Let $\phi:X\rightarrow Y$ be a generically smooth projective surjective morphism of algebraic varieties over $k=\bar k.$ Is it possible for $R^1\phi_*(\mathbb Z/l)$ to be supported on a divisor of $Y$ (where $1<l\neq char(k)$)?
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$\begingroup$ I'm not sure I understand the question. It's certainly possible: e.g. take the identity. $\endgroup$ – Donu Arapura Sep 11 '15 at 15:48

$\begingroup$ Thank you for answering; maybe I should change my question. Actually, I have a generically smooth projective surjective morphism with $\mathbb P^5$ as generic fiber so that $R^1\phi_*\mathbb Z/l$ is zero on an (dense) open subscheme of $Y$. I wanted to conclude that this sheaf is zero so the problem is whether or not (and why) there could be an obstruction to the conclusion. $\endgroup$ – user3001 Sep 11 '15 at 16:29

$\begingroup$ Isn't $R^1\phi_*$ morally a family of $H^1$'s of the fibres? So what about if $X$ is a family of genus 0 curves over $Y$? Then again won't $R^1$ vanish? $\endgroup$ – eric Sep 12 '15 at 17:43
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Let's work over $\mathbf{C}$. Take a K3 surface $S$ with a fix point free involution $\sigma$ and let $X = S \times \mathbf{C}/\langle (\sigma, 1)\rangle$ and $Y = \mathbf{C}/\langle 1 \rangle \cong \mathbf{C}$. Take $l = 2$ and compute using proper base change that your $R^1$ is not zero and supported in $0$.