Milnor K-theory

• Milnor K-theory
• The tame residue map
• Computation of the motivic cohomology

The goal for today is to prove that $$ H^n(\Spec F, \mathbb{Z}(n)) \cong K_n^\mathrm{M}(F). $$ What is the left hand side? Recall that we had defined $$ \mathbb{Z}(q)^i = C_{q-i} \mathbb{Z}_ \mathrm{tr}(\mathbb{G}_ m^{\wedge q}), $$ and then $$ H^{p,q}(X, \mathbb{Z}) = \mathbb{H}_ \mathrm{Zar}^p(X, \mathbb{Z}(q)) = H^\bullet(\Gamma(\Spec F, \mathbb{Z}(q))). $$ So if you unravel the definition, we are just looking at the cokernel of $$ \mathbb{A}^1 \text{-homotopy} \to \frac{Z_0((\mathbb{A}^1 - \lbrace 0 \rbrace)^n)}{\mathrm{faces}}. $$

Milnor K-theory

On the other hand, what is Milnor K-theory? We consider the tensor ring $$ T^\bullet(F^\times) = \mathbb{Z} \oplus F^\times \oplus (F^\times \otimes_\mathbb{Z} F^\times) \oplus \dotsb. $$

Definition 1. We define the Milnor K-theory ring as the quotient $$ K_\bullet^\mathrm{M}(F) = T^\bullet(F^\times) / \langle a \otimes (1-a) : a \in F^\times - \lbrace 1 \rbrace \rangle. $$ Instead of writing $a_1 \otimes \dotsb \otimes a_n$ we will write $\lbrace a_1, \dotsc, a_n \rbrace$.

It turns out that $\lbrace a, -a \rbrace = 0$ and then $\lbrace a, a \rbrace$ is 2-torsion, so that $\lbrace a, b \rbrace = -\lbrace b, a \rbrace$.

If $E/F$ is finite, it induces a map of schemes $$ (\mathbb{A}_ E^1 - \lbrace 0 \rbrace)^n \to (\mathbb{A}_ F^1 - \lbrace 0 \rbrace)^n $$ and hence an induced norm map $$ N_{E/F} \colon H^{n,n}(\Spec E, \mathbb{Z}) \to H^{n,n}(\Spec F, \mathbb{Z}). $$

The tame residue map

On Milnor K-theory, this map is a bit more complicated. For $F$ a field and $v \colon F^\times \to \mathbb{Z}$ a discrete valuation with residue field $f_v$, we have an induced map $$ K_1^\mathrm{M}(F) \to K_0^\mathrm{M}(f_v); \quad \lbrace a, b \rbrace \mapsto (-1)^{v(a) v(b)} (a^{v(b)} / b^{v(a)} \bmod{\mathfrak{m}}). $$ There are similar maps $K_n^\mathrm{M}(F) \to K_{n-1}^\mathrm{M}(f_v)$.

We let $$ L = K_\bullet^\mathrm{M}(f_v)[\Pi] / \langle \lbrace \Pi, \Pi \rangle - \lbrace -1, \Pi \rbrace \rangle, $$ where $\Pi$ is a free variable in degree 1. It turns out that after choosing a uniformizer $\pi \in F^\times$, we get an induced map $$ D \colon K_\bullet^\mathrm{M}(F) \to L. $$ Looking at the degree $n$ piece, we have $$ D_n \colon K_n^\mathrm{M}(F) \to L_n = K_n^\mathrm{M}(f_v) \oplus K_{n-1}^\mathrm{M}(f_v). $$ This defines two maps $$ \lambda \colon K_n^\mathrm{M}(F) \to K_{n-1}^\mathrm{M}(f_v), \quad \partial_v \colon K_n^\mathrm{M}(F) \to K_{n-1}^\mathrm{M}(f_v). $$ The second map $\partial_v$ is the tame residue map an has the formula $$ \partial_v(\lbrace u \pi^i, u_2, \dotsc, u_n \rbrace) = i \lbrace \bar{u}_ 2, \dotsc, \bar{u}_ n \rbrace. $$

Theorem 2. There exists a split exact sequence $$ 0 \to K_n^\mathrm{M}(F) \to K_n^\mathrm{M}(F(t)) \xrightarrow{\bigoplus \partial_v} \bigoplus_{\mathfrak{p} \subseteq F[t]} K_{n-1}^\mathrm{M}(F[t] / \mathfrak{p}) \to 0. $$

We want a Weil reciprocity law, which states that for all $x, y \in F(t)^\times$ we have $$ \prod_{v \in \lvert \mathbb{P}_ F^1 \rvert} N_{K(v)/F} \partial_v(\lbrace x, y \rbrace) = 1. $$ What we can now do is to look at the valuation at infinity $v_\infty \colon F(t)^\times \to \mathbb{Z}$ and the composition $$ K_n^\mathrm{M}(F) \to K_n^\mathrm{M}(F(t)) \to K_{n-1}^\mathrm{M}(F) $$ is zero. So it uniquely factors through $$ \partial_{v_\infty} \colon \bigoplus_{\mathfrak{p} \subseteq F[t]} K_{n-1}^\mathrm{M}(F[t]/\mathfrak{p}) \to K_{n-1}^\mathrm{M}(F). $$

Definition 3. We define the norm $N_{(F[t]/\mathfrak{p}) / F}$ so that $$ \partial_{v_\infty}(\alpha) = \sum_{\mathfrak{p}} N_{(F[t]/\mathfrak{p})/F}(\partial_v \alpha). $$

Proposition 4. For $F/E$ finite, $N_{E/F}$ is well-defined (i.e., independent of the choice of generators), and it also satisfies the expected relation $N_{E/F} = N_{K/F} \circ N_{E/K}$.

Computation of the motivic cohomology

Recall we had this map $$ F^\times \to H^1(\Spec F, \mathbb{Z}(1)); \quad a \mapsto [a] \in Z_0(\mathbb{A}_ F^1 - \lbrace 0 \rbrace) / \mathbb{Z}[1]. $$ But why is this a homomorphism? Remember we had to also quotient by $\mathbb{A}^1$-homotopy. If we look at the variety $$ Z = V(x^2 - t(a+b) x - (1-t) 1 + ab) x + ab \subseteq \mathbb{A}_ t^1 \times (\mathbb{A}_ x^1 - \lbrace 0 \rbrace), $$ then this is finite flat over $\mathbb{A}_ t^1$, and its fiber over $t = 0$ is $\lbrace 1, ab \rbrace$ and its fiber over $t = 1$ is $\lbrace a, b \rbrace$. This shows the relation $$ [1] + [ab] \sim [a] + [b]. $$ By a similar argument, if we have any closed point $p \in \mathbb{A}_ F^1 - \lbrace 0 \rbrace$, what we can do is to multiply all its Galois conjugates and look at the norm $N_{F(p)/F}(x_p)$ and then get an element of $F^\times$. Now the point is that this quantity does not depend on $\mathbb{A}^1$-homotopy, and therefore this is an isomorphism.

In general, we will first construct a map $$ \lambda_F \colon K_n^\mathrm{M} \to H^{n,n}(\Spec F, \mathbb{Z}); \quad \lbrace x_1, \dotsc, x_n \rbrace \mapsto [(x_1, \dotsc, x_n)]. $$ Why is this well-defined? First we have already checked that multiplication correspond to addition.

Claim 5. The point $[(a, 1-a)]$ is $\mathbb{A}^1$-homotopic to something inside $\lbrace 1 \rbrace \times (\mathbb{A}^1 - \lbrace 0 \rbrace) \cup (\mathbb{A}^1 - \lbrace 0 \rbrace) \times \lbrace 1 \rbrace$.

What we do is to first look at the subscheme $$ Z = V(x^3 - t(a^3 + 1) x^2 + t (a^3 + 1) x - a^3) \subseteq \mathbb{A}_ t^1 \times (\mathbb{A}_ x^1 - \lbrace 0, 1 \rbrace). $$ This has the property that its is finite flat over $\mathbb{A}_ t^1$ as long as $a \neq 0$ and $a^3 \neq 1$. The fiber over $t = 0$ is $[a] + [\omega a] + [\omega^2 a]$ (where we interpret $\omega = 1$ when $\operatorname{char} F = 3$). The fiber over $t = 1$ is $[a^3] + [-\omega] + [-\omega^2]$. But the point is that this is a homotopy on $\mathbb{A}^1 - \lbrace 0, 1 \rbrace$.

Now we embed this as $$ \mathbb{A}^1 - \lbrace 0, 1 \rbrace \to (\mathbb{A}^1 - \lbrace 0 \rbrace)^2; \quad x \mapsto (x, 1-x). $$ Then we have $$ \begin{align} [a] \cup [1-a] &+ [\omega a] \cup [1-\omega a] + [\omega^2 a] \cup [1-\omega^2 a] \br &= [a^3] \cup [1-a^3] + [-\omega] \cup [1+\omega] + [-\omega^2] \cup [1+\omega^2]. \end{align} $$ Then we have $$ [a] \cup [1-a^3] = 3 [a] \cup [1-a^3] $$ up to 6-torsion, and so $4 [a^3] \cup [1-a^3] = 0$. That is, if $b \neq 0, 1$ is a cube then $4 [b] \cup [1-b] = 0$. If $b$ is not a cube, then we work in $F(\sqrt[3]{b})$ and we get $12 [b] \cup [1-b] = 0$.

Lemma 6. If there exists an $n$ such that for all $F$ and $x \in F^\times - \lbrace 1 \rbrace$ we have $n [x] \cup [1-x] = 0$, then the same holds for $n = 1$.

Claim 7. The map $\lambda_F$ is injective.

Suppose that we have $$ \lambda_f \colon \sum_i \lbrace x_{i,1}, \dotsc, x_{i,n} \rbrace \mapsto 0. $$ Then there is an $\mathbb{A}^1$-homotopy that pushes these things into the union of the faces. Let this be $\sum n_k Z_k \subseteq \mathbb{A}_ t^1 \times (\mathbb{A}^1 - \lbrace 0 \rbrace)^n$, finite flat over $\mathbb{A}_ t^1$. We consider the fiber over the generic point $t \in \mathbb{A}_ t^1$ and write it as $$ \sum_j n_j [f_{j,1}, \dotsc, f_{j,n}] $$ Now we consider the Milnor K-theory class $$ \varphi_j = \biggl\lbrace \frac{t}{t-1}, f_{j,1}, \dotsc, f_{j,n} \biggr\rbrace \in K_{n+1}^\mathrm{M}(\operatorname{Frac}(Z_j)) $$ and apply Weil reciprocity to these curves $Z_j$. Then we get that $$ 0 = \sum_j n_j \lbrace f_{j,1}(0), \dotsc, f_{j,n}(0) \rbrace = \sum_j n_j \lbrace f_{j,1}(1), \dotsc, f_{j,n}(1) \rbrace = \sum_i \lbrace x_{i,1}, \dotsc, x_{i,n} \rbrace $$ inside $K_n^\mathrm{M}$.

For surjectivity, we just need compatibility of norms.

Proposition 8. For $E/F$ finite, the diagram $$ \begin{CD} K_n^\mathrm{M}(E) @>{\lambda_E}>> H^{n,n}(E) \br @V{N_{E/F}}VV @V{N_{E/F}}VV \br K_n^\mathrm{M}(F) @>{\lambda_F}>> H^{n,n}(F). \end{CD} $$ commutes.

Then any element in $H^{n,n}(F)$ can be written as a linear combination of a norms of a rational points in $H^{n,n}(E)$ for some $E$ (where $E$ might change) and so we will be done.

Here is the idea of proof. Suppose that $[E:F] = \ell$ is prime. We reduce to the case when $F$ has no extensions of degree prime to $\ell$.

Lemma 9 (Bass–Tate). Let $Y/X$ be a field extension where $Y = X(\alpha)$ and $[Y:X] = d$. Then $K_\bullet^\mathrm{M}(Y)$ is generated as a left $K_\bullet^\mathrm{M}(X)$-module by symbols $$ \lbrace p_1(\alpha), p_2(\alpha), \dotsc, p_m(\alpha) \rbrace $$ where $p_i \in X[t]$ are monic irreducible polynomials satisfying $\deg p_1 \lt \dotsb \lt \deg p_m \le d-1$.

In our case, this implies that $K_n^\mathrm{M}(E)$ is generated by $K_1^\mathrm{M}(E)$ as a left $K_n^\mathrm{M}(F)$-module. Then $K_n^\mathrm{M}(E)$ is generated as an abelian group by $\lbrace a_1, \dotsc, a_{n-1}, b \rbrace$ where $a_1, \dotsc, a_{n-1} \in F$. We can now directly check that $\lambda_F \circ N$ and $N \circ \lambda_E$ both sends this to $$ [a_1] \cup \dotsb [a_{n-1}] \cup [N_{E/F}(b)]. $$