2
$\begingroup$

Let $A$ be a commutative ring. If $A$ is an excellent ring, is the reduced ring $A/\sqrt{(0)}$ also an excellent ring?

Improve this question
$\endgroup$
2
  • 7
    $\begingroup$ I clicked on this question from the SE main page thinking it was an April fool’s joke due to the “excellent” property. I was deeply disappointed. $\endgroup$
    – mhdadk
    Commented Apr 1, 2023 at 19:02
  • $\begingroup$ It surprises me that this question became a Hot Network Question. $\endgroup$
    – Timothy Chow
    Commented Apr 2, 2023 at 20:10

2 Answers 2

Reset to default
10
$\begingroup$

On page 260 of Matsumura, "Commutative Ring Theory" (CUP, 1986), he states that "One can prove that the classes of rings satisfying (1), (2) and (3) are closed under localisation, finitely generated extensions and passing to quotients." Excellent rings are defined as those satisfying all three conditions. I think this answers your question.

Improve this answer
$\endgroup$
1
  • 1
    $\begingroup$ He refers the reader to EGA IV (3). $\endgroup$
    – Dave Benson
    Commented Apr 1, 2023 at 10:30
3
$\begingroup$

Finite type extensions of excellent rings are excellent (in particular quotients). A more accessible (modern) reference is [Tag 07QW] in the Stacks project.

Improve this answer
$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .