= Solution
An <integral domain> $A$ is an <integrally closed domain> if every element of its <fraction field> which is an <integral element> over $A$ belongs to $A$. Explicitly, if $z\in\operatorname{Frac}A$ satisfies
$$
z^d+a_{d-1}z^{d-1}+\cdots+a_0=0,\qquad a_i\in A,
$$
then $z\in A$. Thus $\boxed{A\text{ is integrally closed exactly when its integral closure in its fraction field is }A}$.
Suppose $A$ is a <unique factorization domain> and $z=a/b$ is integral, with $a,b\in A$, $b\ne0$, and with no common <irreducible element> as a factor. Multiplying its monic equation by $b^d$ gives
$$
a^d+a_{d-1}a^{d-1}b+\cdots+a_0b^d=0.
$$
Hence $b$ divides $a^d$. Every <irreducible element> dividing $b$ is a <prime element>, so it would divide $a$, contrary to the choice of the fraction. Therefore $b$ is a <unit>, and $z\in A$. We have proved
$$
\boxed{\text{Every unique factorization domain is integrally closed}.}
$$
The <going-down theorem> is the following. Let $A\subseteq B$ be an <integral extension> of <integral domains> with $A$ an <integrally closed domain>. If
$$
\mathfrak p_1\subseteq\mathfrak p_2\quad\text{in }A,\qquad
\mathfrak q_2\cap A=\mathfrak p_2\quad\text{for a prime ideal }\mathfrak q_2\subset B,
$$
then
$$
\boxed{\text{there is a prime ideal }\mathfrak q_1\subseteq\mathfrak q_2\text{ with }\mathfrak q_1\cap A=\mathfrak p_1.}
$$
We first prove this for a <module-finite ring extension> and then remove that extra hypothesis. We use the <Lying-over theorem>, the <going-up theorem>, and the <incomparability theorem for integral extensions>: primes exist over every base prime, prime chains can be extended upwards, and distinct primes over the same base prime cannot contain one another. The <going-up theorem> follows by applying the <Lying-over theorem> to the quotient by the lower prime. These results concern arbitrary <integral extensions>; none assumes the <going-down theorem>.
For the finite case, let $K=\operatorname{Frac}A$ and $L=\operatorname{Frac}B$. Since $B$ is module-finite, $L/K$ is a <finite field extension>. Take a finite <normal field extension> $E/K$ containing $L$, obtained as a <splitting field> of the minimal polynomials of finitely many field generators. The extension $E/K$ need not be a <separable field extension>. Let $C$ be the <integral closure> of $A$ in $E$. Then $B\subseteq C$, and $C$ is integral over both $A$ and $B$.
We need one auxiliary fact: the <finite group> $G=\operatorname{Aut}_K(E)$ acts transitively on the <prime ideals> of $C$ over any fixed <prime ideal> $\mathfrak p$ of $A$. First observe that an element $w\in E$ fixed by all of $G$ is a <purely inseparable algebraic element> over $K$. Indeed, all $K$-embeddings of $K(w)$ into an <algebraic closure> extend to $K$-<automorphisms> of the <normal field extension> $E$. Hence the <minimal polynomial of an algebraic element> $w$ has only one distinct root. In <characteristic zero> this says $w\in K$; in characteristic $p>0$ it says $w^{p^e}\in K$ for some $e\ge0$.
Now let $Q,Q'$ be primes of $C$ over $\mathfrak p$, and suppose $Q$ is different from every $\sigma(Q')$, $\sigma\in G$. By the <incomparability theorem for integral extensions>, $Q$ is contained in none of these conjugate primes. The <prime avoidance> lemma supplies
$$
x\in Q\setminus\bigcup_{\sigma\in G}\sigma(Q').
$$
The product
$$
h=\prod_{\sigma\in G}\sigma(x)
$$
is fixed by $G$, is integral over $A$, and lies in $Q$ because one factor is $x$. Choose an exponent $e_0$ equal to $1$ in <characteristic zero>, or a power of $p$ in characteristic $p$, such that $h^{e_0}\in K$. Since $A$ is an <integrally closed domain>, $h^{e_0}\in A$. It also lies in $Q\cap A=\mathfrak p$, hence in $Q'$. Because $Q'$ is a <prime ideal>, some factor $\sigma(x)$ lies in $Q'$. This means $x\in\sigma^{-1}(Q')$, contradicting our choice. Thus $Q=\sigma(Q')$ for some $\sigma$, proving transitivity. The power $e_0$ accounts for the possible <purely inseparable field extension> and makes the argument valid in every characteristic.
Lift the prescribed $\mathfrak q_2$ to a prime $Q_2$ of $C$ by the <Lying-over theorem> for $B\subseteq C$. Separately, choose a prime $Q'_1$ over $\mathfrak p_1$ and use the <going-up theorem> to find $Q'_2\supseteq Q'_1$ over $\mathfrak p_2$. Transitivity gives $\sigma\in G$ with $\sigma(Q'_2)=Q_2$. Then
$$
Q_1=\sigma(Q'_1)\subseteq Q_2,
\qquad Q_1\cap A=\mathfrak p_1.
$$
Contracting to $B$ gives $\mathfrak q_1=Q_1\cap B\subseteq\mathfrak q_2$ with the required contraction. This proves the <going-down theorem> for the module-finite case.
For a general <integral extension> $A\subseteq B$, consider $B_{\mathfrak q_2}$ and the <ideal>
$$
J=\mathfrak p_1B_{\mathfrak q_2}.
$$
We claim that $J$ is disjoint from $A\setminus\mathfrak p_1$. Otherwise, clearing the finitely many denominators in an expression for an element $a\in J\cap(A\setminus\mathfrak p_1)$ gives
$$
s a=\sum_{i=1}^{h}a_i b_i,\qquad
s\in B\setminus\mathfrak q_2,\quad a_i\in\mathfrak p_1,\quad b_i\in B.
$$
The <subalgebra> $B_0=A[s,b_1,\ldots,b_h]$ is a <module-finite ring extension> of $A$, since all its generators are <integral elements>. Apply the finite case to $\mathfrak q_2\cap B_0$: there is a prime $Q_1\subseteq\mathfrak q_2\cap B_0$ contracting to $\mathfrak p_1$. The displayed identity puts $sa$ in $Q_1$, but $a\notin Q_1$. Thus $s\in Q_1\subseteq\mathfrak q_2$, a contradiction.
Localize $B_{\mathfrak q_2}$ further at the <multiplicative subset> $A\setminus\mathfrak p_1$. The extension of $J$ is a proper <ideal>, by the disjointness just proved. Choose a <maximal ideal> containing it and contract back to a <prime ideal> of $B_{\mathfrak q_2}$. This prime contains $J$ and avoids $A\setminus\mathfrak p_1$. Its contraction $\mathfrak q_1$ to $B$ lies inside $\mathfrak q_2$ by the <prime ideal correspondence for localization>, and its contraction to $A$ is exactly $\mathfrak p_1$. This completes the proof of the <going-down theorem> without any finite generation assumption on $B$.
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