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.


Edit:


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.



Reducedness of a ring with prime nilradical
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 $(S\_1)$.


