# Cohen-Macaulay fibers

Let $Y$ be a set of points in $\mathbb{P}^n$. Then we can write a resolution $$0\rightarrow P_n \rightarrow \cdots \rightarrow P_0\rightarrow \mathcal{O}_Y$$ where each $P_i=\bigoplus_j\mathcal{O}_{\mathbb{P}^n}(-a_{ij})$ (and the length of the resolution is $n$ as a zero dimensional ideal is Cohen-Macaulay).

Now suppose we start with a morphism $\varphi\colon X\rightarrow \mathbb{P}^n$ such that $\varphi_*\mathcal{O}_X$ is a locally free sheaf of rank $d$. Then we have $X\hookrightarrow \mathbb{P}(\varphi_*\mathcal{O}_X)\twoheadrightarrow \mathbb{P}^n$.

If $\varphi_*\mathcal{O}_X$ is not a sum of line bundles then the ring of sections of $X$ is not Cohen-Macaulay but as $\varphi_*\mathcal{O}_X$ is locally free it will be Cohen-Macaulay after localization (restriction to the fibres).

Question: Can I still write a resolution like the first one but now with $P_i=\bigoplus_j\mathcal{O}_{\mathbb{P}(\varphi_*\mathcal{O}_X)}(-a_{ij})$?

• I do not understand your hypotheses. Is the domain $X$ of the morphism $\phi$ supposed to be finite? If so, then $\phi_*\mathcal{O}_X$ will not be a locally free sheaf. However, unless $\phi$ is finite, there is no reason that the induced morphism from $X$ to $\mathbb{P}(\phi_*\mathcal{O}_X)$ should be an immersion. Could you please clarify your hypotheses? – Jason Starr Mar 19 '15 at 18:44
• sorry, the second $X$ is not the same as the first (I changed it). it was just to compare both ideas... what I want is the fiber of $\varphi$ to be a set of $d$ points for all $p\in\mathbb{P}^n$ and the question is if the knowledge of the resolution of each fiber will go for the global scheme (because its a scheme over a base scheme and not over a point). – Bajouca Mar 19 '15 at 19:15