# Regularity of $\mathbb Z/ n \mathbb Z$.

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?

• 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. Oct 17, 2021 at 18:34

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.
• 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? Oct 17, 2021 at 19:16
• @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. Oct 17, 2021 at 19:26