For which values $n$ is the ring $\mathbb Z/ n\mathbb Z$ regular? We know that when $n$ is prime, this is regular. When $n$ is not square-free, it's not regular because it is not reduced. However, what if $n$ is square-free?
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$\begingroup$ I suppose, the definition of a regular ring you are using is this one? Then also for $n=0$ it is regular, although $n$ is not squarefree. $\endgroup$– Dietrich BurdeOct 17, 2021 at 18:34
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Suppose $n$ is squarefree and let $p_1 \cdots p_n$ be a prime factorization of $n$. The Chinese Remainder Theorem implies $\mathbb{Z}/n\mathbb{Z} \cong \prod_{i=1}^n \mathbb{Z}/p_i\mathbb{Z}$ as rings. Each $\mathbb{Z}/p_i\mathbb{Z}$ is a field and thus is regular. Products of regular rings are regular, and so $\mathbb{Z}/n\mathbb{Z}$ is regular.
answered Oct 17, 2021 at 19:12
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$\begingroup$ Thanks a lot. That's a clean proof. Do you by any chance know what the localization at each prime ideals look like? I suspect they are all fields because it's quite obvious that the unique maximal ideal after localization would be zero. $\endgroup$ Oct 17, 2021 at 19:14
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$\begingroup$ By a theorem of Serre, this implies that $\mathbb Z/ n\mathbb Z [x_1, \dots, x_d]$ has global dimension equaling $d$. Is that right? $\endgroup$ Oct 17, 2021 at 19:16
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$\begingroup$ @Singularity Yes that's correct on both accounts. Regularity for a ring an Artinian ring is equivalent to being a direct product of fields, which is equivalent to saying each localization at a prime (and thus maximal) ideal is a field. And yes, $\mathbb{Z}/n\mathbb{Z}[x_1,\dots,x_d]$ has global dimension $d$ as you say, assuming $n$ is squarefree. $\endgroup$ Oct 17, 2021 at 19:26
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