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(I have asked the question The commutativity of minimal extension $\cdots$ and I simplify this question to the next simple question:)

Let $X$ be a rational variety over $\mathbb{C}$, $\phi : \hat{X} \rightarrow X$ be the blow-up of one point $\{p\}$, and $M$ be a simple (holonomic) $D_{\hat{X}}$-module. Then

Is it true that the direct image $\int_{\phi}M (=\phi_+M)$ is also simple (holonomic) $D_X$-module ?

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    $\begingroup$ I'm not a $D$-module expert but I'm pretty sure the answer is no. If you take a $D$-module $M$ supported on the special fiber you should get the simple module supported at $p$ tensored with the cohomology of $M$, and already for $X$ a smooth surface, so the exceptional divisor is $\mathbb P^1$, the cohomology can be arbitrarily large. $\endgroup$
    – Will Sawin
    Commented May 27, 2018 at 19:54

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No. I’ll provide an example for what WillSawin suggested. Let $X$ be $\Bbb C^2$. Let $M$ be the irreducible holonomic module on $\hat X$ supported on the special fiber whose restriction to the special fiber is the structure sheaf. Then the direct image of $M$ is just the de Rham complex of $\Bbb P^1$, or rather the direct image of this complex via the (exact) functor $H^0i_+$, where $i\colon \{0\}\to X$ is inclusion. In particular, $\phi_+M$ has cohomology in more than one degree.

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  • $\begingroup$ Thank you very much. I am a beginner of $D$-module, so I cannot understand what do you mean by ''$\phi_+M$ has cohomology in more than one degree", and why this implies that $\phi_+M$ is not simple... I would be grateful if you would give me more explanation. $\endgroup$
    – Y. M.
    Commented May 31, 2018 at 14:13
  • $\begingroup$ @YukiMatsubara This means that it’s not a module, it’s a complex. It has cohomology in more than one degree (I don’t know how else to phrase that part). In order for it to be a simple module it needs to actually be, well, a module. $\endgroup$
    – Avi Steiner
    Commented Jun 2, 2018 at 4:02

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