## K_2(O_F) for number fields F

Belabas and Gangl have a nice paper ( Generators and relations for $K_2({\mathcal{O}}_F)$, which can be found here) where they compute $K_2({\mathcal{O}}_E)$ for a large number of quadratic fields $E$. There main result is a method for proving upper bounds for $K_2({\mathcal{O}}_E)$ in a rigorous and computationally efficient way. Tate had previously computed these groups for small imaginary quadratic fields by hand;  — the problem is finding an efficient way to do this in general. (Brownkin and Gangl had previously found a non-rigorous way of computing these groups using $K_3({\mathcal{O}}_E)$ and regulator maps, but more on that later.) A good analogy to keep in mind is the problem of computing the class groups of imaginary quadratic fields. In the latter case, however, there are rigorous ways to determine whether an element in the class group is non-trivial, and this is missing from the computation of $K_2({\mathcal{O}}_E)$.  To produce lower bounds, [BG] use theorems of Tate and Keune to relate the $p$-primary part of $K_2({\mathcal{O}}_E)$ to class groups of $E(\zeta_p)$, which they can then compute in some cases. One nice example they give is

$K_2\left({\mathbf{Z}} \displaystyle{\left[\frac{1 + \sqrt{-491}}{2} \right]} \right) = {\mathbf{Z}}/13 {\mathbf{Z}}.$

Akshay and I used this as one of the examples in our paper; in our context, it implies that the order of the group

$H_1(\Gamma_0({\mathfrak{p}}),{\mathbf{Z}})$

is always divisible by $13$ where $\Gamma_0({\mathfrak{p}})$ is the congruence subgroup of $\mathrm{PGL}_2({\mathcal{O}}_E)$ for $E = {\mathbf{Q}}(\sqrt{-491})$ and ${\mathfrak{p}}$ is any prime — even though the group $H_1(\Gamma,{\mathbf{Z}}) = ({\mathbf{Z}}/2{\mathbf{Z}})^{26}$ is not so divisible. (Because we are talking about $\mathrm{PGL}$ rather than $\mathrm{PSL}$, the cusps are quotients of tori by involutions, so only contribute $2$-torsion to $H_1.$ This group is occasionally infinite; we use the convention that $\infty$ is divisible by $13.$) It’s always nice to see a theoretical argument come to life in an actual computation — fortunately, Aurel Page was kind enough to compute a presentation for $\Gamma$ in order for us to do this. Now that I think about it, this and many other interesting things didn’t make it into the submitted version of the paper; you’ll have to read the “directors cut” to learn about it.

Alexander Rahm pointed out to me that the computation of $K_2({\mathcal{O}}_E)$ we used was annotated with an asterisk in [BG], meaning that what was proved was only an upper bound. The issue is as follows. Let $p = 13$, and let $F = E(\zeta_p)$, let $G = {\mathrm{Gal}}(F/E) = ({\mathbf{Z}}/p {\mathbf{Z}})^{\times}$, and let ${\mathrm{Cl}}(F)$ denote the class group of $F$. What is required is to show, in light of Tate’s work on $K_2$, is that

$({\mathrm{Cl}}(F)[p])^{G = \chi^{-1}} \ne 0,$

where $\chi: G \rightarrow {\mathbf{F}}^{\times}_p$ is the cyclotomic character. The problem is that $F$ has degree $24$, and it is difficult to compute class groups explicitly in such cases. Let $H = {\mathrm{Gal}}(F/{\mathbf{Q}})$, so there is a canonical decomposition $H = G \times {\mathbf{Z}}/2{\mathbf{Z}}$. There are two extensions of $\chi$ to $H$, given (with some abuse of notation) by $\chi$ and $\chi \eta$, where $\eta$ is the non-trivial character of ${\mathrm{Gal}}(F/{\mathbf{Q}})$. The main conjecture of Iwasawa Theory (Mazur–Wiles) allows one to easily compute minus parts of class groups in terms of $L$-values without actually computing with explicit number fields. However, we should not expect this to help us here. Namely, it’s not hard to show that there is an isomorphism $({\mathrm{Cl}}({\mathbf{Q}}(\zeta_p))[p])^{G = \chi^{-1}} \simeq ({\mathrm{Cl}}(F)[p])^{H = \chi^{-1}}.$ However, the former is trivial by Herbrand’s theorem, because $B_2 = 1/6$ is not divisible by $13$. That leaves us with the problem of proving that $({\mathrm{Cl}}(F)[p])^{H = \chi^{-1} \eta} \ne 0,$ which is a statement about the class group of a totally real cyclotomic extension. Since $\chi \eta^{-1}$ is an even character, we get some savings by working in the totally real subfield $F^{+}$ of degree $12$. Now $\text{\texttt{pari}}$ happily tells me via $\text{\texttt{bnfinit}}$ and $\text{\texttt{bnfclgp}}$ that the class group of this field is ${\mathbf{Z}}/13 {\mathbf{Z}}$, so it looks like we are in good shape. However, $\text{\texttt{pari}}$ has the habit when computing class groups of assuming not only GRH but something stronger. What information does $\text{\texttt{bnfinit}}$ actually contain? It certainly gives, inter alia:

1. The Galois automorphisms of $F^{+}$, using $\text{\texttt{gal:=nfisisom(nf,nf)}}$.
2. A finite index subgroup $V$ of the unit group $U:={\mathcal{O}}^{\times}_{F^{+}}$, using $\text{\texttt{bnfinit[8][5]}}$.

Let me show how, just with this data, one can prove that the relevant part of the class group we are interested in is non-zero. BTW, if you tell $\text{\texttt{pari}}$: can you confirm this answer is really correct? (using $\text{\texttt{bnfcertify}}$) it complains, and says the following:

*** bnfcertify: Warning: large Minkowski bound: certification will be VERY long.
*** bnfcertify: not enough precomputed primes, need primelimit 59644617.

A rough guess (in part) as to what it might be doing: to compute all the invariants necessary for class field theory, one needs to know the full unit group. To do this, one can take the units $V$ found so far and saturate them in the entire unit group $U$. For each prime $q$, one can do this by taking representatives in $V/q V$ and determining whether or not they are perfect $q$th powers. By taking enough primes, on either rules out the existence of such an element, or finds a candidate $v \in V$ and then checks whether it is a $q$th power. On the other hand, from $V$, one can compute a pseudo-regulator $R_V$, which is related to the actual regulator $R_U$ by the unknown index. So to make this computation finite, it suffices to have some a priori bound on the regulator (to give an upper bound on the index), which will ultimately come down to some a priori bound on an $L$-value at one, which GRH probably tells you something useful about.

One can identify the automorphims of $F^{+}$ computed by $\text{\texttt{pari}}$ with the elements of the Galois group given by the corresponding quotient of $H = G \times {\mathbf{Z}}/2{\mathbf{Z}}$ by $(-1,1)$. This group is generated by the image of $\sigma = (2,1) = \mathrm{Frob}_2$, so it is enough to find the automorphism $\sigma$ such that $\sigma \theta - \theta^2 \equiv 0 \mod 2.$ View $\chi \eta$, a character of degree $12$, as being valued in ${\mathbf{F}}^{\times}_{13}$. Now choose a random unit, say $\text{\texttt{u:=bnf[8][5][6]}}$ (Warning! I have a feeling that $\text{\texttt{bnfinit}}$ does something different each time you run it, which means you might have to tweak the choice of index $6$ above if you are doing this at home. And by “you,” I really mean “me” in six months time. I guess I should also tell myself that the relevant $\text{\texttt{pari}}$ file is $\text{\texttt{\~{}fcale/Zagier/BG491}}$.) We may write down a second unit as follows:

$\displaystyle{{\epsilon} = \prod_{i=0}^{12} (\sigma^i(u))^{\chi \eta (\sigma^i)} \in ({\mathcal{O}}^{\times}_F/{\mathcal{O}}^{\times p}_F)^{H = \chi^{-1} \eta}}$

What we have done is apply the appropriate projector in the group ring ${\mathbf{F}}_{13}[H]$ to $u$. Naturally enough, we can lift ${\epsilon}$ to an actual unit in $F^{+}$.

Now choose an auxiliary prime $q$ which splits completely in $F$, say $q = 38299$. I chose this because it actually splits completely in ${\mathbf{Q}}(\zeta_{13 \cdot 491})$, which will make a computation below slightly easier. We reduce ${\epsilon}$ modulo a prime ${\mathfrak{q}}$ above $q$ in ${\mathcal{O}}_{F^+}$ and we find that

${\epsilon}^{(q-1)/13} \not \equiv 1 \mod {\mathfrak{q}}.$

What this last computation proves is that ${\epsilon}$ actually generates
$({\mathcal{O}}^{\times}_F/{\mathcal{O}}^{\times p}_F)^{H = \chi^{-1} \eta}$, which has dimension one by Dirichlet’s theorem. Note also that the inequality above does not depend on the choice of ${\mathfrak{q}}$ — any other choice is conjugate to ${\mathfrak{q}}$ which replaces ${\epsilon}$ by $\sigma {\epsilon}$ and the latter is a non-zero scalar multiple of the former modulo $13$th powers by construction.

On the other hand, let $\zeta$ be a primitive $13 \cdot 491$th root of unity. Then we may consider the projection of $1 - \zeta$ modulo $13$th powers to the $\chi^{-1} \eta$ eigenspace (the latter is naturally also a character on $F(\zeta_{491})$). Remember this eigenspace is generated by ${\epsilon}$. Take $q = 38299$ again, so $q - 1 = 13 \cdot 491 \cdot 6$. Then $2^{6}$ is a primitive $13 \cdot 491$th root of unity modulo $q$. On the other hand,

$\displaystyle{\left(\prod_{({\mathbf{Z}}/13 \cdot 491 {\mathbf{Z}})^{\times}} (1 -2^{6n})^{n \left(\frac{n}{491}\right)} \right)^{(q-1)/13} \equiv 1 \mod q}$

(The exponent of $(1 - 2^{6n})$ is just the value of $\chi \eta(n)$ — remember that the character gets inverted in the projection formula — and that $\eta^{-1} = \eta$.) This implies that the projection of $(1 - \zeta)$ to the $\chi^{-1} \eta$-eigenspace of units modulo $p = 13$ is trivial, because the image of ${\epsilon}^{(q-1)/13}$ computed above was not $1 \mod q$. The same is trivially true for the units in ${\mathbf{Q}}(\zeta_{491})$ and ${\mathbf{Q}}(\zeta_{13})$, because the projection of any unit in a subfield of $F$ can only be an eigenvalue for a character of the correponding quotient of the Galois group. In particular, if $C$ denotes the group of circular units, we have shown that the map

$(C/13 C)^{\chi^{-1} \eta} \rightarrow ({\mathcal{O}}^{\times}_F/{\mathcal{O}}^{\times 13}_F)^{\chi^{-1} \eta}$

is the zero map. This proves that the index of the circular units in the entire units is divisible by $13$. This is enough to prove that $13$ divides $h^{+}_F$, but even better, by the Gras conjecture (also proved by Mazur–Wiles, following Greenberg) it follows that the $\chi^{-1} \eta$-part of the class group is non-zero, and hence, given the previous upper bound, this gives a proof that

$K_2({\mathcal{O}}_E) = {\mathbf{Z}}/13 {\mathbf{Z}}.$

Further Examples: Let’s now look at other examples in [BG]. Consider the following example:

$\displaystyle{K_2\left( {\mathbf{Z}} \left[ \frac{1 + \sqrt{-755}}{2} \right] \right) =^{?} {\mathbf{Z}}/41 {\mathbf{Z}} \oplus {\mathbf{Z}}/2 {\mathbf{Z}}}.$

Let $F = E(\zeta_{41})$, and let $F^{+}$ be the totally real subfield of $F$ of degree $40$.  Well we certainly won’t be able to say so much about the class group of $F^{+}$. On the other hand, we can do the latter part of the computation, namely, testing that the $\chi^{-1} \eta$-eigenspace in the circular units looks like it has index divisible by $p$ in the entire units. For example,  if $q = 123821 = 1 + 41 \cdot 755 \cdot 4$, we can compute that

$\displaystyle{\left(\prod_{({\mathbf{Z}}/41 \cdot 755 {\mathbf{Z}})^{\times}} (1 -2^{n(q-1)/(41 \cdot 755)})^{n \left(\frac{n}{755}\right)} \right)^{(q-1)/13} \equiv 1 \mod q}$

For good measure, the same congruence holds for the next seven primes which split completely in $F(\zeta_{755})$. (One also has to check that the multiplicative order of $2$ for all these primes is co-prime to $41 \cdot 755$.) But, although this is compelling numerically, it doesn’t prove anything. If ${\epsilon}$ is a generator of $({\mathcal{O}}^{\times}_F/{\mathcal{O}}^{\times p}_F)^{\chi^{-1} \eta}$, it might be the case that ${\epsilon}^{(q-1)/p} \equiv 1 \mod {\mathfrak{q}}$ for ${\mathfrak{q}}$ above the first thousand primes of norm $q \equiv 1 \mod p$.  This would simply correspond to a certain ray class group being divisible by $p$. By Cebotarev, we know that we can find some prime $q$ for which this congruence does not hold, but explicit Cebotarev bounds tend to be rubbish in practice.

If we re-think our original computation, what we really want is a “generic” unit of $F^{+}$ in order to project. Since $F$ is abelian, we actually know how to compute a finite index subgroup of the unit group, namely, by projecting (via the norm map) the group of circular units from some cyclotomic overfield. Of course, this exactly won’t be good enough to find a candidate unit ${\epsilon}$. One approach is to take our lattice $V \subseteq U = {\mathcal{O}}^{\times}_{F^+}$ and saturate it. Now we only have to saturate it at $p = 41$. In fact, we only need to saturate the $\chi^{-1} \eta$ eigenspace, which is one dimensional. That is, it suffices to show that

$e_{\chi^{-1} \eta} N_{F(\zeta_{41})/F^{+}} (1 - \zeta)$

is a $p$th power in $F^{+}$. (Before taking the norm, the element is already in $F^{+}$ up to $p$th powers, and $[F(\zeta_{41}):F^{+}]$ has order prime to $41$.) But if I ask $\text{\texttt{pari}}$ to compute the following:

$\displaystyle{\left(\prod_{({\mathbf{Z}}/41 \cdot 755 {\mathbf{Z}})^{\times}} (1 -\zeta^n)^{n \left(\frac{n}{755}\right)} \right)^{(q-1)/13} \mod \Phi_{41 \cdot 755}(\zeta)}$

it complains and conks out. Well, probably René Schoof could do this computation, but let’s think about these things a little differently.

Higher Regulators: So far, we’ve been relying on the fact that the fields $E$ we are considering are abelian, in order to be able to explicitly write down some finite index subgroup of the full unit group using circular units. But what if we want to compute $K_2({\mathcal{O}}_E)$ for non-abelian fields $E$?  For this, I want to talk about an earlier paper of Gangl with Brownkin ( Tame and wild kernels of quadratic imaginary number fields…  oh bugger, this should also be cited as [BG].) Their approach is through the study of higher regulators. Borel constructs a higher regulator map for odd $K$-groups (the even ones are trivial after tensoring with ${\mathbf{Q}}$). For imaginary quadratic fields and $K_3$, this amounts to a map

$K_3(\mathcal{O}_E) \rightarrow (2 \pi i)^2 \cdot \mathbf{R},$

where the co-volume of the image is a rational multiple of $\zeta_E(2)$. The Quillen–Lichtenbaum conjecture predicts that the covolume differs exactly from $\zeta_E(2)$ by a factor coming from the torsion in $K_3({\mathcal{O}}_E)$, which has order dividing $24$, some slightly mysterious powers of $2$ which I will ignore, and — the most relevant term for us — the order of $K_2({\mathcal{O}}_E)$. Now the Quillen–Lichtenbaum conjecture is true. So how does this help to compute anything? Well, first one has to ask how to compute $K_3({\mathcal{O}}_E)$.  As an abelian group, it is easy to compute, but this is not enough to compute the regulator map. One could give explicit classes in $\pi_3(\mathrm{BGL}({\mathcal{O}}_E)^{+})$, of course, but that may not be the most practical approach. It turns out that the group $K_3$ is computable in a natural way because of its relation to the Bloch group $B(E)$, due to theorems of Bloch and Suslin. (That is, via the Hurewicz map we get classes in $H_3(\mathrm{GL}_N({\mathcal{O}}_E),\mathbf{Z})$ which turn out to be seen by $\mathrm{GL}_2.$) To recall, the Bloch group is defined as the quotient of the pre-Bloch group:

$\displaystyle{ \sum n_i [x_i], x_i \in E^{\times}, \ \text{such that} \ \sum n_i (x_i \wedge (1 - x_i)) = 0 \in \bigwedge^2 E^{\times}}$

by the $5$-term relation

$\displaystyle{[x] - [y] + \left[\frac{y}{x} \right] - \left[ \frac{1-y}{1-x} \right] + \left[ \frac{1 - y^{-1}}{1 - x^{-1}} \right] = 0, x,y \in E^{\times} \setminus 1}.$

Now the Bloch group admits a very natural regulator map

$B(E) \rightarrow {\mathbf{R}}^{r_2}$

(where $E$ has signature $(r_1,r_2)$) given by (under the various complex embeddings) the Bloch–Wigner dilogarithm

$D(z) = \mathrm{Im} (\mathrm{Li}_2(z)) + \arg(1-z) \log |z| \in \mathbf{R}.$

Now all of this is (almost) very computable. Namely, one can replace $E^{\times}$ by the $S$-units of ${\mathcal{O}}_E$ for some (as large as you can) set $S$, compute the pre-Bloch group, then do linear algebra to find the quotient. Since (roughly) $K_3({\mathcal{O}}_E) = {\mathbf{Z}}^{r_2} \oplus T$ for an easy to understand finite group $T$ which has order dividing $24$, as soon as one has a enough indepdenent elements in the Bloch group (which can be detected by computing $D(z)$) you can compute a group $B_S(E)$ which has finite index in $B(E)$. Moreover, the dilogarithm is also easy to compute numerically, and so one can compute a regulator $D_S(E)$ coming from the Bloch group. Now this regulator map is known to be rationally the same as Bloch’s regulator map (by Suslin and Bloch). Assuming this is also true integrally, we expect there to be a formula:

$\displaystyle{\frac{3 |d_E|^{3/2}}{\pi^2 D(E)} \cdot \zeta_E(2) =^{?} K_2({\mathcal{O}}_F)},$

at least for primes $p > 3$.  (The $3$ is coming from the torsion of $K_3$, and this formula is probably only true up to powers of $2$ — this formulation above comes from Brownkin–Gangl.) For $S$ big enough, $D_S(E)$ should stabilize to $D(E)$, which gives a method of computing the order of $K_2({\mathcal{O}}_E)$. This is what Brownkin and Gangl do. There are two issues which naturally one has to worry about. The first is that it’s not known that the regulator map coming from dilogarithms is the same on the nose as Bloch’s map. However, even granting this (and it should be true), this algorithm will not certifiably end, because one can never be sure that $D_S(E) = D(E)$. If you compare this to the computation that $\text{\texttt{pari}}$ is doing with the class group, the problem is that there is no a priori bounds on the size of the corresponding regulators. Well, I guess this algorithm can sometimes end, namely, when one can be sure if the indicated upper bound for $K_2({\mathcal{O}}_E)$ matches with a known lower bound. However, we are exactly in a situation in which we are trying to prove a lower bound. For example, when $E = {\mathbf{Q}}(\sqrt{-755})$, Brownkin and Gangl predict that $|K_2({\mathcal{O}}_E)| = 2 \cdot 41$ because, for a set of larger and larger primes $S$, the index formula above stabilizes. So, beyond the issue of relating two different higher regulator maps, we have the problem of determining whether a class in $K_3({\mathcal{O}}_E)$ is divisible by a prime $p$ or not. This seems harder than our previous problem of determining whether a unit was divisible or not! (To be fair, however, it seems impossible to find units in $E(\zeta_p)$ once $E$ is non-abelian and $p$ is in any sense large.)

Chern Class Maps: We want to understand whether a class in the Bloch group $B(E)$ or in $K_3({\mathcal{O}}_E)$ is divisible by $p$ or not. Instead of working over ${\mathbf{R}}$, another approach is to work modulo a prime $q$. (It may seem a little strange to work modulo $q$ to detect divisibility by $p$, but bear with me.) Soulé constructed certain Chern class maps, which include a map:

$c_2: K_3({\mathcal{O}}_E) = K_3(E) \rightarrow H^1(E,{\mathbf{Z}}_p(2)).$

These maps are the boundary map in the Atiyah–Hirzebruch spectral sequence for étale $K$-theory. Now compose this maps with the reduction modulo $p$ map. Then, after restricting to $F = E(\zeta_p)$, we may identify ${\mathbf{Z}}_p(2)/p$ with $\mu_p$, and so, by Kummer and Hilbert 90, we get a map:

$c_2: K_3(\mathcal{O}_E)/p \rightarrow H^1(F,\mathbf{Z}_p(2)/p) \simeq H^1(F,\mathbf{Z}_p(1)/p) = F^{\times}/F^{\times p}.$

Keeping track of the various identifications, the image lands in the $\chi^{-1}$ invariant subspace, where $\chi$ is the cyclotomic character of $G = {\mathrm{Gal}}(F/E)$.

Lemma Let $p > 3$ be a prime which is totally ramified in $E(\zeta_p)/E$ and suppose that $p$ does not divide the order of $K_2({\mathcal{O}}_E)$. Then the Chern class map induces an isomorphism

$({\mathbf{Z}}/p {\mathbf{Z}})^{r_2} = K_3({\mathcal{O}}_E)/p K_3({\mathcal{O}}_E) \rightarrow ({\mathcal{O}}^{\times}_F/{\mathcal{O}}^{\times p}_F)^{\chi^{-1}}.$

That is, the image of $c_2$ in $F^{\times}/F^{\times p}$ may be taken to land in the unit group, and the ranks of all the groups are the same and equal to $r_2$, the number of complex places of the field $E.$

This lemma follows from Quillen–Lichtenbaum, but it can also be proved directly from the surjectivity of the Chern class map as proved by Soulé, the known rank of $K_3 \otimes {\mathbf{Q}}$ by Borel, and some knowledge of the torsion of $K_3$ proved by Merkuriev and Suslin.  It turns out that the hypothesis on $K_2({\mathcal{O}}_E)$ is necessary not only for the proof but for the lemma to be true.

To detect whether a class in $K_3$ is divisible by $p$, it suffices to “compute” the Chern class map above and see whether it is zero. If one ever wants to compute anything, it makes sense to work with the Bloch group $B(E)$ instead. On the other hand, it seems hopeless to give a “concrete” map:

$B(E) \rightarrow F^{\times}/F^{\times p}.$

Even though one can write down elements in the first group somewhat explicitly, it’s hard to imagine a recipe that would produce explicit elements in $F^{\times}$ with the correct Galois action.

Instead, what we do is reduce modulo $q$ for some prime $q \equiv -1 \mod p$. That is, we pass from the Bloch group over $E$ (which will be generated by $S$ units for some $S$) to the Bloch group of the field ${\mathbf{F}}_q$. The construction over ${\mathbf{F}}_q$ is just the same. By a theorem Hutchinson, this group will have order $q+1$. The numerology here is intimately related to Quillen’s result that $K_3({\mathbf{F}}_q) = {\mathbf{Z}}/(q^2 - 1){\mathbf{Z}}$. Now there are some commutative diagrams one has to check commute here; I think the key point to keep in mind is that Quillen’s computation of $K_3({\mathbf{F}}_q)$ can already be realized in the cohomology group $H^3({\mathrm{SL}}_2({\mathbf{F}}_q),{\mathbf{Z}})$, and so the map of Bloch groups will be the same as the map on $K$-groups via comparison with the Hurewicz map.

Let’s choose a prime $q \equiv -1 \mod p$ which splits completely in $E(\zeta_p + \zeta^{-1}_p) \subset F = E(\zeta_p)$. So we have a map $B(E) \rightarrow B({\mathcal{O}}_E/{\mathfrak{q}}) \otimes {\mathbf{F}}_p = B({\mathbf{F}}_q) \otimes {\mathbf{F}}_p = {\mathbf{F}}_p.$ The Bloch group can be thought of in terms of (a quotient of a subgroup of) the free abelian group of elements of $\mathbf{P}^1(E)$, so there’s no issue about this reduction map. Moreover, given an element of the Bloch group, we can explicitly compute its image in the latter group. If this image is non-zero, that gives a certificate that the original element is not divisible by $p$. This will be enough to compute $K_2({\mathcal{O}}_E)$ as long as the Bloch regulator map agrees with the dilogarithm map.

This argument is still yoked to real regular maps. Let’s try to work entirely with $c_2$ and finite auxiliary primes $q \equiv - 1 \mod p$. Another manifestation of the map $B(E) \rightarrow B({\mathbf{F}}_q) \otimes {\mathbf{F}}_p$ is the map:

$c_2: K_3({\mathcal{O}}_E) \rightarrow {\mathcal{O}}^{\times}_F/{\mathcal{O}}^{\times p}_F \rightarrow ({\mathcal{O}}_F/{\mathfrak{Q}})^{\times} \otimes {\mathbf{F}}_p = {\mathbf{F}}_p,$

where ${\mathfrak{Q}}$ is a prime above ${\mathfrak{q}}$ in $F$.  Let’s go back to considering the case when $E$ is an imaginary quadratic field. The image of a generator of $K_3({\mathcal{O}}_E)$ will map exactly to a non-zero multiple of the non-trivial element unit ${\epsilon} \in F^{\times}/F^{\times p}$. If $K_2({\mathcal{O}}_E)$ is prime to $p$, it will even land in $({\mathcal{O}}^{\times}_F/{\mathcal{O}}^{\times p}_F)^{\chi^{-1}}$. The latter map is exactly computing (up to a non-zero scalar) ${\epsilon}^{(q-1)/p} \mod {\mathfrak{q}},$ and so, purely using the Bloch group, we can check whether this is trivial or not. In particular, given an element of the Bloch group $B(E)$ which (we think) is a generator, or at least not divisible by $p$, we can find a prime $q \equiv -1 \mod p$ such that the reduction to $B({\mathbf{F}}_q) \otimes {\mathbf{F}}_p$ is non-zero, which will imply that the image of $c_2$ is non-zero, which will imply that

${\epsilon}^{(q-1)/p} \not\equiv 1 \mod p.$

This gives an explicit value of $q$ for which this is true without ever having to compute ${\epsilon}$. For such a prime $q$, we can then check that the circular units project to the identity in this space, which will prove unconditionally that $K_2({\mathcal{O}}_E)$ is divisible by $p$. (Part of this computation assumed that $p$ did not divide $K_2({\mathcal{O}}_E)$, but that’s OK, because to prove that $p$ does divide this group we are allowed make that assumption anyway). Back to our example. We now want a prime $q \equiv -1 \mod 41$, which is also a square modulo $755$. We take $q = 163$. Now this is not the most attractive computation in the world, because the root of unity $\zeta$ of order $37 \cdot 755$ cuts out the extension ${\mathbf{F}}_{q^{300}}$, as we can see by computing the multiplicative order of $q = 163$ modulo $41 \cdot 5 \cdot 151$. Let’s do it in baby steps. By choosing a suitable prime ${\mathfrak{Q}}$ in $E(\zeta_{755})$, we can ensure that

$\zeta^{755} + \zeta^{-755} = \zeta_{41} + \zeta^{-1}_{41} \equiv 4 \mod {\mathfrak{Q}}.$

We write

$\zeta^{1510} - 4 \zeta^{755} + 1 = F(\zeta) G(\zeta) \mod 163,$

where $F(\zeta)$ is any of the four factors of degree $300$ (there are also two factors of degree $150$, and factors of degrees $2$, $4$, and $4$.) Now we want to compute, with $p = 41$, $q = 163$, and $r = 755$,

$\displaystyle{\eta:= \left(\prod_{({\mathbf{Z}}/p r {\mathbf{Z}})^{\times}} (1 -\zeta^n)^{n \left(\frac{n}{r}\right)} \right)^{(q^{300}-1)/p} \mod \mathfrak{Q} = (163,F(\zeta))}$

Of course, one should first reduce the exponents $\chi^{-1} \eta(n) = n (n/r)$ modulo $p = 41$ before taking the powers. (Actually, it’s probably kind of stupid to take a product over $\varphi(pr) = 24000$ different terms, and one can surely set this up much more effeciently, but whatever.) We find (drum roll) that:

$\eta \equiv 1 \mod 163.$

To finish, we have to take an element in the Bloch group $B({\mathcal{O}}_E)$ and show that it doesn’t vanish in $B({\mathcal{O}}_E/{\mathfrak{q}}) \otimes {\mathbf{F}}_p = B({\mathbf{F}}_{163}) \otimes {\mathbf{F}}_{41}$. At this point, I email Herbert (Gangl), and he sends me an email with the following beautiful element of $B(E)$, where $\alpha^2 = - 755$:

$\displaystyle{-8 \left[\frac{3 - \alpha}{10}\right] - 10 \left[\frac{7-\alpha}{10} \right] - 8 \left[\frac{3 - \alpha}{100} \right] + \ldots + 6 \left[\frac{7 \alpha + 221}{972} \right]}.$

(There are $114$ terms in all! This should be a generator of $B(E).$) Into my $\text{\texttt{magma}}$ programme it goes, which cheerily reports that the image of this element is non-zero in $B({\mathbf{F}}_{163}) \otimes \mathbf{F}_{41}$! So $K_2$ is really divisible by $41$. (You might question the veracity of my programme’s output, but more on that below.)

Stark and Beyond: Here are some more general remarks. Let’s still suppose that $E$ is imaginary quadratic. Take the image of a generator $[M]$ of $B(E)$, which is defined up to torsion and up to sign. The image of the Chern class map for some $p > 3$ and $p$ not dividing $K_2({\mathcal{O}}_E)$ gives a canonical unit in ${\mathcal{O}}^{\times}_F/{\mathcal{O}}^{\times p}_F$, where $F = E(\zeta_p)$. Let me a be a bit more careful here: by writing $F$ as $F = E(\zeta_p)$, we are choosing a root of unity (this unit depends on this choice). There’s also an automorphism of ${\mathrm{Gal}}(E/{\mathbf{Q}})$ which acts, but this changes the sign of $[M]$, so that is the same ambiguity we had before. What is this canonical unit? It is not just a circular unit, but a canonical one (modulo $p$th powers). What is it? More generally, when $r_2 = 1$, both $K_3$ and $(\mathcal{O}^{\times}_F/\mathcal{O}^{\times p}_F)^{\chi^{-1}}$ have rank $r_2 = 1$, so if $p$ is prime to $K_2(\mathcal{O}_E)$ we are generating canonical units. It’s tempting here to conjecture some relation to Stark units here, and in particular to the special value of $L(1,E,\chi^{-1})$, but let me say no more about this. When $r_2 > 1$, one is no longer in the Stark world, but there is still a canonical map from the Bloch group to the unit group (the group $\mathbf{Z}^{r_2}$ has no canonical generator when $r_2 > 1$ — but in the manifestation of this group as a Bloch group, one does have explicit elements.)

Actually, I haven’t even explained how to compute $c_2$. So far, I have only explained how to compute whether it is zero or not modulo $p$. To evaluate it exactly requires a further threading of the needle through the previous maps (on the Bloch group), and ultimately uses a test element coming from torsion in $B({\mathbf{Q}}(\zeta_p + \zeta^{-1}_p))$. Although this is somewhat delicate, and I have not yet proved all of the appropriate diagrams commute (blech), one can work with it in practice and it gives many consistency checks on all the computations. (So, for example, once one has the image of $c_2$, one can compute the reduction of the corresponding element in the Bloch group in $B({\mathbf{F}}_{q}) \otimes {\mathbf{F}}_p$ for one prime $q \equiv -1 \mod p$ knowing its image in the corresponding group for another such prime. Generating the same element of $\mathbf{F}_p$ for $p = 13$ and twenty different primes $q$ is pretty convincing.)

In fact, computing this map exactly is exactly the problem that I was thinking about in the first place. I did compute it explicitly for $K = {\mathbf{Q}}(\sqrt{-491})$ and $p = 13$ (and also $K = {\mathbf{Q}}(\sqrt{-571})$ for $p = 5$), and the image of a generator of the Bloch group is not a unit. Instead, it gives a generator of $\frak{a}^{13}$ for a non-trivial ideal in the class group ${\mathrm{Cl}}(F)$ of $F = E(\zeta_p)$, indeed, an element of ${\mathrm{Cl}}(F)[p]^{\chi^{-1} \eta}$. (In particular, it gives, having fixed a root of unity, a canonical element of this class group, which is also somewhat mysterious.) Let me also mention the Coates–Sinnott conjecture, higher Stickelberger elements, and work of Banaszak and Popescu which are closely related to the topics in this post (in particular, using Chern class maps to construct Euler systems generalizing the circular unit Euler system, although not so much question of identifying these elements in some explicit way — especially because much less is known about higher analogues of the Bloch group). But perhaps this is enough for now.

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