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Class field theory defines an isomorphism between the Brauer group of a finite extension of p-adic fields and a cyclic group with a canonical generator. This in turn defines an isomorphism of the Brauer group with $\mathbb{Z}/n\mathbb{Z}$, where $n$ is the degree of the finite extension in question. This latter group has a canonical ring structure.

Question: Is there a natural interpretation of this ring structure at the level of $H^2$ or central simple algebras?

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    $\begingroup$ It is canonical, but not "canonically" canonical. In local class field theory there is always the issue that one can choose geometric or arithmetic Frobenius and this changes the sign of the generator, which suggests that the answer to your question is negative. You similarly ask whether $H^1(\mathbf{F}_p,\mathbf{Z}/n \mathbf{Z}) = \mathbf{Z}/n \mathbf{Z}$ (where the cocycle of the LHS sending Frobenius to $1$ is identified with $1 \in \mathbf{Z}/n \mathbf{Z}$) gives a natural ring structure, and the same meta-argument suggests this is unlikely. $\endgroup$
    – user491858
    Commented Aug 20 at 20:00
  • $\begingroup$ Huh, interesting! I'm actually interested only in the squaring map on the ring actually, which seems like it would be well-defined in that case. But point well-taken that there may not be a clear interpretation of it on the cohomological side of things. Thanks for the answer! $\endgroup$
    – NZK
    Commented Aug 20 at 21:19
  • $\begingroup$ Squaring doesn't commute with negation. $\endgroup$
    – Alexei Entin
    Commented Aug 21 at 18:11
  • $\begingroup$ Sorry, what I meant is that invariance is what matters for me. Good point. $\endgroup$
    – NZK
    Commented Aug 21 at 20:48

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I believe the answer is simpler than I expected.

Since $\mu_n$ has a $\mathbb{Z}/n\mathbb{Z}$ action, the whole $H^2(K, \mu_n)$ inherits the structure of a module over that ring. This is the product structure.

More abstractly, we have an isomorphism $H^2(K, \mu_n) \cong H^0(K, \mathbb{Z}/n\mathbb{Z})$ which I might venture to call the Hodge star in this context. We can then use that $H^2$ is a module over $H^0 \cong \mathbb{Z}/n\mathbb{Z}$.

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    $\begingroup$ This is really a non-answer. Of course if $M$ is an abelian group and one has an isomorphism $\mathbf{Z}/n = M$ this gives a ring structure on $M$, but it is not intrinsically defined on $M$ itself which was the spirit of the question. The $n$th roots of unity $M$ inside $\mathbf{C}$ are $= \mathbf{Z}/n$ by sending $1$ to $\zeta = \exp(2 \pi i/n)$, but there is nothing "natural" about $\zeta^a \star \zeta^b = \zeta^{ab}$. $\endgroup$
    – user491858
    Commented Aug 22 at 12:48

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