I've been reading Demazure and Gabriel to learn the functor of points approach to algebraic geometry. They define a functor $X$ from commutative rings to sets as local if the standard sequence $$X(R) \to \prod_{i \in I}X(R_{f_i}) \overset{\to}{\to} \prod_{\substack{i, j \in I \\ i \neq j}}X(R_{f_if_j})$$ is exact for each commutative ring $R$ and sequence $f_i$ generating the unit ideal. My understanding is that this is the same as the sequence $$\mathrm{Mor}(Y, X) \to \prod_{i \in I}\mathrm{Mor}(Y_i, X) \overset{\to}{\to} \prod_{\substack{i, j \in I \\ i \neq j}}\mathrm{Mor}(Y_i \cap Y_i, X)$$ being exact, where $Y$ is any functor with an open covering $Y_i$. I'm happy using either definition.
A scheme is a local functor that has a covering by affine open subfunctors. As an example that the above condition is not trivial they suggest looking at the subfunctor $X \subseteq \mathrm{Gr}_{n, r}$ of the Grassmannian. Where $\mathrm{Gr}_{n, r}(R)$ is the set of rank $n$ summands of $R^{n + r}$ they define $X(R)$ to be the set of free rank $n$ summands of $R^{n + r}$. It's easy to see that the affine opens that cover $\mathrm{Gr}_{n, r}$ are actually subfunctors of $X$ and cover $X$, but Demazure and Gabriel claim that $X$ is not local whereas $\mathrm{Gr}_{n, r}$ is.
I can't figure out why $X$ is not local. Can anyone explain this to me, or alternatively give an example of a functor $X$ that is covered by affine open subfunctors but which is not local?
I think I understand things now, since a few people favorited the question I'll post what I've figured out. I'll be using Demazure and Gabriels definition of local. First a little lemma:
Lemma. Let $X$ be a subfunctor of $Y$ and let $Y$ be local. Then $X$ is local if and only if for every $R$ and every $f_1, \ldots, f_n \in R$ generating the unit ideal, if $y \in Y(R)$ maps to $X(R_{f_i})$ for each $i$ then $y \in X(R)$.
Proof. The sequence that we have to check is exact for $X$ is just a restriction of the sequence for $Y$, so the first map $X(R) \to \prod_iX(R_{f_i})$ is automatically injective. An element of $\prod_iX(R_{f_i})$ on which the two following maps agree lifts to $Y(R)$ because $Y$ is local and that lift is unique so you get exactness if and only if that lift to $Y(R)$ actually landed in $X(R)$. $\square$
Now we take $R = \mathbb Z[\sqrt{-5}]$. The accepted answer of this question explains why the ideal $\mathfrak a = (3, 1 + \sqrt{-5})$ is a non-free summand of $R^2$. Then take $$f_1 = 1 - \sqrt{-5}$$ $$f_2 = 1 + \sqrt{-5}$$ $$f_3 = 3$$ and note that these generate the unit ideal. Localizing $\mathfrak a$ at $f_1$ we get $1 + \sqrt{-5} = \frac{6}{1 - \sqrt{-5}}$ so the localization $R_{f_1}\mathfrak a = (3)$ is principal, hence free of rank $1$. Since $\mathfrak a$ contains $f_2$ and $f_3$ we get that localizing at either of these elements gives the unit ideal, which is free of rank $1$. Hence $\mathfrak a \in \mathrm{Gr}_{1, 1}(R)$ maps to $X(R_{f_i})$ for each $i$, but $\mathfrak a \notin X(R)$. This proves that the subfunctor $X$ is not local.