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Let $p$ be an odd prime and $\mu_{p}$ be the group of $p^\text{th}$-roots of unity. Then, there exists a cup-product map which maps the product of $p$-copies of $H^{1}(\mathbb{Q}, \mu_{p})$ into $H^{p}(\mathbb{Q}, \mu^{\otimes p})$.

However, using the norm-residue isomorphism theorem, $H^{p}(\mathbb{Q}, \mu^{\otimes p}) \cong K_{p}^{M}(\mathbb{Q})\big/pK_{p}^{M}(\mathbb{Q})$, where $K_{p}^{M}(\mathbb{Q})$ is the $p$-th Milnor $K$-group of $\mathbb{Q}$.

Since the Milnor $K$-group $K_{n}^{M}(\mathbb{Q}) \cong \mu_{2}$ for $n \geq 3$ and $p$ is odd, we have that $pK_{p}^{M}(\mathbb{Q}) = K_{p}^{M}(\mathbb{Q})$ and therefore, $H^{p}(\mathbb{Q}, \mu^{\otimes p}) = 0$.

What does the cup-product do, in this concrete situation, so that every element of the cup-product of $p$ copies of $H^{1}(\mathbb{Q}, \mu_{p})$ ends up getting mapped to zero? This is my main question. Some explanation/help/insights would be greatly appreciated.

This is even more baffling to me because $H^{1}(\mathbb{Q}, \mu_{p})$ is so simply isomorphic to $\mathbb{Q}^{\times}\big/\mathbb{Q}^{\times p}$, in which coset of $a$ being trivial means $a$ is a $p^\text{th}$ power.

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    $\begingroup$ I don't really understand what the question is. Milnor K-theory of a field $k$ is by definition a quotient of the tensor algebra on $k^\times$, so the cup product of elements $a_1,\ldots,a_n$ is by definition the symbol $\{a_1,\ldots,a_n\}$. So whatever argument you use to compute $K_n^M(\mathbf Q)$ will tell you why these classes are trivial... $\endgroup$
    – R. van Dobben de Bruyn
    Commented Nov 7 at 2:30

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