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The motivic coniveau spectral sequence tells us that for a scheme $X/k$, its cohomology $H^2(X,\mathbb{Z}(2))$ is the kernel of the tame symbol $K_2^M(k(X))\to \oplus_{Y} K_1^M(k(Y))$ where $Y$ runs over all divisors in $X$. Thus, all elements of this motivic cohomology group can be generated by linear combinations products of elements in $k(X)^\times$, such that their tame symbols cancel out.

Question: when is it the case that elements in $H^2(X,\mathbb{Z}(2))$ can all be generated by products of actual units on $X$?

Cases where this is known and specific illustrative counterexamples are both welcome. Results ignoring torsion are of use as well. If anything is known for bidegree-$(k,k)$, that would also be interesting.

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  • $\begingroup$ How will you cancel residues if you do not allow at least two pure tensors (with opposite signs)? $\endgroup$
    – Jason Starr
    Commented Jun 24, 2022 at 13:00
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    $\begingroup$ I think I'm misunderstanding something; where am I not allowing multiple pure tensors? $\endgroup$
    – xir
    Commented Jun 24, 2022 at 13:50
  • $\begingroup$ I misunderstood your question. I thought that you were asking about “symbol length”. I now see that you are asking about whether you can choose the units in $k(X)^\times$ to be elements in $\mathcal{O}_X(X)^\times$. $\endgroup$
    – Jason Starr
    Commented Jun 24, 2022 at 16:57
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    $\begingroup$ That should be the case if $X$ is a product of copies of the multiplicative group. $\endgroup$
    – Jason Starr
    Commented Jun 24, 2022 at 16:58
  • $\begingroup$ Can anything be said for curves in general? $\endgroup$
    – xir
    Commented Jun 26, 2022 at 19:36

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