each U_h [16, V.3.7.2], followed by (9), identifies w|_T with the pushforward of \theta_D(t) along \Spec D\hookrightarrow T. By Proposition 3.1, this is precisely the coefficient reduction of c|_T.

Suppose that w|_T=0. The element w|_{U_1} represents this class in the filtered colimit (9). A zero element in that colimit becomes zero at a finite stage, so w|_{U_h}=0 for some h\in\Sigma. Give Z_h=C\setminus U_h its reduced closed structure. Apply Moore coefficients to coherent localization and dévissage at these closed points [11, §7, Proposition 3.2 and paragraph 3.4]. The resulting exact segment is

Consequently there are classes a_q\in K_3(E_q;\Z/5) such that

These are arbitrary coefficient classes supplied by localization. Lemma 4.1 applies to every finite extension E_q/L, so its surjectivity writes each of them as a_q=\theta_{E_q}(u_q) for some u_q\in E_q^\times.

Apply the proper pushforward p_*:G_3(C;\Z/5)\to K_3(L;\Z/5) to (10). For every closed point, the composite p_*q_* is the field transfer. Lemma 5.1 gives 5\mid[k_{E_q}:\F_5], and Lemma 4.2 therefore gives

Thus (10) forces \lambda_L(p_*w)=0. On the other hand, the distinguished point has residue field D, and its parameter satisfies N_{D/L}(t)=\pi. The same transfer formula yields

This contradiction proves w|_T\ne0. Hence the coefficient reduction of c|_T is nonzero, so the integral class c\in K_3(A) is itself nonzero. Its image at the fraction field is zero by Proposition 3.1; together with the properties of A in Lemma 2.1, this proves the theorem.□