# $R^2f_{\operatorname{et},*}\mathbb{G}_m$ vs $R^2f_{\operatorname{Zar},*}\mathbb{G}_m$

Let $S$ be the spectrum of a discrete valuation ring and $f:X\rightarrow S$ be a relative projective curve with generic fiber smooth and special fiber semistable. How much differ the sheaf $R^2f_{\operatorname{et},*}\mathbb{G}_m$ (computed in the étale topology) from $R^2f_{\operatorname{Zar},*}\mathbb{G}_m$ (computed in the Zariski topology, this is zero)? If they are not the same is there a condition on $f$ to ensure that they coincide?

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+1 for the austere efficiency of the title. –  Tom LaGatta Jul 2 '13 at 7:04

Actually the result cited (stating that the étale one vanishes) is more general than suggested: it doesn't require that the morphism be smooth, just that $X$ be regular. Also it's worth pointing out that in characteristic zero this follows easily from the Kummer sequence and proper base change, and doesn't need $X$ to be regular; the meat of the theorem proved by Artin is the vanishing of the $p$-torsion in characteristic $p$. –  Martin Bright Jul 2 '13 at 9:13