6
$\begingroup$

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.

$\endgroup$

1 Answer 1

7
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.