First cohomology of a cochain complex depends on field?

Question

Let $\Delta$ be a simplicial complex on $\{1,\dots,n\}$ and let us consider the cochain complex of $\Delta$ over $\mathbb{K}$

$$0 \rightarrow C^{-1} \xrightarrow{\delta^{-1}} C^{0} \xrightarrow{\delta^{0}} \cdots\rightarrow C^{d}\xrightarrow{\delta^{d}} 0.$$

Let us denote the $i$-th reduced cohomology of $\Delta$ over $\mathbb{K}$ by $H^i (\Delta; \mathbb{K}) := \ker(\delta^{i+1}) / \mathrm{Im} (\delta^{i})$. Elements of $\ker(\delta^{i+1})$ are called $i$-cocycles and elements of $\mathrm{Im} (\delta^{i})$ are called $i$-coboundaries.

Does $H^1 (\Delta; \mathbb{K})$ depend on $\mathbb{K}$? That is, if I find that $H^1 (\Delta; \mathbb{F}_2)$ is not $0$ (showing an element that is a cycle but not a boundary, with respect to $\mathbb{F}_2$), then there is any known result that guarantees that $H^1 (\Delta; \mathbb{K})$ is not $0$ for any field $\mathbb{K}$?

Answer 1

Yes, the cohomology certainly depends on the field. For example consider $\mathbb{RP}^2$ (which can be triangulated, so you can see this as a simplicial complex), then $H^1(\mathbb{RP}^2; \mathbb{F}_2) = \mathbb{F}_2$, however $H^1(\mathbb{RP}^2; \mathbb{K}) = 0$ if the characteristic of $\mathbb{K}$ is not $2$ (e.g. if $\mathbb{K} = \mathbb{Q}$).

You do have the Universal Coefficient Theorem though, which implies for example that if $\operatorname{char} \mathbb{K} = 2$, then $$H^i(\Delta; \mathbb{K}) \cong H^i(\Delta; \mathbb{F}_2) \otimes_{\mathbb{F}_2} \mathbb{K}.$$ But you can't expect information about $H^1(\Delta;\mathbb{F}_2)$ to give you information about cohomology over all the other fields.

Answer 2

Define the chain complex $0 \to \mathbf{Z} \to \mathbf{Z} \to 0$ where the map is multiplication by 2. Now compare what happens if you tensor with $\mathbf{Q}$ with what happens if you tensor with $\mathbf{F}_2$.