Why is the "smallest non-affine scheme" not affine?

Question

Exercise I-25 of Eisenbud and Harris' "Geometry of Schemes" defines the following scheme, which is then claimed to be the smallest non-affine scheme:

Let $X=\{p,q_1,q_2\}$ with the open subsets $\emptyset,\{p\},\{p,q_1\},\{p,q_2\},X$ and define a sheaf $\mathcal O$ of rings on $X$ as $\mathcal O(\{p\})=k(x)$, $\mathcal O(\{p,q_1\})=\mathcal O(\{p,q_2\})=\mathcal O(X)=k[x]_{(x)}$ with the obvious restriction maps.

Now I have been struggling for hours to prove that this is not an affine scheme. I can derive several properties of the ring whose spectrum would be isomorphic to $(X,\mathcal O)$ as a locally ringed space, but so far I haven't found any contradictions. How can I prove this scheme is not affine? Could somebody provide a hint to point me in the right direction?

Answer 1

Hint: an affine scheme $X$ is always isomorphic to Spec $\mathcal O (X)$

Answer 2

$X$ is an open subset of $X$ too. And by the sheaf criterion, we can compute it to be the pullback

$$\begin{matrix} \mathcal{O}(X) &\to& \mathcal{O}(\{p, q_1\}) \\\downarrow & & \downarrow \\ \mathcal{O}(\{p, q_2\}) &\to& \mathcal{O}(\{p\}) \end{matrix} $$

I'm going to assume "the obvious maps" mean that both should send $x$ to $x$; this, we compute that $\mathcal{O}(X) \cong k[x]_{(x)}$. (e.g. both maps are monic, so we just compute the intersection of the subrings)

Now, all that's left is to see if $X \cong \mathop{\mathrm{Spec}} \mathcal{O}(X)$.