Ring-theoretic description of the norm-residue homomorphism

Recently, the norm-residue homomorphism has been the subject of intense discussions in the K-theoretic community following the proof of the Bloch-Kato conjecture by Voevodsky, Suslin and Rost (see Rost’s lecture at this year’s Arbeitstagung.) The goal of this post is to explain the norm-residue homomorphism in a down to earth ring-theoretic language.

Recall that the description of $K_2$ of fields is given by the following theorem.

Matsumoto’s Theorem.
For any field $F$

$K_{2}(F)= \frac{F^{*}\otimes_{\mathbb{Z}} F^{*}}{ <a \otimes (1-a)~|~ a\neq 1>}$

or equivalently in the context of presentations of groups, $K_{2}F$ is the Abelian group with
Generators: $\{x,y\}\quad x,y \in F^{*}$
Relations:

$\{x, 1-x\} =0\; \forall x \in F^{*} x \neq 0$ (The Steinberg relation)
$\{xy,z\} = \{x,z\} +\{y,z\}\; \forall x,y, z \in F^{*}$
$\{x, yz\} =\{x,y\} +\{x,z\}\; \forall x,y, z \in F^{*}.$

Definition. Let $F$ be a field and $A$ be an Abelian group. A Steinberg symbol on $F$ (with coefficients in $A$ ) is a $\mathbb{Z}$ -bilinear map $s:F^{*}\times F^{*}\longrightarrow A$ such that

$s\{x,1-x\}=0\quad \forall x\in F^{*} \quad x \neq 1.$

By Matsumoto’s theorem any Steinberg symbol $s:F^{*} \times F^{*} \longrightarrow A$ gives a unique group homomorphism $\tilde{s}:K_2(F) \longrightarrow A$ such that $s(x,y)=\tilde{s}\{x,y\}.$

Norm Residue Algebras.

Let $F$ be a field which contains a primitive $n$ -th root of unity $\omega$ and let $\alpha,\beta$ be two given elements in $F^{*}$ . The $n^{2}$ dimensional $F$ vector space

$A_{\omega}(\alpha,\beta):= \bigoplus\limits_{0\leq i,j <n} F x^{i}y^{j}$

with the following rules of multiplication:

$x^{n}=\alpha \quad y^{n}=\beta \quad yx=\omega yx$

is a central simple $F$ -algebra and it is called the norm residue algebra.

Theorem 1. Let $A$ be a central simple algebra of degree $n$ and let

$f(x)=x^{n}+a_{n-1}x^{n-1}+\dots +a_{0}$

be the minimal polynomial of $x\in A$ over $F$ . If $f$ splits into distinct linear factors over $F$ , then $A\simeq M_n(F).$

Corollary. Let $\alpha,\beta \in F^{*}$ . If either $\alpha$ or $\beta$ has an $n$ -th root in $F$ , then

$A_{\omega}(\alpha,\beta)\simeq M_n(F).$

As a special case of the above statement we have

$A_{\omega}(\alpha,1)\simeq M_n(F)$ .

Theorem 2. $A_{\omega}(\alpha,1-\alpha)\simeq M_n(F)$

Proof.
We define the non-commutative binomial coefficients

$b_{i}^{n}:=\frac{f_n(c)} {f_i(c)f_{n-i}(c)},$

where

$f_n(c)=\prod\limits_{j=1}^{n}(c^{j}-1) .$

It can be easily checked that $b_{i}^{n}(c)\in \mathbb{Z}[c]$ . Now suppose that $x,y$ are elements of an arbitrary ring $R$ such that $yx=cxy$ for some $c$ in the center of $R$ . Induction on $n$ shows that

$(x+y)^{n}= \sum_{i=0}^{n}b_{i}^{n}(c)x^{i}y^{n-i}$ .

In particular, for the generators $x$ and $y$ of $A_{\omega}(\alpha,1-\alpha)$ , since $b_{0}^{n}(\omega)=b_{n}^{n}(\omega)=1$ and $b_{i}^{n}(\omega)=0$ for all $0<i<n$ , we obtain that

$(x+y)^{n}=x^{n}+y^{n}=\alpha+(1-\alpha)=1.$

Now by the same reason as the previous corollary we have

$A_{\omega}(\alpha,1-\alpha)\simeq M_n(F)$ .

Theorem 3. Let $\alpha, \beta, \gamma$ be in $F^{*}$ . Then

$A_{\omega}(\alpha,\beta)\otimes_F A_{\omega}(\alpha,\gamma)\simeq A_{\omega}(\alpha,\beta \gamma)\otimes_F A_{\omega}(1,\gamma).$

Proof.
Let $x_1, y_1$ be the generators for $A_{\omega}(\alpha,\beta)$ and $x_2, y_2$ be the generators for $A_{\omega}(\alpha,\gamma)$ . Define

$x_3=x_1\otimes 1 \quad y_3=y_1\otimes y_2 \quad x_4=x_{1}^{-1}\otimes x_2 \quad y_4=1\otimes y_2.$

Let $A'$ be the algebra generated by $x_3, y_3$ and $A''$ be the algebra generated by $x_4, y_4$ . Now $x_{3}^{n}=\alpha\otimes 1$ , $y_{3}^{n}=\beta\gamma\otimes 1$ and

$y_3x_3=y_1x_1\otimes y_2=\omega (x_1y_1\otimes y_2)=\omega x_3y_3.$

So $x_3$ and $y_3$ satisfy the relations for $A_{\omega}(\alpha,\beta \gamma)$ , thus $A'\simeq A_{\omega}(\alpha,\beta \gamma)$ . Similarly $A''\simeq A_{\omega}(1,\gamma)$ . Notice that $x_3$ and $y_3$ commute with $x_4$ and $y_4$ , hence we have a natural $F$ -algebra homomorphism

$\varphi: A'\otimes_F A''\longrightarrow A_{\omega}(\alpha,\beta)\otimes_F A_{\omega}(\alpha,\gamma)$ .

Since $A'\otimes_F A''$ is simple, $\varphi$ is injective. Since the dimensions of two sides are equal $n^{4}$ it must be an isomorphism.

Remark. We have already seen that $A_{\omega}(\alpha,1)\simeq M_n(F)$ . So by the above theorem we have

$[A_{\omega}(\alpha,\beta \gamma)]=[ A_{\omega}(\alpha,\beta)] [A_{\omega}(\alpha,\gamma)],$ similarly

$[A_{\omega}(\alpha\beta, \gamma)]=[ A_{\omega}(\alpha,\gamma)][A_{\omega}(\beta,\gamma].$

Here $[A]$ denotes the equivalence class of $A$ in the Brauer Group.

Define

$s : F^{*}\times F^{*} \longrightarrow Br(F) \quad s(\alpha, \beta):= [A_{\omega}(\alpha,\beta)].$

The above remark says that $s$ is $\mathbb{Z}$ -bilinear. By Theorem 3 we observe that $s$ is a Steinberg symbol, hence we get a homomorphism

$\tilde{s}: K_2(F)\longrightarrow Br (F) \quad \tilde{s}\{\alpha,\beta\}= [A_{\omega}(\alpha,\beta)].$

From Corollary it follows that

$[A_{\omega}(\alpha,\beta)]^{n}=[A_{\omega}(\alpha^{n},\beta)]=1,$

which shows that the image of $\tilde{s}$ is contained in

${_n}Br(F):=\{[A]\in Br(F)~|~[A]^{n}=1\}.$

Since $n\{\alpha,\beta\}=\{\alpha^{n},\beta\}$ , the homomorphism $\tilde{s}$ annihilates $nK_2(F)$ , therefore it induces a homomorphism

$R_{n,F}: K_2(F)/n K_2(F) \longrightarrow_n Br(F)$

which is called the norm residue homomorphism.

The following surprising theorem was proved by A. Merkurjev and
A. Suslin in 1982.

The Merkurjev-Suslin Theorem. Let $F$ be a field which contains an $n$ -th primitive root of unity. Then

$R_{n,F}: K_2(F)/nK_2(F)\longrightarrow {_n}Br(F)$

is an isomorphism.

Norm Residue Homomorphism via Galois Cohomology.

The norm residue homomorphism can be described in terms of Galois cohomology. As a preliminary we need to recall the notion of the cup product in the cohomology of groups.

Let $F$ be a field and let $n$ be an integer coprime to char $(F).$ Set

$\mu_{n}=\{ x\in F_{sp} |\quad x^{n}=1 \} .$

The condition $(n,char F)=1$ implies that $\mu_{n}$ has exactly $n$ elements. Assume that $F$ has an $n$ -th primitive root of unity, i.e. $\mu_{n} \subset F$ . Set $G:= Gal(F_{sp}/F)$ and consider the following exact sequence of $G$ -modules:

$1 \stackrel{\mu_{n}}{\rightarrow} \rightarrow F_{sp}^{*} \stackrel{\text{n}}{\rightarrow} F_{sp}^{*} \rightarrow1$

The associated exact cohomology sequence is

$1 \rightarrow H^{0} (G ,\mu_{n}) \rightarrow H^{0} ( G ,F_{sp}^{*}) \stackrel{\text{n}}{\rightarrow} H^{0}(G, F_{sp}^{*}) \rightarrow$

$H^{1} (G , \mu_{n}) \rightarrow H^{1} (G ,F_{sp}^{*}) \stackrel{\text{ n}}{\rightarrow} H^{1}(G, F_{sp}^{*}) \rightarrow$

$H^{2} ( G ,\mu_{n}) \rightarrow H^{2} ( G ,F_{sp}^{*}) \stackrel{\text{n}}{\rightarrow} H^{2}(G , F_{sp}^{*}).$

As $\mu_{n} \subset F$ , the action of $G$ on $\mu_{n}$ is trivial, so $H^{0} (G ,\mu_{n})=\mu_{n}$ . By Hilbert’s Satz $90$ we have $H^{1} ( G ,F_{sp}^{*})=1$ , so the above sequence breaks up to the following exact sequences:

$1 \rightarrow \mu_{n} \rightarrow F^{*} \stackrel{\text{n}} {\rightarrow} F^{*} \stackrel{\delta}{\rightarrow} H^{1}(G,\mu_{n}) \rightarrow 1$

$1 \rightarrow H^{2}(G , \mu_{n}) \stackrel{\lambda} {\rightarrow} H^{2}(G , F_{sp}) \stackrel{\text{n}}{\rightarrow} H^{2}(G , F_{sp}).$

Hence the map $\delta$ induces an isomorphism $H^{1}(G , \mu_{n})\simeq F^{*}/F^{*^{n}})$ , and the map $\lambda$ induces an isomorphism between

$H^{2}(G , \mu_{n})$ and $ker( H^{2}(G , F_{sp}) \stackrel{\text{ n}}{\rightarrow} H^{2}(G ,F_{sp})).$

By using that $Br(F) \simeq H^{2}(G , F_{sp})$ we obtain that $H^{2}(G , \mu_{n})\simeq _nBr(F).$ Since $G$ acts trivially on $\mu_{n}$ it follows that $\mu_{n}^{\otimes^{2}}$ is isomorphic to $\mu_{n}$ as $G$ -module, hence

$H^{2}(G , \mu^{\otimes^{2}} ) \simeq H^{2}(G , \mu_{n}) \simeq{_n}Br(F).$

The composition of the following maps

$F^{*} \times F^{*} \rightarrow F^{*}/F^{*^{n}}\times F^{*}/F^{*^{n}}\simeq H^{1}(G,\mu_{n})\times H^{1}(G,\mu_{n})$ $\stackrel{\cup}{\rightarrow} H^{2}(G, \mu_{n})\simeq {_n}Br(F)$

gives a $\mathbb{Z}$ -bilinear map which can be proved to be a Steinberg symbol, and the induced homomorphism $K_2(F)\longrightarrow {_n}Br(F)$ is the norm residue homomorphism.

References:

1. Kersten, Ina Brauergruppen von Körpern. (German) [Brauer groups of fields] Aspects of Mathematics, D6.1.

2. Milnor, John Introduction to algebraic $K$-theory. Annals of Mathematics Studies, No. 72.

3. Rost, Markus Arbeitstagung 2007 – Norm residue homomorphism.

4. Tate, John Relations between $K\sb{2}$ and Galois cohomology. Invent. Math. 36 (1976), 257–274.

July 2007
M	T	W	T	F	S	S
		Aug »
	1
2	3	4	5	6	7	8
9	10	11	12	13	14	15
16	17	18	19	20	21	22
23	24	25	26	27	28	29
30	31

9 comments

Comments feed for this article

July 12, 2007 at 5:50 pm

davidspeyer

I think you have a typo. Three lines above the statement of Theorem 3, you want to say that (x+y)^n=x^n+y^n=alpha+(1-alpha)=1, not =0, right?

July 12, 2007 at 7:26 pm

Anton

Thank you, I’ve corrected that. And also $b_{n}^{n}(\omega)=1$ instead of $b_{1}^{n}(\omega)=1$ .

July 16, 2007 at 3:17 pm

Jason Starr

This reminds me of another theorem of Merkurjev-Suslin: if F is a perfect field of cohomological dimension 2, then for every central simple algebra A over F the reduced norm map Nm:A^* –> F^* is surjective. With no hypotheses on F, it is not hard to see the cokernel depends only on [A] in Br(F) and factors through F^*/(F^*)^n, where n = order([A]). Thus, given [A] in Br(F)[n]=H^2(F,\mu_n), the cokernel is a quotient of H^1(F,\mu_n). It is reasonable to guess the cokernel is the image of the homomorphism H^1(F,\mu_n) –> H^3(F,\mu_n^2) determined by cup product with [A]. If so, that would explain this MS theorem, since H^3(F,\mu_n^2) vanishes if cd(F)=2. Do you know if this is correct? If so, is the proof related to the MS theorem in your post?

July 17, 2007 at 1:55 pm

Hi,
probably you are right may be you can use the MS theorem above to prove that the image of the norm map in $H^1(F, \mu_n)$ is exactly the part killed by the cup product with $[A]$ . For example, if $A=A_\omega(\alpha, \beta)$ then $1-\alpha$ is the norm of $1-x$ and indeed, $\alpha \cup \beta \cup (1-\alpha)=0$ . But I don’t know much about the norm map, so we should look in the Merkur’jev-Suslin papers.

July 18, 2007 at 11:13 pm

Hi,

I will take a look. I ran across a comment in Serre’s “Galois cohomology” that MS construct a map from the cokernel of the reduced norm map into H^3(F,\mu_n^2) <>. That hypothesis makes me believe there is more to this than meets the eye.

Best,
Jason

July 19, 2007 at 12:20 pm

The hypothesis that was supposed to appear in “” somehow didn’t compile: MS construct the map if n is square-free.

May 27, 2008 at 6:07 am

Ben

PARI is already integrated with Python:

http://www.sagemath.org

May 27, 2008 at 6:10 am

Sorry, my comment above was not meant to be on this post.

August 30, 2009 at 6:43 am

PEN Chentra

Dear Madams/Miss/Mrs

Can you help me to proof (Z/abZ) and (Z/aZ)(Z/bZ) is homomorphisme (gcd(a,b)=1).
Sincerely

	John on Tricks using PARI
	John on Tricks using PARI
	PEN Chentra on Ring-theoretic description of …
	Successful Researche… on Something truly outrageou…
	Jaime Montuerto on Tricks using PARI

Ring-theoretic description of the norm-residue homomorphism

Pages

Links

Recent Comments

Recent Posts

a

Archives

9 comments