# Galois cohomology H^1(Q_p, Z_p(2)) = 0?

For Tate twists Z_p(2), which is defined by the projective limit of \mu_{p^m}(2) over all m>0, I would like to calculate H^1(Q_p, Z_p(2)).

I guess this is zero, but cannot prove it. Is it possible to calculate and prove H^1(Q_p, \mu_{p^m}(2)) = 0 for each m> 0?

Pierre MATSUMI

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Dear Pierre, Galois cohomology over local fields is most easily computed using Tate local duality and the Tate Euler char. formula. A convenient reference is Washington's article in "Modular forms and Fermat's Last Theorem" (Cornell, Silverman, Stevens eds.). In your particular case it will tell you that this $H^1$ is f.g. over $\mathbb Z_p$, of free rank $1$, and torsion free if $p > 3$. Regards, Matthew. –  Emerton Feb 5 '12 at 12:20

Explicitly we have $H^1(\mathbb{Q}_p,\mathbb{Z}_p(2)) = \begin{cases} \mathbb{Z}_p \oplus \mathbb{Z}/p\mathbb{Z} & \text{if } p \le 3 \newline \mathbb{Z}_p & \text{if } p > 3.\end{cases}$

This follows from the Remark following Prop. 7.3.10 of Neukirch et. al (same book as in Timo's answer): $$H^1(\mathbb{Q}_p,\mathbb{Z}_p(2)) \cong H^1(\mathbb{Q}_p,\mathbb{Q}_p/\mathbb{Z}_p(-1))^\vee \overset{7.3.10}{\cong} (\mathbb{Q}_p/\mathbb{Z}_p \oplus \mathbb{Z}/w_p^2 \mathbb{Z})^\vee = \mathbb{Z}_p \oplus \mathbb{Z}/w_p^2 \mathbb{Z}$$ where $\vee$ denotes the Pontryagin dual and $n=w_p^2$ is the maximal $p$-power such the the degree of $\mathbb{Q}_p(\mu_n) \mid \mathbb{Q}_p$ divides $2$.

The same argument can be used to compute $H^1(K,\mathbb{Z}_p(i))$ for all finite extensions $K \mid \mathbb{Q}_p$ and all $i \in \mathbb{Z}$.

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Dear Ralph, There is also torsion in the case when $p = 3$. I find this easiest to think about using the cohomology long exact sequence attached to the short exact sequence $0 \to \mathbb Z_p(2) \to \mathbb Z_p(2) \to \mathbb F_p(2) \to 0,$ but from the point of view you describe, it will be related to the fact that $\mathbb Q_3(\mu_3)$ has degree $2$. Regards, Matthew –  Emerton Feb 6 '12 at 6:03
Yep, thank you very much, I'll edit my answer. –  Ralph Feb 6 '12 at 6:40

Have you tried the Hochschild-Serre spectral sequence $H^p(\mathbf{Q}_p(\mu_{p^m})/\mathbf{Q}_p, H^q(\mathbf{Q}_p(\mu_{p^m}), \mu_{p^m}^{\otimes 2})) \Rightarrow H^{p+q}(\mathbf{Q}_p, \mu_{p^m}^{\otimes 2})$ and the exact sequence of lower terms $0 \to E_2^{1,0} \to E^1 \to E_2^{0,1} \to E_2^{2,0} \to E^2$?

See also Neukirch, Schmidt, Wingberg, Cohomology of Number Fields, Corollary (7.3.8). This reduces the determination of $H^1$ to that of $H^0$, which is trivial, and of $H^2$, which can be treated using the dualising module $\mu$.

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