Mathematics Lecture Notes

Theorem6.5Suslin's Rigidity

Let $k \hookrightarrow K$ be an extension of algebraically closed fields of the same characteristic. Then for any $n \geq 0$ and any $m \geq 1$ :

$K_n(k; \mathbb{Z}/m) \xrightarrow{\;\sim\;} K_n(K; \mathbb{Z}/m)$

is an isomorphism. In particular, $K_n(\overline{\mathbb{F}}_p; \mathbb{Z}/m) \cong K_n(\overline{k}; \mathbb{Z}/m)$ for any algebraically closed field $k$ of characteristic $p$ .

ExampleAlgebraic vs topological K-theory

Combining rigidity with the comparison to topological K-theory:

For $k = \mathbb{C}$ : The map $BGL(\mathbb{C})^{\delta} \to BGL(\mathbb{C})$ from the discrete to the continuous classifying space induces (by Suslin) an isomorphism with finite coefficients:

$K_n(\mathbb{C}; \mathbb{Z}/m) \cong K_n^{\text{top}}(\mathbb{C}; \mathbb{Z}/m) = \pi_n(BU; \mathbb{Z}/m)$

Since $\pi_n(BU) = \begin{cases} \mathbb{Z} & n \text{ even} \\ 0 & n \text{ odd} \end{cases}$ , we get:

$K_n(\mathbb{C}; \mathbb{Z}/m) \cong \begin{cases} \mathbb{Z}/m & n \text{ even} \\ 0 & n \text{ odd} \end{cases}$

By rigidity, this holds for any algebraically closed field of characteristic 0:

$K_n(\overline{\mathbb{Q}}; \mathbb{Z}/m) \cong K_n(\mathbb{C}; \mathbb{Z}/m) \cong \pi_n(BU; \mathbb{Z}/m).$

ExamplePositive characteristic

For algebraically closed fields of characteristic $p > 0$ and $m$ coprime to $p$ :

$K_n(\overline{\mathbb{F}}_p; \mathbb{Z}/m) \cong K_n(\overline{k}; \mathbb{Z}/m)$

for any algebraically closed $k$ with $\operatorname{char}(k) = p$ . Since $K_{2i-1}(\overline{\mathbb{F}}_p) = \varinjlim K_{2i-1}(\mathbb{F}_{p^n}) = \varinjlim \mathbb{Z}/(p^{ni} - 1) = \mathbb{Q}/\mathbb{Z}[\text{prime-to-}p]$ , we get:

$K_{2i-1}(\overline{\mathbb{F}}_p; \mathbb{Z}/m) \cong \mathbb{Z}/m \quad (m \text{ coprime to } p)$

$K_{2i}(\overline{\mathbb{F}}_p; \mathbb{Z}/m) = 0 \quad (m \text{ coprime to } p)$

This matches the topological computation, confirming that algebraic K-theory of algebraically closed fields is "topological" with finite coefficients.

Proof

The proof uses the theory of algebraic cycles and a norm argument.

Step 1: Reduction to extensions of transcendence degree 1. Write $K = \varinjlim K_\alpha$ as a directed union of subfields with $\operatorname{tr.deg}(K_\alpha/k) < \infty$ . Since K-theory commutes with filtered colimits of rings, $K_n(K; \mathbb{Z}/m) = \varinjlim K_n(K_\alpha; \mathbb{Z}/m)$ . It suffices to show $K_n(k) \to K_n(K_\alpha)$ is an isomorphism mod $m$ for each $K_\alpha$ , which reduces to the case $\operatorname{tr.deg}(K/k) = 1$ by induction.

Step 2: The case of transcendence degree 1. Let $K = \overline{k(C)}$ for a curve $C$ over $k$ . Choose a smooth projective model $\bar{C}$ over $k$ with function field $k(C)$ . We have:

$K_n(k) \to K_n(k(C)) \to K_n(K)$

For $m$ invertible in $k$ , the key is to show both maps are isomorphisms mod $m$ .

Step 3: Norm and transfer argument. For a closed point $x \in \bar{C}$ with residue field $k(x)$ (which equals $k$ since $k$ is algebraically closed), the localization sequence gives:

$\cdots \to K_{n+1}(k(C); \mathbb{Z}/m) \xrightarrow{\partial_x} K_n(k; \mathbb{Z}/m) \to \cdots$

The composite $K_n(k; \mathbb{Z}/m) \xrightarrow{\text{pullback}} K_n(k(C); \mathbb{Z}/m) \xrightarrow{\text{norm via } x} K_n(k; \mathbb{Z}/m)$ is multiplication by $\deg(x) = 1$ , hence the identity.

Step 4: Injectivity. The pullback $K_n(k; \mathbb{Z}/m) \to K_n(k(C); \mathbb{Z}/m)$ has a left inverse (the norm via any rational point), so it is injective.

Step 5: Surjectivity. This is the harder direction. Suslin shows that every element of $K_n(k(C); \mathbb{Z}/m)$ can be "specialized" to an element of $K_n(k; \mathbb{Z}/m)$ by choosing appropriate specialization maps $k(C) \to k$ . The key technical ingredient is the Suslin--Voevodsky theorem on relative algebraic cycles: the complex $z^p(C, \bullet)$ computing motivic cohomology satisfies a rigidity property when $k$ is algebraically closed.

Step 6: Passage to algebraic closure. Finally, $K_n(k(C); \mathbb{Z}/m) \to K_n(K; \mathbb{Z}/m)$ is an isomorphism because $K = \overline{k(C)}$ and taking the algebraic closure does not change K-theory mod $m$ (by a Galois cohomology argument: $K_n(\overline{F}; \mathbb{Z}/m)^{G_F} \cong K_n(F; \mathbb{Z}/m)$ is trivial when $G_F$ acts trivially). $\square$

■

RemarkGabber's rigidity and generalizations

Gabber extended Suslin's rigidity to the following setting: for a Henselian local ring $(R, \mathfrak{m})$ with residue field $k$ and $m$ invertible in $R$ :

$K_n(R; \mathbb{Z}/m) \xrightarrow{\;\sim\;} K_n(k; \mathbb{Z}/m).$

This Gabber rigidity theorem is the local analogue of Suslin's result. It implies:

$K$ -theory with finite coefficients is insensitive to Henselian thickenings.
For a smooth scheme $X$ over an algebraically closed field $k$ , $K_n(X; \mathbb{Z}/m)$ depends only on the etale homotopy type of $X$ (when $m$ is invertible in $k$ ).

The combination of Suslin rigidity and Gabber rigidity shows that algebraic K-theory with finite coefficients is, in a precise sense, an etale-local invariant.

Math Notes

Suslin's Rigidity Theorem

Statement

Consequences

Proof sketch

Related results