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

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. $\endgroup$– mhdadkApr 1 at 19:02

$\begingroup$ It surprises me that this question became a Hot Network Question. $\endgroup$– Timothy ChowApr 2 at 20:10
2 Answers
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.

1
Finite type extensions of excellent rings are excellent (in particular quotients). A more accessible (modern) reference is [Tag 07QW] in the Stacks project.