Implications between $(\frak{a} \cap \frak{b})$ = $\frak{a} \frak{b}$ and $(\frak{a})$ + $(\frak{b})$= $(1)$ In a general commutative ring, $(\frak{a} \cap \frak{b})$ = $\frak{a} \frak{b}$ does not imply ($\frak{a}$) + ($\frak{b}$) = ($1$); whereas ($\frak{a}$) + ($\frak{b}$) = ($1$) does imply $(\frak{a} \cap \frak{b})$ = $\frak{a} \frak{b}$.
E.g., $(2) \cap (X)$ = $(2X)$ = $(2)(X)$, but $(2) + (X)$ = $(2,X) \ne (1)$.
But in $\mathcal{O}_K$, apparently the implication works in both directions.
I would appreciate help understanding what is happening in the general ring and in $\mathcal{O}_K$ that determines and distinguishes how these implications work or don't work.
Thanks.
 A: Let $\mathcal{O}_K$ be the ring of algebraic integers in a number field $K$. The nice property you mention is due to the fact that $\mathcal{O}_K$ is a Dedekind domain. In particular, every ideal in $\mathcal{O}_K$ factors uniquely as a product of prime ideals and every non-zero prime ideal is maximal.
Let $a_1,\dotsc,a_s,b_1,\dotsc,b_r\geq1$ such that $\frak{a}=\prod\frak{p}_1^{a_1}\dotsm\frak{p}_s^{a_s}$ and $\frak{b}=\prod\frak{q}_1^{b_1}\dotsm\frak{q}_r^{b_r}$ with $\frak{p}_i\neq \frak{p}_j$ and $\frak{q}_i\neq \frak{q}_j$ for $i\neq j$. Observe that $\frak{a}\frak{b}=\frak{a}\cap\frak{b}$ implies (check it!) that $\frak{p}_i\neq\frak{q}_j$ for $i\neq j$.
To prove that $\frak{a}+\frak{b}=(1)$ we will use the radical of an ideal $\newcommand{\radical}[1]{\operatorname{r}\left(#1\right)} \radical{\cdot}$ and some of its properties.
First of all, note that
$$
\radical{\frak{a}+\frak{b}}=
\radical{\radical{\frak{a}}+\radical{\frak{b}}}=
\radical{\bigcap \frak{p}_i + \bigcap \frak{q}_j}
$$
hence if we can prove that $\bigcap \frak{p}_i + \bigcap \frak{q}_j=(1)$ we will have that $\radical{\frak{a}+\frak{b}}=\radical{1}=(1)$, which implies $\frak{a}+\frak{b}=(1)$.
By contradiction, suppose $\bigcap \frak{p}_i + \bigcap \frak{q}_j\neq(1)$. Then $\frak{m}\supseteq \bigcap \frak{p}_i + \bigcap \frak{q}_j$ for some maximal ideal $\frak{m}$. In particular, this means that $\frak{m}\supseteq \bigcap \frak{p}_i$, therefore $\frak{m}\supseteq \frak{p}_i$ for some $1\leq i \leq s$. Similarly we have that $\frak{m}\supseteq \frak{q}_j$ for some $1\leq j \leq r$. Now recall that both $\frak{p}_i$ and $\frak{q}_j$ are prime, hence maximal. Thus $\frak{p}_i=\frak{m}=\frak{q}_j$, contradicting our hypothesis that $\frak{a}\frak{b}=\frak{a}\cap\frak{b}$.

Remark: Let $\frak{a},\frak{b}$ be ideals in a commutative ring $R$ with $1$. In general, "every prime ideal in $R$ is maximal" is necessary to have $\frak{a}\frak{b}=\frak{a}\cap\frak{b}$ imply $\frak{a}+\frak{b}=(1)$, while we don't actually need $R$ to be a domain.
Indeed, as you already observed, $\Bbb Z[X]$ is a Noetherian UFD (hence integrally closed) where $(2)(X)=(2)\cap(X)$, but $(2)+(X)=(2,X)\neq(1)$.
On the other hand, consider $R=k\times k$, where $k$ is a field. Then the only two non-trivial ideals in $R$ are $\frak{p}_1=\mathit{k}\times {0}$ and $\frak{p}_2={0}\times \mathit{k}$, which are clearly coprime. Hence for every $\frak{a},\frak{b}\subset \mathit{R}$ ideals we have that $\frak{a}\frak{b}=\frak{a}\cap\frak{b} \Rightarrow \frak{a}\neq\frak{b} \Rightarrow \frak{a}+\frak{b}=(1)$.

Claim. Let $\frak{a},\frak{b}$ be ideals in a commutative ring $R$ with $1$. Then "every prime ideal in $R$ is maximal" is a necessary, but not sufficient, condition for "$\frak{a}\frak{b}=\frak{a}\cap\frak{b} \Rightarrow \frak{a}+\frak{b}=(1)$".
For the first part, suppose that $\mathfrak{p}\subset R$ is a non-maximal prime contained in some maximal ideal $\mathfrak{m}$. Pick $a\in \frak m \setminus p$ and let $\mathfrak{a}=(a)$. Then for any $b\in\frak a\cap p$ we have $b=ac\in\mathfrak{p}$, hence $c\in\mathfrak{p}$ since $\mathfrak{p}$ is prime and $a\notin\mathfrak{p}$. Therefore $b\in\frak ap$.
A friend of mine helped me find the following example, which shows that the condition is not sufficient. Let $R'$ be the localisation of $\Bbb R[X,Y]/(X^3-Y^2)$ at $(x,y)$, where $x,y$ are the images of $X,Y$ in the quotient ring. Observe that it is a domain of Krull dimension $1$, since $X^3-Y^2$ is irreducible in $\Bbb R[X,Y]$ and $\Bbb R[X,Y]$ has dimension 2. Now consider the ideals $\mathfrak{a}=(x,y^2),\mathfrak{b}=(x^3,y)$ of $R'$ and define $R=R'/(\frak a\cap b)$. Finally, observe that in $R$ we have $\frak\bar{a}\neq\bar{b}$ since $\frak a\neq b$ in $R'$, $\frak \bar{a}\cap \bar{b}=\{0\}=\bar{a}\bar{b}$, and $\bar{\mathfrak{a}}+\bar{\mathfrak{b}}=(\bar{x},\bar{y})\neq R$.
Note that this is coherent with what we proved before, because $R$ is a Noetherian local domain of dimension 1, but it isn't a Dedekind domain since it isn't integrally closed. You can see this, for example, as $R'$ is the local ring of the curve $X^3-Y^2=0$ at the origin, where it isn't smooth. Alternatively, you could prove that the maximal ideal $(x,y)$ of $R'$ isn't principal.
