Norm residue isomorphism theorem

• Construction of the norm residue map
• Some reductions
• Hilbert 90

Theorem 1 (Voevodsky, Rost). Let $\ell$ be prime and $k$ be a field in which $\ell \neq 0$. Then there exists a natural isomorphism $$ \mathrm{NR}_ {k,n} \colon K_n^\mathrm{M}(k) \otimes_\mathbb{Z} \mathbb{Z}/\ell\mathbb{Z} \xrightarrow{\cong} H_\mathrm{et}^n(\Spec k, \mu_\ell^{\otimes n}). $$

Before constructing this map, let us look at some example. For $n = 0$, we have $$ K_0^\mathrm{M}(k) = K_0(k) = \mathbb{Z}, \quad H_\mathrm{et}^0(\Spec k, \mu_\ell^{\otimes 0}) = \mathbb{Z}/\ell\mathbb{Z}. $$ For $n = 1$, we have $$ K_1^\mathrm{M}(k) = K_1(k) = k^\times, \quad H_\mathrm{et}^1(k, \mu_\ell) \cong k^\times / (k^\times)^\ell $$ by Kummer theory. So the norm residue map should be the boundary map associated to $0 \to \mu_n \to \mathbb{G}_ m \to \mathbb{G}_ m \to 0$.

In the previous talks we constructed a motivic cohomology group for smooth varieties $X/k$, namely $$ H^{p,q}(X, \mathbb{Z}/\ell\mathbb{Z}) = H^p(X, \mathbb{Z}/\ell\mathbb{Z}(q)). $$ Some time ago, we proved the isomorphism $$ H^{p,p}(K, \mathbb{Z}/\ell\mathbb{Z}) \cong K_p^\mathrm{M}(k) / \ell. $$ We also have a natural graded ring map $$ H^{p,q}(X, \mathbb{Z}/\ell\mathbb{Z}) \to H_\mathrm{et}^\ast(X, \mu_\ell^{\otimes q}). $$

Theorem 2. For $X/k$ smooth, the maps $$ H^{p,q}(X, \mathbb{Z}/\ell\mathbb{Z}) \to H_\mathrm{et}^p(X, \mu_\ell^{\otimes q}) $$ are isomorphisms for $p \le q$.

Remark 3. What was proved last time was that $H_L^{p,q}(X, \mathbb{Z}/\ell\mathbb{Z}) \cong H_\mathrm{et}^p(X, \mu_\ell^{\otimes q})$. This is different from $H^{p,q}$ because there we were looking at the étale hypercohomology.

It is clear that this implies that norm residue isomorphism theorem. It turns out that they are equivalent.

Construction of the norm residue map

Let us now define $K_n^\mathrm{M}(k)/\ell \to H_\mathrm{et}^n(\Spec k, \mu_\ell^{\otimes n})$. Recall that $$ K_n^\mathrm{M}(K) = \wedge^n(k^\times) / \langle \lbrace a, 1-a \rbrace \rangle. $$ Now there is an obvious map given by $$ \lbrace a_1, \dotsc, a_n \rbrace \mapsto [a_1] \cup \dotsb \cup [a_n]. $$ So to define this norm residue map, it remains to verify the Steinberg relation.

Lemma 4 (Tate). We have $[a] \cup [1-a] = 0$ in $H^2(k, \mu_\ell^{\otimes 2})$ for $a \neq 0, 1$.

Some reductions

The rest of this talk will be on how the proof generally works. We will reduce to the case when

• $k$ is perfect,
• $k$ contains $\mu_\ell$,
• $k$ is $\ell$-special, i.e., it contains no finite extensions of degree prime to $\ell$.

Lemma 5. Let $k^\prime / k$ be a finite extension of order prime to $\ell$. Then $\mathrm{NR}_ {k^\prime,n}$ being an isomorphism implies that $\mathrm{NR}_ {k,n}$ is an isomorphism.

This is because there is a diagram $$ \begin{CD} K_n^\mathrm{M}(k)/\ell @>>> K_n^\mathrm{M}(k^\prime)/\ell @>{\mathrm{tr}}>> K_n^\mathrm{M}(k)/\ell \br @VVV @VVV @VVV \br H^n(k, \mu_\ell^{\otimes n}) @>>> H^n(k^\prime, \mu_\ell^{\otimes n}) @>{\mathrm{cores}}>> H^n(k, \mu_\ell^{\otimes n}) \end{CD} $$ where both horizontal compose to multiplication by $[k^\prime : k]$.

Lemma 6. If $\mathrm{NR}_ {k,n}$ is an isomorphism for all $\operatorname{char}(k) = 0$, then $\mathrm{NR}_ {k,n}$ is an isomorphism for all $\ell \in k^\times$.

When $\operatorname{char}(k) = p$, we can use $W(k)$ the ring of Witt vectors. Then there are specialization maps $$ \begin{CD} K_n^\mathrm{M}(W(k)[p^{-1}])/\ell @>>> H_\mathrm{et}^n(W(k)[p^{-1}], \mu_\ell^{\otimes n}) \br @VV{\mathrm{sp}}V @VV{\mathrm{sp}}V \br K_n^M(k)/\ell @>>> H_\mathrm{et}^n(k, \mu_\ell^{\otimes n}) \end{CD} $$ that are both slit surjections. (We really need compatible splittings, but this can be done.) Now we are reduced to the case when $k$ has characteristic zero.

Hilbert 90

Recall that we had $\mathbb{Z}(i)$ a complex of sheaves, and we had $\mathbb{Z}(0) = \mathbb{Z}$ and $\mathbb{Z}(1) = \mathscr{O}^\times[-1]$. We have $H^{n+1}(\Spec k, \mathbb{Z}(n)) = 0$ and therefore $$ K_n^\mathrm{M}(k)/\ell = H^n(\Spec k, \mathbb{Z}(n)) / \ell = H^n(\Spec k, \mathbb{Z}/\ell\mathbb{Z}(n)). $$ On the other hand, we have seen last week that $$ H_\mathrm{et}^n(X, \mathbb{Z}/\ell\mathbb{Z}(i)) = H_\mathrm{et}^n(X, \mu_\ell^{\otimes i}) $$ for all $n$ and $i$.

Definition 7. We say that $H90(n)$ holds if for all $k \ni \ell^{-1}$ we have $H_\mathrm{et}^{n+1}(k, \mathbb{Z}_{(\ell)}(n)) = 0$.

Note that when $n = 1$, this is just saying that $$ H_\mathrm{et}^2(k, \mathbb{Z}(1)) = H_\mathrm{et}^2(k, \mathbb{G}_ m[-1]) = H_\mathrm{et}^1(k, \mathbb{G}_ m) = 0, $$ which is true by usual Hilbert 90.

Proposition 8. For fixed $n$ and $\ell$, we have $H90(n)$ if and only if $\mathrm{NR}_ {k,n}$ is an isomorphism for all $k \ni \ell^{-1}$.

So how do we prove this $H90(n)$? The proof uses induction on $n$ and has two big inputs.

Theorem 9. If $H90(n-1)$ holds, and $k$ is an $\ell$-special field, then $K_n^\mathrm{M}(k)/\ell = 0$ implies $H_\mathrm{et}^n(k, \mu_\ell^{\otimes n}) = 0$. In this case, we moreover have $H_\mathrm{et}^{n+1}(k, \mathbb{Z}_ {(\ell)}(n)) = 0$.

Theorem 10. If $H90(n-1)$ holds and $k$ is in characteristic zero, then for all $0 \neq a \in K_n^\mathrm{M}(k)/\ell$ there exists a smooth projective variety $X_a / k$ with function field $K_a = K(X_a)$ such that

• the map $K_n^\mathrm{M}(k)/\ell \to K_n^\mathrm{M}(K_a)/\ell$ sends $a$ to zero,
• the map $H_\mathrm{et}^{n+1}(k, \mathbb{Z}_ {(\ell)}(n)) \hookrightarrow H_\mathrm{et}^{n+1}(K_a, \mathbb{Z}_ {(\ell)}(n))$ is injective.

Given these facts, let’s prove that $H90(n-1)$ implies $H90(n)$. We first well-order the elements of $K_n^\mathrm{M}(k)/\ell$ and choose the corresponding varieties so that we can find a field $k^{(1)} \supseteq k$ such that $K_n^\mathrm{M}(k)/\ell \to K_n^\mathrm{M}(k^{(1)})/\ell$ is zero and $H_\mathrm{et}^{n+1}(k, \mathbb{Z}_ {(\ell)}(n)) \hookrightarrow H_\mathrm{et}^{n+1}(k^{(1)}, \mathbb{Z}_ {(\ell)}(n))$ is injective. We iterate this procedure and write $$ L = \bigcup_{m \ge 1} k^{(m)}. $$ Then we see that $K_n^\mathrm{M}(L)/\ell = 0$, and moreover $H_\mathrm{et}^{n+1}(k, \mathbb{Z}_ {(\ell)}(n)) \hookrightarrow H_\mathrm{et}^{n+1}(L, \mathbb{Z}_ {(\ell)}(n))$. Then the first theorem implies that $H_\mathrm{et}^{n+1}(k, \mathbb{Z}_ {(\ell)}(n)) = 0$.