# How to think about infinite generatedness of motivic cohomology

In this question I previously asked how to think about the motivic complex $\mathbf{Z}(1)_{\mathcal{M}}$, whose Zariski hypercohomology should morally be the "singular cohomology" $H^*((-)\wedge S^{2n},\mathbf{Z})$, or the "singular cohomology of a pair" $H^*((-)\times\mathbf{P}^n_k,(-)\times\mathbf{P}^{n-1}_k)$.

Denis Nardin's answer explains a suggestive analogy, and here is a followup.

It happens that several motivic cohomology groups are infinitely generated, even if $X$ is a smooth projective variety over a field (even for $k = \mathbf{Q}$ and $X = \text{Spec}(k)$).

How to think about this? Is there an analogy with infinite generatedness of the singular cohomology of a pair, for instance?

In other words, what should be the "moral reason" why this infinite generatedness occurs? Any analogy coming from algebraic topology would be greatly clarifying.

• For example, $H^{n,n}_\mathrm{mot}(k,\mathbf{Z}) = K^M_n(k)$ is Milnor $K$-theory of the field $k$, which is usually not finitely generated. – TKe Feb 12 '18 at 11:56
• I love the fact that this question, for someone with no context, is probably almost indistinguishable from the ravings of a lunatic. It's really fun to imagine a non mathematician reading this. :D – msouth Feb 12 '18 at 12:03
• Other examples are provided by the Beilinson-Lichtenbaum conjecture which relates motivic with étale cohomology. – TKe Feb 12 '18 at 12:20

Let us fix the ground field to be $\mathbb{C}$. The Picard group of a variety is the group of isomorphism classes of line bundles with multiplication given by the tensor. It turns out to be a very "homotopical" invariant. We can consider various variants of it: topological Picard group, analytic Picard group, algebraic Picard group..., depending on what kind of line bundles we are interested in.
It is well known that the topological Picard group of a space $X$ is just $H^2(X;\mathbb{Z})$ (this comes from the identification of $\mathbb{CP}^\infty$ with $K(\mathbb{Z},2)$).
In the algebraic world, something similar happens. The Picard group is equivalent to $H^2_{mot}(X;\mathbb{Z}(1))$. But in this case we know the Picard group cannot be finitely generated: there is a map from the algebraic Picard group to the topological Picard group (called the first Chern class) but, for example, if $X$ is a curve its kernel is given by the points of an abelian variety, the Jacobian of $X$. This is a massively nonfinitely generated abelian group, in fact it is isomorphic to $(\mathbb{R}/\mathbb{Z})^{2g}$ where $g$ is the genus of $X$.