The Stacks project

Lemma 97.27.6. In Situation 97.27.1 if there exists an algebraic space $X$ locally of finite type over $S$ and a functorial bijection $F(V) = \mathop{\mathrm{Mor}}\nolimits _ S(V, X)$ on the category of locally Noetherian schemes $V$ over $S$, then $X$ is a solution.

Proof. We have to construct a proper morphism $f : X' \to X$, a closed subset $T \subset |X|$, and an isomorphism $a : X_{/T} \to W$ with properties (1), (2), (3) listed just below Situation 97.27.1.

The discussion in this proof is a bit pedantic because we want to carefully match the underlying categories. In this paragraph we explain how the adventurous reader can proceed less timidly. Namely, the reader may extend our definition of the functor $F$ to all locally Noetherian algebraic spaces over $S$. Doing so the reader may then conclude that $F$ and $X$ agree as functors on the category of these algebraic spaces, i.e., $X$ represents $F$. Then one considers the universal object $(T, u', \hat x)$ in $F(X)$. Then the reader will find that for the triple $X'' \to X$, $\hat x'$, $x'$ witnessing the compatibility between $u'$ and $\hat x$ the morphism $x' : X'' \to X'$ is an isomorphism and this will produce $f : X' \to X$ by inverting $x'$. Finally, we already have $T \subset |X|$ and the reader may show that $\hat x$ is an isomorphism which can served as the last ingredient namely $a$.

Denote $h_ X(-) = \mathop{\mathrm{Mor}}\nolimits _ S(-, X)$ the functor of points of $X$ restricted to the category $(\textit{Noetherian}/S)_{\acute{e}tale}$ of Section 97.25. By Limits of Spaces, Remark 69.3.11 the algebraic spaces $X$ and $X'$ are limit preserving. Hence so are the restrictions $h_ X$ and $h_{X'}$. To construct $f$ it therefore suffices to construct a transformation $h_{X'} \to h_ X = F$, see Lemma 97.25.2. Thus let $V \to S$ be an object of $(\textit{Noetherian}/S)_{\acute{e}tale}$ and let $\tilde x : V \to X'$ be in $h_{X'}(V)$. Then we get an element $(Z, u', \hat x)$ in $F(V)$ as follows

  1. $Z \subset V$ is the inverse image of $T'$ by $\tilde x$,

  2. $u' : V \setminus Z \to U'$ is the restriction of $\tilde x$ to $V \setminus Z$,

  3. $\hat x : V_{/Z} \to W$ is the composition of $x_{/Z} : V_{/Z} \to X'_{/T'}$ with $g : X'_{/T'} \to W$.

This triple satisfies the compatibility condition: first we always obtain $V' \to V$ and $\hat x' : V'_{/Z'} \to X'_{/T'}$ for free (see discussion preceding Lemma 97.27.4). Then we just define $x' : V' \to X'$ to be the composition of $V' \to V$ and the morphism $\tilde x : V \to X'$. We omit the verification that this works.

If $\xi : V \to X$ is an ├ętale morphism where $V$ is a scheme, then we obtain $\xi = (Z, u', \hat x) \in F(V) = h_ X(V) = X(V)$. Of course, if $\varphi : V' \to V$ is a further ├ętale morphism of schemes, then $(Z, u', \hat x)$ pulled back to $F(V')$ corresponds to $\xi \circ \varphi $. The closed subset $T \subset |X|$ is just defined as the closed subset such that $\xi : V \to X$ for $\xi = (Z, u', \hat x)$ pulls $T$ back to $Z$

Consider Noetherian schemes $V$ over $S$ and a morphism $\xi : V \to X$ corresponding to $(Z, u', \hat x)$ as above. Then we see that $\xi (V)$ is set theoretically contained in $T$ if and only if $V = Z$ (as topological spaces). Hence we see that $X_{/T}$ agrees with $W$ as a functor. This produces the isomorphism $a : X_{/T} \to W$. (We've omitted a small detail here which is that for the locally Noetherian formal algebraic spaces $X_{/T}$ and $W$ it suffices to check one gets an isomorphism after evaluating on locally Noetherian schemes over $S$.)

We omit the proof of conditions (1), (2), and (3). $\square$


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