It is not true that for an integrally closed domain $A$ and any prime ideal $\mathfrak p$, the quotient $A/\mathfrak p$ is integrally closed as well. For example $\mathbb Z[x]$ is integrally closed, as it is a UFD, but $\mathbb Z[x]/(x^2+5) \cong \mathbb Z[\sqrt{-5}]$ is not. So there must be a mistake in my 'proof' of the statement:
Write $[c]$ for the class in $A/\mathfrak p$ of some element $c \in A$.
Let $b := \displaystyle \frac{[c]}{[d]} \in \text{Frac}(A/\mathfrak p)$ satisfying the monic equation
$$
b^n + [a_{n-1}]b^{n-1} + \cdots + [a_0] = 0,
$$
for $a_i \in A$. Then
$$
\left(\frac{c}{d}\right)^n + a_{n-1}\left(\frac{c}{d}\right)^{n-1} + \cdots + a_0 \in \mathfrak p,
$$
and
$$
\left(\frac{c}{d}\right)^n + a_{n-1}\left(\frac{c}{d}\right)^{n-1} + \cdots + a_0 – a = 0,
$$
for some $a \in \mathfrak p$. As $A$ is integrally closed, we have $\displaystyle \frac{c}{d} \in A$ and so $b \in A/\mathfrak p$ as $\displaystyle \left[\underbrace{\frac{c}{d}}_{\in A}\right] = \frac{[c]}{[d]}$.
Does anyone see what I am apparently not? Thank you very much!
Best Answer
As you stated yourself it does not hold in general that quotients of integrally closed domains are integrally closed. I did not check very carefully, but i think the problem occurs with your first equation. Since $b \in \text{Frac}(A/\mathfrak{p})$, the equation $b^n + \dots = 0$ makes sense in the field of fractions and not in $A/\mathfrak{p}$. Thus you cannot lift to the situation in $A$.