Geometric genus of projective bundle over a scheme? This is Hartshorne's AG, exercise III.8.4(d):
Let $Y$ be a noetherian scheme and $\mathscr{E}$ be a locally free $\mathscr{O}_Y$-module of rank $n+1$. Let $X=\mathbb{P}(\mathscr{E})=\underline{\mathrm{Proj}}_Y(\mathrm{Sym}(\mathscr{E}))$. Let $\pi:X\to Y$.
(d) Show that the geometric genus $p_g(X)=0$ (Hint: Use II.8.1, the first exact sequence of kahler differentials).

(I) We defined the geometric genus only over the $k$-schemes where $k$ be a field, so here we may assume that $X,Y$ are all $k$-schemes.
(II) The hint is effective now only when $Y$ is smooth. Indeed for now we get
$$0\to\pi^*\Omega_{Y/k}\to \Omega_{X/k}\to\Omega_{X/Y}\to0.$$
Hence we have $$\omega_{X/k}=\pi^*\omega_{Y/k}\otimes\omega_{X/Y}.$$
The (b) of this exercise told us that
$\Omega_{X/Y}=(\pi^*\bigwedge^{n+1}\mathscr{E})(-n-1)$, hence
$$\omega_{X/k}=\pi^*(\omega_{Y/k}\otimes\bigwedge^{n+1}\mathscr{E})(-n-1).$$
But why $p_g(X)=\dim_k\Gamma(X,\omega_{X/k})=0$?
(III) Actually consider an affine cover $U_i\subset Y$ such that $\mathscr{E}|_{U_i}$ are all trivialized, then we have $V_i:=\pi^{-1}(U_i)\cong\mathbb{P}_{U_i}^n$. So we take any $s\in\Gamma(X,\omega_{X/k})$, we have $s|_{V_i}=0$ as over $U_i$,
$\omega_{X/k}|_{V_i}=O_{V_i}(-n-1)$. Hence by the axiom of sheaves, we must have $s=0$! Hence $p_g(X)=0$.
Is this right? If this is right, the hint in Hartshorne's book of this (d) is meaningless...
Thank you for your help!
 A: I think the exactness of II.8.11 on the left even for non-singular $Y$ should be an application of Exercise II.8.3. There one shows that this sequence splits in case of a product scheme. The projective bundle $X$ is locally a product, which still means that the sequence should be exact on the left (because exactness can be checked locally). However I'm not sure if we get an isomorphism $\omega_{X/k} \cong \pi^* \Omega_{Y/k} \otimes \omega_{X/Y}$ if $Y$ is singular, in which case those sheaves will not be locally free...
Now apply Exercise III.8.3. to your formula for $\omega_{X/k}$ to see
$$\pi_* \omega_{X/k} = \omega_{Y/k} \otimes \Lambda^{n+1} \mathcal E \otimes \pi_* \mathcal O(-n-1) = 0,$$
where the last equality follows from part (a): $\pi_* \mathcal O(l) = 0$ for $l < 0$. On the other hand you get
$$p_g(X) = H^0(X, \omega_{X/k}) = H^0(Y, \pi_* \omega_{X/k}) = 0$$
by definition of $\pi_*$.

I think your solution works as well, if you also take the $U_i$ such that $\omega_{Y/k}|_{U_i} \cong \mathcal O_{U_i}$.
