Let $A$ be a regular ring and $\mathfrak q$ be an ideal, such that $\sqrt{\mathfrak q}$ is prime. Further assume that $\mathfrak q$ is locally principal (i.e. $\mathfrak q$ is an irreducible divisor which is not necessarily reduced).

Now we assume that $(A/\mathfrak q)_{\sqrt{\mathfrak q}}$ is regular.

Under these hypotheses, we can deduce that $A/\mathfrak q$ is reduced (equivalently $\mathfrak q$ itself is prime) because it is a generically reduced (since $(A/\mathfrak q)_{\sqrt{\mathfrak q}}$ is a zero-dimensional regular local ring, hence a field) Cohen-Macaulay-Ring (since $A$ is regular and $\mathfrak q$ is locally generated by non-zerodivisors).

The geometric interpretation is: If an irreducible component of a divisor $D$ in a regular variety contains a regular point of $D$, then it is a reduced component.

My 2 questions:

1) Can anyone come up with a direct proof (for example not using words like Cohen-Macaulay), which is more elementary?

2) What if we drop the assumption that $\mathfrak q$ is locally principal? Then the question comes down to: Let $B$ be a ring (which is a quotient of a regular local Ring) with prime nilradical such that the localization at the nilradical is a field. Does this imply that $B$ is reduced, hence a domain? If not, can someone come up with a counterexample?

All rings can assumed to be Noetherian.


Ok, $R=k[x,y]/(x^2,xy)$ should be a counterexample to question 2). We have $nil(R)=(x)/(x^2,xy)$ and $R/nil(R) = k[y]$, so $nil(R)$ is prime. Since we have $xy=0$ in $R$, we deduce $x=0$ in $R_{nil(R)}$, thus the maximal ideal of $R_{nil(R)}$, which is generated by $x$, is zero.

Question 1) is now less interesting because there is no hope to come up with a proof that shows that the "locally principal"-assumption is not necessary at all.


1 Answer 1


A Noetherian ring is reduced if and only if it is $(R_0)$ and $(S_1)$. (See, for example, http://stacks.math.columbia.edu/tag/031R ) This condition can be used to provide a simple proof that your ring $A/\mathfrak{p}$ is reduced.

A ring $R$ is $(R_k)$ iff $A_P$ is a regular local ring for all primes $P$ of height at most $k$. So $R$ is $(R_0)$ iff $R_P$ is regular for all minimal primes $P$. This is a simple condition and clearly holds for your ring $R = A/{\mathfrak q}$ since it has a unique minimal prime and its localization at that prime is a field. The assumption that $\mathfrak{q}$ is locally principal is not required here.

A ring $R$ is $(S_k)$ iff $PR_P$ contains a regular sequence in $R_P$ of length at least $\min(k,\mbox{ht} P)$ for every prime $P$ of $R$. So $R$ is $(S_1)$ iff $PR_P$ contains a nonzerodivisor in $R_P$ for every non-minimal prime $P$ of $R$. This condition also clearly holds for your ring $R = A/\mathfrak{q}$. In general the condition required beyond $(R_0)$ is precisely the condition $(S_1)$.

  • $\begingroup$ Thank You for your answer. Actually I am aware of this criteria for a Noetherian ring to be reduced and it was precisely what I had in mind when giving the proof. $(R_0)$ is just another term for "generically reduced" and $(S_1)$ is kind a weaker version of Cohen-Macaulay. Actually in my counterexample $A/\mathfrak q$ was one-dimensional, hence Cohen-Macaulay and $(S_1)$ coincide in this ring. Maybe this is the remaining the question now: Is there another sufficient condition in terms of the ideal $\mathfrak q$, that suffices to show that $A/\mathfrak q$ is reduced? $\endgroup$
    – MooS
    Jan 22, 2015 at 8:09
  • $\begingroup$ If you're looking for a condition on $\mathfrak{q}$, then one such necessary and sufficient condition is that $\mathfrak{q}$ be primary. (Indeed, $\mathfrak{q}$ is primary if and only if the map $A/\mathfrak{q} \longrightarrow (A/\mathfrak{q})_{\mathfrak{p}}$ is injective, which since the image is a field implies that $\mathfrak{q}$ is prime.) Under the assumption that $A/\mathfrak{q}$ is $(R_0)$, the ideal $\mathfrak{q}$ is prime iff it is primary iff it is radical. I'm not sure what else can be said. $\endgroup$ Jan 22, 2015 at 10:32
  • $\begingroup$ This clarifies everything. Thanks a lot! $\endgroup$
    – MooS
    Jan 22, 2015 at 13:16

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