# Unicity in Hartshorne Corollary II.7.15

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?

• The inverse image of the ideal sheaf isn't zero. I'd recommend starting there. Commented Feb 23, 2023 at 15:55
• @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? Commented Feb 23, 2023 at 16:32
• @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. Commented Feb 23, 2023 at 16:37
• 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 $1\notin (x,y)$. Commented Feb 23, 2023 at 16:39
• @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... Commented Feb 23, 2023 at 16:46

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.