Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $K$ be a local field with $K^{ur}$ the maximal unramified extension and $K^{tr}$ the maximal tamely ramified extension. Assume that the characteristic of the residue field of $K$ is $q$.

If $p$ is a prime not equal to $q$, then we know that the cohomological dimension of $Gal (K^{ur}/K) = 1$ so $H^2 (Gal(K^{ur}/K), \mu_p) = 0$ as $Gal(K^{ur}/K) = \widehat{Z}$. Do we know if $H^2(Gal(K^{tr}/K), \mu_p) =0$ as well?

share|improve this question
3  
Never. Let $M = \mu_p$, $D = {\rm{Gal}}(K^{\rm{tr}}/K)$, and $I = {\rm{Gal}}(K^{\rm{tr}}/K^{\rm{ur}})$. In $E_2^{i,j} = H^i(D/I,{\rm{H}}^j(I,M)) \Rightarrow H^{i+j}(D,M)$, $E_2^{i,j} = 0$ for $i > 1$ & for $j > 1$ since $I$ has c.d. 1. So cohom. of interest identifies with $H^1(D/I,{\rm{H}}^1(I,M))$; same size as $H^0(D/I,H^1(I,M))$ since $D/I$ has trivial Herbrand qts. By sp. seq., its coker of inj. $H^1(D/I,M) \rightarrow H^1(D,M))$. By Herbrand, $h^1(D/I,M) = h^0(D,M) = \#\mu_p(K)$. Since $H^1(D,M) = H^1(K,M)^P$ ($P=$ wild inertia) is $K^{\times}/(K^{\times})^p$ by Kummer, sizes differ. –  BCnrd Sep 14 '10 at 15:27
    
To clarify, the "Never" means "Never 0", not "We never know" (quite the opposite...). –  BCnrd Sep 14 '10 at 18:44

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.