Pullback of ideal sheaf defining closed immersion

Question

Let $Z\to X$ be a closed immersion with ideal sheaf $\mathcal{I}$. To me the exact sequence $0\to\mathcal{I}\to \mathcal{O}_X\to i_*\mathcal{O}_Z\to 0$ says that the functions in $\mathcal{I}$ are precisely those functions that vanish when we pull them back along $Z\to X$. Based on this interpretation, I would expect that $i^*\mathcal{I}=0$, since we pull back functions that vanish when we pull them back.

However, $i^*\mathcal{I}$ is actually the conormal sheaf, and hence not zero. I can do computations to see that $i^*\mathcal{I}$ is indeed the conormal sheaf in the sense that the dimensions of the fibers are indeed the codimension of $Z$ in $X$, so it works out.

What I do not understand is why my first reasoning fails. It fails because its not true, but it suggests that I'm not thinking about the ideal sheaf exact sequence in a correct way. So I hope that someone can point out the flaw in the first reasoning.

Answer 1

The problem is that when you form $i^*\mathcal{I}$, you are just considering $\mathcal{I}$ as a sheaf on its own and don't know how it's embedded in $\mathcal{O}_X$. Your argument shows the image of the canonical map $i^*\mathcal{I}\to i^*\mathcal{O}_X$ is $0$, but that canonical map may not be injective and so this does not imply $i^*\mathcal{I}=0$.

It is illuminating to consider a concrete example. Let $X=\mathbb{A}^1$ and $Z=\{0\}$, so on global sections we have $\mathcal{O}_X=k[x]$ and $\mathcal{I}=(x)$. The pullback functor $i^*$ is then $k \otimes_{k[x]}-$ where $k=k[x]/(x)$ is a $k[x]$-module by letting $x$ act trivially. But notice that as a $k[x]$-module, $(x)$ is isomorphic to $k[x]$ itself, since it is freely generated by $x$. So $k\otimes_{k[x]} (x)$ is isomorphic to $k\otimes_{k[x]}k[x]$ and in particular is not $0$, even though the induced map $k\otimes_{k[x]}(x)\to k\otimes_{k[x]}k[x]$ is $0$.

Your argument wants to say that $(x)$ should become $0$ when you tensor with $k$, since you're modding out $x$ and every element of $(x)$ is divisible by $x$. But not every element of $(x)$ is divisible by $x$ within the module $(x)$: instead, that is only true if you consider $(x)$ as sitting inside $k[x]$. In geometric terms, in this case $\mathcal{I}$ is just some line bundle on $X$ (in fact, a trivial line bundle), so its pullback to any subscheme $Z$ will again be a line bundle on $Z$, not $0$. The fact that $\mathcal{I}$ consists of functions that vanish on $Z$ is not an intrinsic property of the sheaf $\mathcal{I}$, but rather a property of the embedding $\mathcal{I}\to\mathcal{O}_X$.

Ultimately, in algebraic terms, this is just the fact that tensor products are not left-exact (i.e., not all modules are flat), if you find that helpful.