### Algebraic K-groups as Galois modules

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Thus we can reduce to the case in which G is a finite l-group. However, we reduced to the case in which G is an l-group. This completes the proof of Proposition 2. In the above proof, the assumption that A is a D. Suppose we assumed only that A is of finite index in a D. Then flat finitely generated 3m -modules are free. The argument used to prove 2. We say F is construcible resp. Suppose A and S satisfy Hypothesis 2. Let D be the derived category of the homotopy category of complexes of A[G]-modules which are bounded below. With the notations of Proposition 2. Proof of Proposition 2.

Then as in [M1, p. In the statement of [M1, Lemma 8. The argument of [M1, Lemma VI. We recall some notations from Section 2. LEMMA 3. Let Gv be the decomposition group of a place v of N, and let SR be i Z r is naturally isomorphic to the set of real places of N. Repeating the above argument completes the proof of Lemma 3. Remark 3. Cases b and c were incorrectly interchanged in [CKPS1, p. Let l be a prime. Hli Zl r is finitely generated over Z resp. This completes the proof. Thus ker d2 is finite, and coker d1 is finite by Lemma 3. Hence sequence 3. Suppose M and N are finitely generated G-modules.

By Theorem 3. However, Lemma 3. Lemma 3. Since the right and left vertical homomorphisms in 3. Statement c is implied by the fact that the extension classes of the top and bottom rows of 3. Finally the assertions at the end of the statement of Proposition 3.

## (PDF) Galois Structure of K-Groups of Rings of Integers | Victor Snaith - syvufile.tk

Via Corollary 3. Recall from Proposition 3. Since the isomorphism 3. One can now construct 3. We will prove the following result in Section 4.

## Algebraic K-Groups as Galois Modules

From 3. LEMMA 4. By Proposition 3. Let S be fixed. As in Corollary 3. Suppose t is a non-zero integer. The right vertical homomorphism in 4. This m must be prime to G, and 4. As shown in the proof of Proposition 3. The first isomorphism in 4. Suppose t in 4. So we have 4. This in turn shows that there is a diagram of the form 4.

Diagrams 4. Hence from 4. Thus if t is divisible by G times this index, we can conclude from 4. To prove Proposition 4. Because of condition a of Lemma 4. Therefore Lemma 4. Hence to prove Proposition 4. Lemma 4. It follows that the groups in 4. With the notation of Lemma 4. Proof of Proposition 4. In terms of 4. Thus 4. Pushing out 4. By Lemma 4. Hence, [C1, Prop. Thus, 4. If Gv is such an inertia group, then Gv has order 1 or two, and all finite Gv modules of order prime to Gv have trivial class in K0 Z[Gv ]. The equality 4. The following Proposition can be proved in the same way as Proposition 1 and Lemme 5 in Section 3.

For the convenience of the reader, we will include a proof. Suppose S is the union of S with a finite G-stable set of places of N. Let Sl be the union of S with the set of places of N over l. By [M1, Prop. By [M1, Cor.

- Algebraic K-Groups as Galois Modules | Victor P. Snaith | Springer.
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Let Kwh be the fraction field of Ow h. It follows from [M2, p.

Now arguing as on p. Then 4. A superscript indicates an object constructed using S in place of S. The top and middle rows are exact sequences of the kind used in Proposition 3. The proof of this Proposition will be given after Corollary 4. With the notation of Proposition 4. The isomorphism 4. To show 4. In view of Lemma 3. Lemma 2. Because of 4. The exact sequence 4. Proposition 6. The long exact cohomology sequence of 4.

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It was shown in the course of the proof of Lemma 4. Thus ker d1 maps isomorphically to ker d1 under restriction. Proposition 3. In view of Corollary 4.

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We see now from 4. This makes 4. Hence M must be cohomologically trivial, and Proposition 4. Suppose w is a finite place of N. Let Iw resp. Gw be the inertia group of w in G. We will also regard Mw r as a Gw -module by inflation. The following Lemma is shown in exactly the same as [C3, Lemma 4. Let w be a finite place of N. This follows on inflating the sequence 4. Completion of the proof of Proposition 3. In view of Lemmas 4. From diagram 4. Diagram 4. The formula 4. From 4. LEMMA 5. The first row of diagram 3. Combining 5. Substituting 5. Statement a is shown in Appendix 2 of [K1].

One can now deduce d from c by the argument used to prove 5. From 5. This finishes the proof. Proof of Proposition 5. From Proposition 5. Hence Corollary 4. This will lead to a proof of Theorem 1. From the top row of diagram 3. The statements concerning 6. The Borel regulator in 6. This follows from the isomorphism in 6.

This follows from the fact that the terms of the top row of 6. Let V be a complex representation of G, with dual V. The formula 6. Since the right hand side of 6. The final assertion in 6. Remark 6. Dividing the left-most two terms of 6. This is clear from Corollary 6. Let E be a number field with the properties stated in Definition 6. Equation 6. The rest of the Proposition is proved in the same way as Proposition 3.

In view of Proposition 6. Suppose V is a symplectic representation of G. If Conjecture 6. We now establish the statements concerning the case of totally real fields which were made in the Remark following the statement of Theorem 1 of Section 1. Then conjecture 6.

## Mixed Tate motives, algebraic K-theory and multiple zeta values

In the notation of Conjecture 6. Under the hypotheses of Proposition 6. In the rest of this section, we will prove Proposition 6. The argument given in [C1, Remark 2. LEMMA 6. The truth of Conjecture 6. David White David White There's not so much specifically on K Z though That's the perfect introduction to the subject for me.

It's by one of my favorite authors and contains all of my favorite things, with just enough K-theory mixed in for me to actually learn something. Thanks for the link! By the way, now that things have settled down a bit I guess I owe you an email about those semi-model categories. It will be coming soon. See folk. I'm somewhat amazed that the Langlands program is involved with algebraic K-theory, but you explain this point very well. And that connection to Galois cohomology groups is great.

Also, welcome to MathOverflow! Could you expand on your four analogies? The unstable cohomology comes from Jason Polak Jason Polak 1, 1 1 gold badge 12 12 silver badges 25 25 bronze badges. Still, it sounds like Kurihara's paper might contain some further applications, so I'll look into that paper soon.

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