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What am I missing in the following:

Let $X=\mathbb{A}^2_k$ be the affine plane over an algebraically closed field $k$, and let $O$ be the origin. Let $\tilde{X}$ be the blow-up of $O$. If $\mathcal{I}$ is the ideal sheaf of $Y=\{O\}$ (with reduced induced structure), then the inverse image ideal sheaf $\mathcal{J}$ of $\mathcal{I}$ under the closed immersion $f:Y\to X$ is $0$. Hence the blow-up of $Y$ with respect to $\mathcal{J}$ is just $Y$ itself, i.e. the strict transform $\tilde{Y}$ is just $\tilde{Y}=Y$. But aren't there then plenty of morphisms $\tilde{f}:\tilde{Y}\to\tilde{X}$ such that $\pi_X\circ\tilde{f}=f\circ\pi_Y$, where $\pi_X:\tilde{X}\to X$ resp. $\pi_Y:\tilde{Y}\to Y$ are the natural maps? Because we can just send $\tilde{Y}$ to any point on the exceptional curve $E\subseteq \tilde{X}$ over $O$.

But then, this would contradict the unicity in Corollary II.7.15 of Hartshorne, so I'm sure I'm going wrong somewhere. But where?

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  • $\begingroup$ The inverse image of the ideal sheaf isn't zero. I'd recommend starting there. $\endgroup$
    – KReiser
    Commented Feb 23, 2023 at 15:55
  • $\begingroup$ @KReiser The two generators $x,y$ of $\mathcal I$ certainly are mapped to zero in $k[Y] = k[x,y] / (x,y) \cong k$, right? $\endgroup$ Commented Feb 23, 2023 at 16:32
  • $\begingroup$ @KReiser It isn't? The map $f$ corresponds to $k[x,y]\to k[x,y]/(x,y)$, and thus $f^{*}\mathcal{I}=((x,y)\otimes_{k[x,y]}k[x,y]/(x,y))^{\sim}$ which I'm pretty sure is $0$. Then $f^{-1}\mathcal{I}\cdot\mathcal{O}_Y$ is the image of $f^{*}\mathcal{I}$ under the natural map $f^{*}\mathcal{I}\to\mathcal{O}_X$, so it should be $0$ as well. And even if it wasn't, then it would be $\mathcal{O}_Y$, giving $\tilde{Y}=Y$ as well. $\endgroup$
    – imtrying46
    Commented Feb 23, 2023 at 16:37
  • $\begingroup$ The global sections of $f^*\mathcal{I}$ are $(x,y)\otimes_{k[x,y]} k[x,y]/(x,y)\cong (x,y)/(x,y)^2$, which is a two-dimensional vector space on the generators $x,y$. This is a classic misstep - you can't say $x\otimes_{k[x,y]} 1 = 1\otimes x = 0$ since $1\notin (x,y)$. $\endgroup$
    – KReiser
    Commented Feb 23, 2023 at 16:39
  • $\begingroup$ @KReiser Sure, $f^* \mathcal I$ is a two-dim vector space. But the inverse image ideal sheaf is the ideal generated by the map $f^* \mathcal I \to \mathcal O_Y$, which is zero. I mean, there are not that many ideals in $k$ anyway... $\endgroup$ Commented Feb 23, 2023 at 16:46

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As you correctly pointed out, the inverse image ideal sheaf $J$ is zero. But this means $$\tilde Y = \operatorname{Proj} \bigoplus_{d=0}^\infty J^d = \operatorname{Proj}(k \oplus 0 \oplus 0 \oplus \dots )= \emptyset$$ because the only homogeneous ideal in $k$ is the negligible one.

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  • $\begingroup$ Ah yes of course, I forgot about the negligible ideal! $\endgroup$
    – imtrying46
    Commented Feb 23, 2023 at 16:51

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