TITLE: Irreducible of finite Krull dimension implies quasi-compact?
QUESTION [6 upvotes]: Let $X$ be the underlying space of a scheme. 

If $X$ is irreducible of finite Krull dimension, is it necessarily
quasi-compact? 
Is it necessarily Noetherian?
What if we assume not
only that Krull dimension is finite but also that it is 1?

REPLY [10 votes]: The answer is no for all these questions. Take the line with infinite origins: the scheme obtained by gluing an infinite amount of copies of $\mathbb{A}^1$ along the open subsets $\mathbb{G}_m$. This has Krull dimension 1 (there are only closed points and the unique generic point) and it is irreducible (the only proper nonempty closed subsets are the closed points) but it is not quasi-compact (it contains an infinite discrete set), and so in particular not Noetherian.