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.