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In commutative algebra, a regular ring is a commutative Noetherian ring, such that the localization at every prime ideal is a regular local ring: that is, every such localization has the property that the minimal number of generators of its maximal ideal is equal to its Krull dimension.

In view of this definition, is it true to say the ring $0$ is regular?

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  • $\begingroup$ Someone who is more of an expert than I am can surely give a better answer, but it would be strange to me to say that zero ring is regular. The spectrum of the zero ring is empty, and so it has no localizations at primes to consider. But the zero ring is not a local ring either (although it is semilocal), so it doesn't really make sense to consider regularity for the zero ring itself. In my mind, this makes your question one of convention, and I don't see any persuasive reason to call the zero ring regular. Again, this is just my nonexpert opinion, so a more authoritative answer would be good. $\endgroup$
    – Alex Wertheim
    Jul 12, 2021 at 9:42
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    $\begingroup$ Dear Alex. Thanks for your comment. But I think the zero is regular, exactly because its spectrum is empty. $\endgroup$
    – Yasin
    Jul 12, 2021 at 22:33
  • $\begingroup$ The arguments mentioned in the other comment are common misconceptions about the empty set. I strongly recommend to read mathoverflow.net/questions/45951/…. $\endgroup$
    – Martin Brandenburg
    Jul 12, 2021 at 22:59

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The zero ring is regular by vacuous truth. There is no prime ideal which witnesses non-regularity. Geometrically, the empty scheme is regular. More generally, any smooth scheme is regular, and the empty scheme is also smooth.

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