Product of Coprime Ideals equals Intersection

Theorem
Let $A$ be a commutative ring with unity.

Let $\mathfrak a, \mathfrak b \subseteq A$ be coprime ideals.

Then their product equals their intersection:
 * $\mathfrak a \mathfrak b = \mathfrak a \cap \mathfrak b$

Proof
By Intersection of Ideals of Ring contains Product, $\mathfrak a \mathfrak b \subseteq \mathfrak a \cap \mathfrak b$.

It remains to show that $\mathfrak a \mathfrak b \supseteq \mathfrak a \cap \mathfrak b$.

Let $c \in \mathfrak a \cap \mathfrak b$.

Because $\mathfrak a$ and $\mathfrak b$ are coprime, there exist $x \in \mathfrak a$, $y \in \mathfrak b$ with $x+y = 1$.

Then $c = cx + cy$.

Because:
 * $c \in \mathfrak b$ and $x \in \mathfrak a$
 * $c \in \mathfrak a$ and $y \in \mathfrak b$

we have $c \in \mathfrak a \mathfrak b$.