Let $S$ be a separable irreducible Noetherian scheme and let $X$ be a projective smooth curve over $S$. Let $\mathcal F$ be a coherent sheaf on $X$ which is flat over $S$. Suppose the restriction $\mathcal F\mid\_{X\_s}$ of $\mathcal F$ on the fiber $X\_s$ is locally free for some point $s\in S$.


**Question:** is $\mathcal F$ also locally free?



Is a coherent and flat sheaf locally free?
As pointed out in the comments, being a vector bundle at a point is (by definition) an open property and for coherent sheaves flat over a base, being a vector bundle at a point can be checked on fibers. This means,


**Prop** Let $f : X \to Y$ be a flat map of schemes and $\mathcal{F}$ a coherent $\mathcal{O}\_{X}$-module flat over $Y$. Suppose that $\mathcal{F}|\_{X\_y}$ is a vector bundle on $X\_y$ for some $y$. Then there is an open $U \subset X$ containing $X\_y$ such that $\mathcal{F}|\_U$ is a vector bundle.


If $f : X \to Y$ is flat and proper and $\mathcal{F}$ is coherent and flat over $Y$, this implies there is actually an open of the base over which $\mathcal{F}$ is a vector bundle.


Now, in your question I take that $X \to S$ is a *relative curve*. Then there exist degenerations of a vector bundle to a non-locally-free coherent sheaf on a family $X \to S$ which is as nice as possible.


Indeed consider


Let $\pi\_1 : X = \mathbb{P}^1 \times \mathbb{P}^1 \to \mathbb{P}^1 = S$ be the projection. Let $x = X$ be a point and $\mathcal{I} \subset \mathcal{O}\_{X}$ the ideal sheaf of $x = (0,0) \in X$. For each fiber $X\_t$ with $t \neq 0$ we have $\mathcal{I}|\_{X\_t} = \mathcal{O}\_{X\_t}$ is a vector bundle. However, $\mathcal{I}$ is not a vector bundle. I claim that $\mathcal{I}$ is $\pi\_1$-flat. This is clear on $X \setminus \{ x \}$ so I we consider the local structure around $x$. On a dense open we have the following algebra problem,


$$ A = k[x]\_{(x)} \to k[x,y]\_{(x,y)} = B \quad \text{ with the ideal } \quad I = \mathfrak{m}\_B = (x,y) \subset k[x,y]\_{(x,y)} $$


I claim that $I$ is flat over $A$. There is an exact sequence,


$$ 0\to B \xrightarrow{(y \; -x)} B^2 \xrightarrow{(x \; y)} I \to 0 $$


Then applying <https://stacks.math.columbia.edu/tag/00MK> we just need to show that $B/\mathfrak{m}\_A B \to (B / \mathfrak{m}\_A B)^2$ is injective which is true because $y$ is a non zero-divisor on $B / \mathfrak{m}\_A B$. Thus $I$ is $A$-flat. Furthermore, we get the local structure,
$$ I / \mathfrak{m}\_A I \cong k \oplus k[y]\_{(y)} $$
but its image in $B / \mathfrak{m}\_A B$ is just $(y)$ which is locally free. This we see that $\mathcal{I}|\_{X\_0} \cong \mathcal{O}\_{X\_0}(-1) \oplus \iota\_\* k$ which has degree zero as it must because $\mathcal{I}|\_{X\_t} \cong \mathcal{O}\_{X\_t}$ for $t \neq 0$ and degree is constant in flat families.


Another example is given by considering the relative cotangent bundle of the flat projective family,


$$ f : X = \mathrm{Proj}{(k[t][X,Y,Z]/(XY - t Z^2))} \to \mathrm{Spec}{(k[t])} = S $$


