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 $12 \ \text{(b)}$