Let $D$ be a degree $3$ division algebra over a field $k$ of char not 2 and 3.
Any such division algebra is cyclic. I am interested in knowing the cases when the reduced norm map $Nrd : D^\* \rightarrow k^\*$ is surjective. Of course, this happens over $\bar k$ and finite field etc. Here is my explicit question.


I want to relate surjectivity of reduced norm to the finiteness of $k^\*/(k^\*)^3$. To me it looks like not having enough of degree 3 field extensions is somehow responsible. I would appreciate examples, counterexamples or any reference in this direction.


Thanks a lot. 



reduced norm from degree 3 division algebra
The Merkurjev-Suslin Theorem says that an element $x \in k^\*$ is a norm from $D$ if and only if $[D] \cup (x)$ is zero in the Galois cohomology group $H^3(k, \mathbb{Z}/3)$. Therefore, for the reduced norm to be surjective for every division $k$-algebra of degree 3, it suffices to assume that $H^3(k, \mathbb{Z}/3) = 0$.


Conversely, if $k$ contains a primitive 3rd root of unity, then $H^3(k, \mathbb{Z}/3)$ is isomorphic to $K\_3(k)/3$, i.e., degree 3 Milnor $K$-theory mod 3, and in particular is generated as an additive group by symbols of the form $[D] \cup (x)$. So, in this case, for the reduced norm to be surjective for every division $k$-algebra of degree 3, it is also necessary that $H^3(k, \mathbb{Z}/3) = 0$. 


