# Pull back of indecomposable sheaves

Suppose $$f:X\rightarrow Y$$ is a morphism of schemes and $$\mathcal{F}$$ is an indecomposable coherent sheaf on $$Y$$ (suppose both schemes are as nice as possible so that Krull-Schmidt theorem holds for coherent sheaves). If $$f$$ is some really nice morphisms, say finite-étale for instance, can we show that $$f^*\mathcal{F}$$ is indecomposable?

Although I can't remember the exact statement, I believe I have seen a result that says that if $$A\rightarrow B$$ is a "nice" morphism of rings and $$0\rightarrow M' \rightarrow M \rightarrow M''\rightarrow 0$$ is an exact sequence of "nice" $$A$$-modules, then the base change of this exact sequence to $$B$$ splits if and only if it originally splits. I was hoping this would help, but I am not really sure!

• You may know this, but it is clearly wrong if you choose $X = \bigsqcup U_i$ where the $U_i \subset Y$ form a trivializing Zariski cover of $M', M$ and $M''$. Then $X \to Y$ is étale, but not finite. Nov 18, 2022 at 10:55
• Yes. This definitely makes sense. But I want to consider a really nice condition like finite-étale Nov 18, 2022 at 11:00

Let $$f \colon X \to Y$$ be an etale double cover and let $$L$$ be a line bundle on $$X$$ which is not isomorphic to the pullback of a line bundle on $$Y$$ (an example of such can be easily found if $$Y$$ is a curve of genus $$g \ge 2$$). Set $$F := f_*(L).$$ Then $$F$$ is indecomposable, but $$f^*(F) \cong L \oplus \tau^*(L)$$, where $$\tau$$ is the involution of $$X$$ over $$Y$$.
Indeed, the isomorphism for $$f^*(F)$$ follows easily from base change because $$X \times_Y X \cong X \sqcup X$$, where the first component sits in $$X \times X$$ diagonally and the second is the graph of $$\tau$$.
On the other hand, $$F$$ is locally free of rank $$2$$ and if $$F \cong L_1 \oplus L_2$$ then $$L \oplus \tau^*(L) \cong f^*(F) \cong f^*(L_1) \oplus f^*(L_2),$$ and since a maximal direct sum decomposition of a sheaf is unique, we must have $$L \cong f^*(L_1)$$ or $$L \cong f^*(L_2)$$, which contradicts the assumption about $$L$$.
• In this case $\dim(J(Y)) = g < 2g - 1 = \dim(J(X))$. Nov 18, 2022 at 12:54