What is the field of de Rham cohomology vector space over?

On Tu's an introduction to Manifold, 2nd edition, p275 (please see the image below) It said that all the closed $$k$$-form and exact $$k$$-forms on a manifold are both vector space. I think the vector space are both over $$\mathbb{R}$$. My question is: Given a manifold $$M$$ of dimension $$n$$ and a coordinate of $$M:x_1,x_2\cdots,x_n$$ I think the $$k$$-form on $$M$$ is some thing like $$f(p)\,x_{i_1}\wedge x_{i_2} \wedge x_{i_3} \cdots x_{i_k}$$, where $$1\le{i_1} <{i_2}<\cdots {i_k} \le n$$, $$p \in M$$ . Now it look like the vector space should be over $$f(p)$$, all the smooth function on $$M$$, because multiply a real number seem not give us all the $$k$$-form on $$M$$. Nevertheless, if $$\omega$$ is a closed form, $$f(p)\omega$$ is not necessary a closed form. It makes me confused.

Futhermore, other sources use the term "generator" of a de Rham cohomology. Does it means the basis of the vector space? Or we just see the de Rham cohomology as a group, an the generator means group generator? If so, what is the group operation of de Rham cohomology group?

• Yes, they are vector spaces over $\mathbf R$. Smooth functions do not form a field. – Asal Beag Dubh Jan 7 at 16:03
• To make sense of calling them vector space, it must be $\mathbb{R}$. However, they space should be all the closed forms (or exact forms) on $M$. I have difficulty to see how to generate all the closed forms on $M$. On second thought, it think $B^k(M)$ and $Z^k(M)$ are both infinite dimensional vector space. A more clear explanation is welcomed. – Rikeijin Jan 7 at 16:43