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Denis Nardin
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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.

If you want an example that's not even locally Noetherian you replace $\mathbb{A}^1$ with the nilpotent thickening obtained by adding an infinite amount of square-zero elements.

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.

If you want an example that's not even locally Noetherian you replace $\mathbb{A}^1$ with the nilpotent thickening obtained by adding an infinite amount of square-zero elements.

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.

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Denis Nardin
  • 16.5k
  • 2
  • 69
  • 103

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.

If you want an example that's not even locally Noetherian you replace $\mathbb{A}^1$ with the nilpotent thickening obtained by adding an infinite amount of square-zero elements.