Can a non-affine scheme admit an fpqc covering by an affine scheme?

Question

Suppose a scheme $X$ admits an fpqc covering $\operatorname{Spec} R \to X$. Is $X$ affine? I'd be particularly interested in 'nice' counterexamples (say when the covering is in fact fppf, or when $X$ is Noetherian).

Answer 1

Thanks to Ariyan Javanpeykar for clearing up my confusion. The answer to the main question is: a non-affine scheme can indeed be covered by a single affine scheme, and in fact, the standard covering of $\operatorname{Spec} k[x] \sqcup \operatorname{Spec} k[y]$ onto $\mathbb{P}^1$ will do.

When I thought about this question myself, I initially discarded these types of examples. I would not expect this morphism to be flat, because the sizes of the pre-images of this covering can 'jump' between 1 and 2, which contradicts the intuition that flat morphisms should have continuously varying fibres.

To clarify where this intuition goes wrong, let me simply remark that an open immersion is flat. Here, too, there is a 'jump' in the size of the fibres, but this 'jump' occurs, for the lack of better terms, on the boundary of an open set. And indeed a similar thing happens in the case of the aforementioned covering, as one can regard the images of $\operatorname{Spec} k[x]$ and of $\operatorname{Spec} k[y]$ as being 'open' in $\mathbb{P}^1$.

An additional layer of confusion for me was that the covering of $\mathbb{P}^1$ seemed, to me, to be a finite morphism, but I was aware of the theorem saying that a scheme admitting a finite surjection from an affine scheme, is in fact affine (see Stacks Project, Tag 01ZT, or the exercise in Hartshorne's book as mentioned by Alex Youcis). My reasoning was that finite morphisms roughly correspond to having finite fibres, and this is clearly the case for our covering.

The intuition is flawed because, as Ariyan mentions, a morphism is finite if and only if it is quasi-finite (meaning of finite type and having finite fibres) and proper. It is the properness that fails for our covering. Using the intuition that proper schemes should be a 'compactness' criterion, one can predict why this indeed fails: viewing $\operatorname{Spec} k[x]$ and $\operatorname{Spec} k[y]$ as infinite lines, they clearly aren't compact, yet we're trying to map them onto a proper (compact) space $\mathbb{P}^1$. To be more mathematically precise, the morphism $\operatorname{Spec} k[x] \sqcup \operatorname{Spec} k[y] \to \mathbb{P}^1$ fails to be universally closed, as can be seen by taking the fibre product along the inclusion $D(x) \to \mathbb{P}^1$.