Let $K$ be an algebraically closed field. Denote by $\mathcal F_d$ the category of extensions $K\to F$ of transcendent degree $d$.(Objects are pairs $F,j$ consisting of a field $F$ and the embedding $j\colon K\to F$, such that transcendent degree is equal to $d$, morphisms between $(F_1,j_1)$ and $(F_2,j_2)$ are finite extensions $\psi\colon F_1\to F_2$ such that $\psi\circ j_1=j_2$). Consider the category of (covariant) functors $Fun(\mathcal F_d, \mathcal{Vect})$ from this category to the category of vector spaces over $\mathbb Q$. What is known about its cohomological dimension?

Especially we have a functor which associate to a field extension $K\to F$ the multiplicative group $(F^\times)\otimes \mathbb Q$. Is it true that all higher cohomology of this functor vanishes?

  • $\begingroup$ $F^{\times}$ is not a vector space... $\endgroup$
    – abx
    Feb 29 at 13:39
  • $\begingroup$ Yes, of course. I edited the question. Or we can study the same question with abelian groups instead of vector spaces. $\endgroup$ Feb 29 at 13:45
  • $\begingroup$ I don't understand the category $\mathcal F_d$. How can finite extensions be morphisms between extensions? $\endgroup$
    – Nandor
    Feb 29 at 13:46
  • $\begingroup$ I edited question. Is it clear now? $\endgroup$ Mar 1 at 19:46


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