Let $X \rightarrow$ Spec$A$ be a scheme over a ring $A$. I know that if $X$ is a projective scheme over $A$, and $A$ is Noetherian. $H^p(X, \mathcal{O}_X)$ is afinitely generated A-module for all $p \geq 0$.
If $A$ is not a field, $H^p(X, \mathcal{O}_X)$ not necesarilly has to be a free $A$-module, but I can not find an example of this.
Thus, does anyone know any projective $A$-scheme $X$, with $A$ Noetherian, such that $H^p(X, \mathcal{O}_X)$ is not a free module for some $p \geq 1$?
If, because of some result that I do not know, $H^p(X, \mathcal{O}_X)$ happens to be free always in this case, I would like to ask this question in general without assuming that $A$ is Noetherian or $X$ is projective: does anyone know an $A$-scheme $X$ such that $H^p(X, \mathcal{O}_X)$ is not a free $A$ module for some $p \geq 1$?
Thanks in advance.