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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!

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  • $\begingroup$ 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. $\endgroup$ Nov 18, 2022 at 10:55
  • $\begingroup$ Yes. This definitely makes sense. But I want to consider a really nice condition like finite-étale $\endgroup$ Nov 18, 2022 at 11:00

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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 $L$.

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  • $\begingroup$ cool answer,a question: what is a nice way to see that "can be easily found for g>1"? $\endgroup$
    – Simonsays
    Nov 18, 2022 at 12:23
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    $\begingroup$ In this case $\dim(J(Y)) = g < 2g - 1 = \dim(J(X))$. $\endgroup$
    – Sasha
    Nov 18, 2022 at 12:54
  • $\begingroup$ Thanks! This really helps! $\endgroup$ Nov 18, 2022 at 14:18

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