Closed subschemes and pulling back the structure sheaf via the inclusion map

Question

I would just like a clarification related to closed subschemes.

If $(X,{\cal O}_X)$ is a locally ringed space and $A\subset X$ is any subset with the subspace topology then $i^{-1}{\cal O}_X$ will be a sheaf of rings on $A$ where $i:A\rightarrow X$ is the inclusion map. (Recall that the inverse image $i^{-1}{\cal O}_X$ is the sheafification of the presheaf $U \mapsto \lim_{V\supset i(U)} {\cal O}_X(V)$ for $U\subseteq A$ open, where the inductive limit is over all open subsets $V$ of $X$ containing $U$.)

Is the reason why we don't do this (and instead start talking about closed subschemes, etc. etc.) just that $(A,i^{-1}{\cal O}_X)$ need not be a scheme even when $X$ is?

Put differently: given any closed subset of a scheme there will be many ways to make it a closed subscheme. What is the relation between the locally ringed spaces on a closed subset making it a closed subscheme and the locally ringed space I have described above which we obtain by pulling back the structure sheaf via the inclusion map.

Answer 1

It might help to consider the extreme case when $x$ is a closed point of $X$, and $i$ is the inclusion $\{x\} \hookrightarrow X$. The pullback $i^{-1}\mathcal O_X$ is then the stalk of $\mathcal O_X$ at $x$, i.e. the local ring $A_{\mathfrak m}$, if Spec $A$ is an affine n.h. of $x$ in $X$, and $\mathfrak m$ is the maximal ideal in $A$ corresponding to the closed point $x$.

Now a single point, with a local ring $A_{\mathfrak m}$ as structure sheaf, is not a scheme (unless $A_{\mathfrak m}$ happens to be zero-dimensional).

Moreover, the restriction map from sections of $\mathcal O_X$ over $X$ to section of $i^{-1}\mathcal O_X$ over $x$ is not evaluation of functions at $x$ (which corresponds to reducing elements of $A$ modulo $\mathfrak m$), but is rather just passage to the germs of functions at $x$.

The idea in scheme theory is that sections of $\mathcal O_X$ should be functions, and restriction to a closed subscheme should be restriction of functions. In particular, restriction to a closed point should be evaluation of the function (if you like, the constant term of the Taylor series of the function), not passage to the germ (which is like remembering the whole Taylor series).

If you bear this intuition in mind, and think about the case of a closed point, you will soon convince yourself that the general notion of closed subscheme is the correct one: If we restrict functions to the locus cut out by an ideal sheaf $\mathcal I$, or (in the affine setting) by an ideal $I$ in $A$, then two sections will give the same function on this locus if they coincide mod $\mathcal I$ (or mod $I$ in the affine setting), and so it is natural to define the structure sheaf to then be $\mathcal O_X/\mathcal I$ (or to take its global sections to be $A/I$ in the affine settin), rather than $i^{-1}\mathcal O_X$.

Answer 2

If $J \subseteq \mathcal{O}_X$ is an ideal, the corresponding zero set is $V(J) = supp \mathcal{O}_X / J$. It is a closed subset of $X$, let $i$ be the inclusion. Then the associated closed sub-locally ringed space of $X$ is defined to be $(V(J),i^{-1} (\mathcal{O}_X/J))$. It has the desired universal property, maps to $V(J)$ are just maps to $X$ such that the sheaf map vanishes on $J$. If $X$ is a scheme and $J$ is quasi-coherent, it turns out that $V(J)$ is a scheme.

If we take $i^{-1} \mathcal{O}_X$ instead of $i^{-1}(\mathcal{O}_X/J)=i^{-1} \mathcal{O}_X / i^{-1} J$, the universal property does not hold anymore and if $X$ is a scheme and $J$ quasi-coherent, in general, this is not a scheme anymore. The reason is simply that here points and functions are not really connected with each other.

If you like to think about closed subsets, remark that every closed subset $A$ arises as $V(J)$. Simply take $J(U) = \{f \in \mathcal{O}_X(U) : f_x \in \mathfrak{m}_x ~~ \forall x \in U \cap A\}$.

For example if $X$ is a scheme and $x$ is closed point of $X$, take $A = \{x\}$. Then $(A,i^{-1} \mathcal{O}_X)$ is the locally ringed space on $A$ with sections $\mathcal{O}_{X,x}$, which is only a scheme if $\mathcal{O}_{X,x}$ has only one prime ideal.