Prove that if a flat coherent sheaf has vector bundle fibres over each point, then the entire sheaf is a vector bundle | Step-by-Step Solution

Hello! This is a beautiful problem in algebraic geometry that connects local and global properties of sheaves.

What We're Solving:

We need to prove that if we have a flat family of coherent sheaves $\mathcal{F}$ over a base scheme, and each fiber (what we get when we restrict to individual points) is a vector bundle, then the entire sheaf $\mathcal{F}$ must itself be a vector bundle.

The Approach:

The key insight is that flatness gives us excellent "compatibility" between the local and global structure. We'll use flatness to show that the local freeness of fibers "lifts" to local freeness of the whole sheaf. Think of it like this: if every slice of a cake is perfectly layered, and the cake was made with a "flat" process that preserves structure, then the whole cake must be perfectly layered too!

Step-by-Step Solution:

Step 1: Set up the notation Let $f: X \to S$ be a morphism of schemes, and let $\mathcal{F}$ be a coherent sheaf on $X$ that is flat over $S$. For each point $s \in S$, assume the fiber $\mathcal{F}_s = \mathcal{F}|_{X_s}$ is a vector bundle on $X_s$.

Step 2: Use the local nature of the problem Since being a vector bundle is a local property, we can work locally on $S$. Fix a point $s \in S$ and choose an affine neighborhood $U = \text{Spec}(A)$ around $s$.

Step 3: Apply the key theorem about flat families Here's where flatness becomes crucial! There's a fundamental result: if $\mathcal{F}$ is flat over $S$ and locally free on each fiber, then $\mathcal{F}$ is locally free in a neighborhood of each fiber. This uses the fact that flatness preserves the "freeness" property under base change.

Step 4: Show local freeness The technical heart is this: since $\mathcal{F}_s$ is locally free of some rank $r$ and $\mathcal{F}$ is flat over $S$, we can find:

• A neighborhood $V$ of $s$ in $S$
• An open cover of $f^{-1}(V)$ where $\mathcal{F}$ is free of rank $r$

This follows from the flatness condition and semicontinuity theorems.

Step 5: Conclude globally Since this works for every point $s \in S$, and being locally free is a local property, we can cover all of $X$ with open sets where $\mathcal{F}$ is free of constant rank. This is precisely the definition of $\mathcal{F}$ being a vector bundle!

The Answer:

The sheaf $\mathcal{F}$ is indeed a vector bundle. The proof relies on the fundamental principle that flatness allows local properties of fibers (being vector bundles) to "lift" to global properties of the total space.

Memory Tip:

Remember: "Flatness preserves freeness!" When you have a flat family and each slice is free (locally free = vector bundle), then the whole family must be free. Think of flatness as a "structure-preserving" condition that doesn't let nice properties get "twisted" as you move around the base.

This is a perfect example of how algebraic geometry lets us go from pointwise information to global conclusions - the flatness hypothesis is exactly what we need to make this leap!