Is $R/bR$ reduced if $R$ is reduced and $\operatorname{Ann}(b)$ is a minimal prime of $R$? Let $R$ be a noetherian, reduced ring. Let $P = \operatorname{Ann}_R(b)$ with $b \in R$ be a minimal prime of $R$.

Does this imply that $R/bR$ is reduced?

Of course, the property in question is equivalent to $bR$ being radical.
Considering coordinate rings of affine curves over a field $k$, we have for instance $R = k[x,y]/(fgh)$ where $f,g,h \in k[x,y]$ are irreducible. Then $R$ is reduced. The minimal primes of $R$ are given by
$$P_1 = (f) = \operatorname{Ann}_R(gh),\quad P_2 = (g) = \operatorname{Ann}_R(fh),\quad P_3 = (h) = \operatorname{Ann}_R(fg).$$
Now for $P = P_1= (f)$ we have $b = gh$ and $R/bR \cong k[x,y]/f$ which is reduced since $f$ was irreducible.
Now I am curious whether this generalizes. For me it's like one of those general questions which goes like What can we tell in general about such $b \in R$ with $\operatorname{Ann}_R(b)$ a minimal prime of $R$? For now, I am concerned about the above question.
Thank you in advance!

If $R$ is a domain then for all $b \in R$, $b\ne 0$, $\operatorname{Ann}_R(b) = 0$ which is a minimal prime of $R$, but not every factor ring of a domain is reduced.
Specifically, let $R = \Bbb Z$, $b = 4$. Then $\operatorname{Ann}_R(b) = 0$, but $\Bbb Z/4\Bbb Z$ is not reduced.
