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:

$c = c x + c y$

Because:

$c \in \mathfrak b$ and $x \in \mathfrak a$
$c \in \mathfrak a$ and $y \in \mathfrak b$

we have:

$c \in \mathfrak a \mathfrak b$

$\blacksquare$

Converse theorem


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In a Dedekind domain, the conditions

$\mathfrak a + \mathfrak b = \ideal 1$

and

$\mathfrak a \cap \mathfrak b = \mathfrak a \mathfrak b$

are equivalent, both expressing the fact that $\mathfrak a$ and $\mathfrak b$ have no common ideal factors except $\ideal 1$.

In a general commutative ring with unity, these conditions are not equivalent.

For example, in the polynomial ring $\Z\sqbrk T$ we have

$\ideal 2 + \ideal T = \ideal {2, T} \ne \ideal 1$

but

$\ideal 2 \cap \ideal T = \ideal {2T} = \ideal 2 \ideal T$

Sources

  • 2005: Serge Lang: Undergraduate Algebra (3rd ed.): Chapter $\text {III}$, $\S 3$ Exercises: Problem $12 \ \text{(b)}$
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