A vanishing condition for cup products in Galois cohomology

Question

Let $k$ be a field of characteristic $\neq 2$. For a non-zero element $a \in k^*$, let us write $[a] \in H^1(k,\mathbb{Z}/2)$ for the Galois cohomology class corresponding to the quadratic extension $k(\sqrt{a})/k$. If $a,b \in k^*$ are non-zero elements then the class $[a] \cup [b] \in H^2(k,\mathbb{Z}/2)$ is trivial if and only if $b$ is a norm from $k(\sqrt{a})/k$. In particular, if $ab = -1$ then $b=-a$ is a norm from $k(\sqrt{a})/k$ and hence $[a] \cup [b] = 0$. I have a reason to believe the following generalization is true:

Claim: Let $a_1,...,a_n \in k^*$ be such that $\prod_i a_i = -1$. Then $[a_1] \cup [a_2] \cup ... \cup [a_n] = 0$.

Is this claim true?

One way to approach this problem is via quadratic forms. Let us denote by $\left<b_1,...,b_n\right>$ the isomorphism class of the quadratic form $\sum_i b_i x_i^2$. The Witt ring $W(k)$ of $k$ is the ring generated by isomorphism classes of non-degenerate quadratic forms over $k$, modulu the relation $\left<1,-1\right> \sim 0$ (more precisely, the addition in $W(k)$ is determined by the direct sum operation and multiplication by tensor product). Let $I \subseteq W(k)$ be the ideal generated by forms of even rank. A deep result in the theory of quadratic forms (closely related to Milnor's conjecture), is that there is a natural isomorphism of graded rings $$ \oplus_n I^n/I^{n+1} \stackrel{\cong}{\longrightarrow} H^*(k,\mathbb{Z}/2) .$$ In particular, the class of $\left<1,-a\right> \in I$ (mod $I^2$) is sent to the class $[a] \in H^1(k,\mathbb{Z}/2)$, and hence the class of $\left<1,-a_1\right> \otimes ... \otimes \left<1,-a_n\right>$ is sent to $[a_1] \cup ... \cup [a_n]$. A possible strategy is then to show that if $\prod_i a_i = -1$ then the quadratic form $\left<1,-a_1\right> \otimes ... \otimes \left<1,-a_n\right>$ (which has rank $2^n$) is trivial (in the sense that it is isomorphic to a direct sum of $2^n$ copies of $\left<1,-1\right>$).

Answer 1

After realizing that the claim is wrong (thanks to the answers above), I managed to find a weaker statement, that turned out to be what I needed anyway. So just in case someone else happens to be interested in this question, here is the idea.

For each $l$, let $\mu_l$ denote the Galois module of $l$-roots of unity. In particular, let us use $\mu_2 = \{1,-1\}$ instead of $\mathbb{Z}/2$. For each $n$ we have an inclusion of Galois modules $\mu_l \longrightarrow \mathbb{G}_m$. Now indeed it is not true that if $\prod_i a_i = -1$ then $[a_i] \cup ... \cup [a_i] = 0$. However, it is true that in this case the image of $[a_1] \cup ... \cup [a_n]$ in $H^n(k,\mathbb{G}_m)$ is $0$ (which happens to be sufficient for my original application).

To see this, consider the short exact sequence of Galois modules $$ 0 \longrightarrow \mu_2 \longrightarrow \mu_4 \longrightarrow \mu_2 \longrightarrow 0 .$$ We then obtain a long exact sequence in homology groups, and in particular a boundary map $$ H^{n-1}(k,\mu_2) \longrightarrow H^n(k,\mu_2) .$$ Unwinding the definitions, one can check that this boundary map sends a class $\alpha \in H^{n-1}(k,\mu_2)$ to the class $\alpha \cup [-1] \in H^n(k,\mu_2)$. Now, as explained in the answer above, it is sufficient to consider the case where $n > 1$ is odd. Let $a_1,...,a_n$ be such that $\prod_i a_i = -1$. Then, again as explained above, we have $$ [a_1] \cup ... \cup [a_n] = [a_1] \cup ... \cup [a_{n-1}] \cup [-1] $$ and so $[a_1] \cup ... \cup [a_n]$ is in the image of the boundary map $H^{n-1}(k,\mu_2) \longrightarrow H^n(k,\mu_2)$. It follows that the image of $[a_1] \cup ... \cup [a_n]$ in $H^n(k,\mu_4)$ is $0$, and hence the image in $H^n(k,\mathbb{G}_m)$ is $0$ as well.