This is a follow-up question to these two: 1 2.
I am studying a paper about the degeneration of the de Rham-Hodge spectral Sequence. The following data are given: A field $k$ and a smooth proper $k$-scheme $X\longrightarrow\operatorname{Spec}(k)$, a finitely generated $\mathbb{Z}$-algebra $A$ and a smooth proper scheme $\mathfrak{X}\longrightarrow A$ so that $\mathfrak{X}_{k}\cong X$. The author claims:
After possibly localizing $A$, we can assume that the coherent $A$-modules $H^{n}_{\operatorname{dR}}(\mathfrak{X}/A)$ and $H^{n}_{\operatorname{Hodge}}(\mathfrak{X}/A)$ are locally free.
Why?
I am by no means an expert on the topic, so please be as détailed as possible in your answers; a reference might also help.
this is a general property of coherent modules. First it is a very good idea to look at the paper of Illusie:Frobenius and Hodge degeneration.which gives much more detail than the paper you are reading now.(and written by the person who prove the theorem!)
about your question, look at the generic point $\eta$ of $A$(the point associated to the prime ideal 0) local ring of $A$ at this point is just your fraction field. now assume that your coherent sheaf is given by the module $M$,then $M_\eta$ is finite free over $A_\eta=Frac(A)$.let $(m_1/a_1,m_2/a_2,...,m_n/a_n)$ be a basis for the module $M_\eta$. consider generators for $M$,$(k_1,...,k_t)$ and write $$k_i=\Sigma \frac{b_{ij}}{c_{ij}}\frac{m_j}{a_j}$$. now if invert all the $a_i,c_{ij}$ in $A$ and call the ring $A_S$. now you have a surjective map $A_S^n\to M_S$ by sending $e_i\to \frac{m_i}{a_i}$.call the kernel of this map $N$, so you have $$0\to N\to A_S^n\to M_S\to 0$$ but you know that $A_\eta^n\to M_\eta$ is an isomorphism and $A_S\to A_\eta$ is flat,so from the above exact sequence we have $N_\eta=0$. because $N$ is finitely generated(by coherence) there should be an element $a\in A$ such that $aN=0$. so in summary after inverting $a$ the map between $A^n\to M$ become an isomorphism and hence $M$ would be free.
now back to your problem you know that $R^if\Omega_{X/A}$ is coherent and so after going to a localisation $A_S$,${(R^if\Omega_{X/A})}_S$ is free. but this is ${R^if\Omega_{X_S/A_S}}$(because R^if is always quasi cohehrent) so we are done.