A functor $(Sch)^{op}\to(Sets)$ can not represented by a scheme?

Question

This is M. Olsson's book "Algebraic Spaces and Stacks". Exercise 1.D.(b):

1.D.(a) Let $$\mathbb{A}^n-\{0\}:(Sch)^{op}\to(Sets)$$ be a functor sending a scheme $Y$ to the set of $n$-tuples $(y_1,...,y_n)$ of sections of $\Gamma(Y,\mathscr{O}_Y)$ such that for all $y\in Y$, the image of $y_i$ in field $\kappa(y)$ are not all zero. Show that $\mathbb{A}^n-\{0\}$ is representable;

(b) Let $$\mathbb{A}^n-\{0\}/\mathbb{G}_m:(Sch)^{op}\to(Sets)$$ be a functor sending a scheme $Y$ to the quotient of the set $(\mathbb{A}^n-\{0\})(Y)$ by equivalence relation $$(y_1,...,y_n)\sim(y_1',...,y_n')$$ if there exists $u\in\Gamma(Y,\mathscr{O}_Y^*)$ s.t. $y_j=uy_j'$ for all $j$. Show that $\mathbb{A}^n-\{0\}/\mathbb{G}_m$ is NOT representable.

I have solved (a) but I don't know how to prove a functor is not representable. Like (b).

Thank you for your help!!

Answer 1

Here's an alternative to Cranium Clamp's (correct) answer; the moral here is that representable functors must be sheaves for the Zariski topology. (In fact this is true for most other topologies that people care about.)

Suppose your functor is representable by a scheme $Z$. As in Cranium Clamp's answer, observe that $Z$ comes with a natural monomorphism $\phi: Z \to \mathbb P^{n-1}_{\mathbb Z}$. Now suppose we are given a morphism $f: Y \to \mathbb P^{n-1}_{\mathbb Z}$ for some $Y$. By the description of the functor represented by $\mathbb P^{n-1}_{\mathbb Z}$, this corresponds to a tuple $(\mathcal L, s_1, \dots, s_n)$, where $\mathcal L$ is a line bundle on $Y$ and $s_1, \dots, s_n$ are sections of $\mathcal L$, not all vanishing at any point--modulo isomorphism of all this data.

The morphism $f$ factors through $Z$ (necessarily uniquely) if and only if the line bundle $\mathcal L$ is trivial. Of course, $\mathcal L$ may not in fact be trivial, but it is certainly locally trivial; that is, there exists an open cover $\{U_i\}$ of $Y$ such that $\mathcal L|_{U_i} \cong \mathcal O_{U_i}$ for each $i$. This translates to the statement that for all $i$, the restriction $f|_{U_i}$ (which corresponds to $(\mathcal L|_{U_i}, s_1|_{U_i}, \dots, s_n|_{U_i})$) factors through $\phi$. Call the factoring maps $g_i: U_i \to Z$.

We now claim that the $g_i$ glue to a morphism $g: Y \to Z$ which factors $f$ as $\phi \circ g$. In order to glue them, we must only show that they agree on pairwise intersections. But this follows from the fact that $\phi$ is a monomorphism: the maps $g_i|_{U_{ij}}$ and $g_j|_{U_{ij}}: U_{ij} \to Z$ both factor $f|_{U_{ij}}: U_{ij} \to \mathbb P^n_{\mathbb Z}$ through $Z$, so they must agree. Thus we get a map $g: Y \to Z$, and we have $f = \phi \circ g$ because this is true on each $U_i$ by construction.

At this point, we have proved that every morphism to $\mathbb P^{n-1}_{\mathbb Z}$ factors through the monomorphism $\phi$. But this means that $Z = \mathbb P^{n-1}_{\mathbb Z}$, which we know is false. Thus the functor is not representable.

Answer 2

First we recall a standard fact: $ \mathbb{P}^{n-1}_{\mathbb{Z}} $ represents the functor which sends a scheme $ Y $ to the data $ (\mathcal{L}, s_1, \cdots , s_n) $ where $ \mathcal{L} $ is a line bundle on $ Y $ and $ (s_1, \cdots, s_n) $ are sections generating $ \mathcal{L} $. The equivalence relation is given by $ (\mathcal{L}, s_1, \cdots , s_n) \sim (\mathcal{L}' , s_1', \cdots , s_n') $ iff there is an isomorphism $ \beta : \mathcal{L} \rightarrow \mathcal{L}' $ such that $ \beta(s_i) = s_i' $ for all $ i = 1, \cdots, n $.

Suppose now that the functor in question is represented by a scheme $ Z $. In view of the above remark we see that there is an obvious morphism of functors $ \eta : \mathbb{A}^n - \{ 0 \}/\mathbb{G}_m \rightarrow \mathbb{P}^{n-1}_{\mathbb{Z}} $ such that $ \eta(Y) $ is an injection for every scheme $ Y $. By the Yoneda lemma, $ \eta $ corresponds to a morphism of schemes $ \phi : Z \rightarrow \mathbb{P}^{n-1}_{\mathbb{Z}} $which is a monomorphism simply by the above remark.

Furthermore, for any field $ k $ and for any local ring $ R $, the maps $ Z(k) \rightarrow \mathbb{P}^{n-1}_{\mathbb{Z}}(k) $ and $ Z(R) \rightarrow \mathbb{P}^{n-1}_{\mathbb{Z}}(R) $ are bijections. This is because every projective module over $ k $ or $ R $ is free. (Line bundles correspond to projective modules)

The previous remark about fields and local rings, applied to a valuation ring (and its field of fractions) immediately shows that the valuative criteria for properness holds for $ \phi $.

So $ \phi $ is a proper monomorphism, hence a closed immersion. (See the stacks project tag 04XV for a reference to this fact.) Noting the fact that $ \phi $ is obviously surjective (being a bijection on field valued points) and that $ \mathbb{P}^{n-1}_{\mathbb{Z}} $ is reduced, we conclude that $ \phi $ is an isomorphism. (Closed immersion + surjection onto a reduced target implies isomorphism, easy to check locally)

In summary, we get $ Z = \mathbb{P}^{n-1}_{\mathbb{Z}} $ which is obviously wrong as there are many schemes with non-trivial line bundles having a generating set of size $ n $. So the functor in question is not representable by a scheme.