Let $R$ be a euclidean domain. For ideals $I, J\subseteq R$, prove that if $IJ = I ∩ J$, then $I + J = R$ Let $R$ be a euclidean domain. For ideals $I, J\subseteq R$, define $IJ$ to be the set 
$$
\{a_1b_1+\dots+a_nb_n:n∈N; a_i∈I; b_j∈J\}.
$$
Prove that if $IJ = I ∩ J$, then $I + J = R$.
Attempt:  
$d(I) \le d(IJ) = d(I ∩ J)$
$d(J) \le d(IJ) = d(I ∩ J)$
I tried to derive an equal relation between $I + J$ and $R$ but I got stuck here.
 A: Theorem: If $R$ is a Euclidean Domain and $I, J$ ideals in $R$, then $IJ = I \cap J$ means that $I + J = (1)$.
Proof: Since $R$ is a Euclidean Domain and $I, J$ are ideals in $R$, then $I = (x)$ and $J = (y)$ for some $x, y \in R$ (since ED $\implies$ PID).
Note that
\begin{align*}
I + J &= \big( \gcd(x, y) \big)\\
IJ &= (xy) \\
I \cap J &= \big(\operatorname{lcm}(x, y) \big)
\end{align*}
Being a Euclidean Domain guarantees us existence of $\gcd$ and $\operatorname{lcm}$, which we'll denote by $g$, $l$ respectively from now on.
Suppose $IJ = I \cap J$, then $(xy) = (l)$ and thus $l = xy \cdot w$ for some $w \in R$.
But by property of $l, g$, we have $lg = xy$ (this is one of the big properties of how the $\gcd$ and $\operatorname{lcm}$ are connected) which means $lg = xy = xywg$.
Since $R$ is a Euclidean Domain, it's an Integral Domain, so we can cancel out and get $wg = 1$, hence $g$ is a unit. Thus $I + J = (g) = (1)$.

The converse also holds, but we can loosen the hypothesis a bit from Euclidean Domain to just Commutative Ring with identity:
Theorem: If $R$ is an Commutative Ring with identity and $I, J$ ideals in $R$, then $I + J = (1)$ means $IJ = I \cap J$.
Proof: Observe that $IJ \subseteq I \cap J$ always, so it suffices to show that $I \cap J \subseteq IJ$ with the given hypothesis.
Since $I + J = (1)$, there exist $i \in I, j \in J$ such that $i + j = 1$. Let $x \in I \cap J$, then $$x = x \cdot 1 = x \cdot (i + j) = ix + xj \in IJ = \bigg\{ \sum_i x_i, y_i : x_i \in I, y_i \in J\bigg\}$$
and we can conclude that $IJ = I \cap J$.
