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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.

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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 $H^3(k, \mathbb{Z}/3) = 0$.

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