Condition of coprime ideal

Question

Let $A$ is a commutative ring with unit 1. Let $\alpha,\beta$ be ideals of $A$.

I have proved that $\alpha \cap \beta=\alpha \beta$ if $\alpha+\beta=(1)$,

I wonder if the converse is true or not?

Answer 1

Concerning your question, it may be interesting to see counterexamples not involving the zero ideal. So here are some:

Let $A = \mathbb{Z}[x]$, $\alpha = (2)$ and $\beta = (x)$. Then one easily checks that $\alpha \beta = \alpha \cap \beta$, but $ \alpha + \beta = (2,x)$ is a maximal ideal and therefore proper.

Analoguous examples can be constructed in any factorial domain $A$ that has two distinct prime ideals of height greater than $1$ containing two distinct height $1$ primes.

A phenomenon from which we also can construct examples is the following:

Let $A$ be a commutative ring with unity. There is a general notion of ideals $\alpha \subseteq A$ with the property that for all ideals $\beta \subseteq A$, we have $\alpha \cap \beta = \alpha \beta$. Such ideals are called pure. We have the following characterization of pure ideals, that you can find in any book on multiplicative ideal theory:

Theorem: The following assertions are equivalent for an ideal $\alpha \subseteq A$:

1. $\alpha$ is pure.
2. For all $x \in \alpha$ we have that $x\alpha = xA$.
3. $\alpha$ is an idempotent ideal and locally principal.
4. For every maximal ideal $M \subseteq A$ it holds that $\alpha_M = (0)_M$ or $\alpha_M = A_M$ in the localization $A_M$.

Pure ideals $\alpha$ are very special in the sense that they satisfy the property you were asking, no matter which ideal $\beta$ you take. We can easily construct examples of non-zero pure ideals, using properties 3. and 4. from above:

Examples:

1. Let $I = (x^2 - x)$ the ideal of $\mathbb{Z}[x]$ generated by $x^2 - x$ and let $A = \mathbb{Z}[x]/I$. The ideal $\alpha = (x + I)$ in $A$ is generated by an the idempotent element $x + I$ and is therefore idempotent. As a principal ideal it is locally principal, hence it is pure by property 3. in the theorem above.

2. Let $A$ be an infinite product of fields and $M \subseteq A$ a maximal ideal. Then the canonical map $A \to A_M$ with $a \mapsto \frac{a}{1}$ has kernel $M$ (you can find a proof of this here). Therefore $M_M = (0)_M$. For every other maximal ideal $N \subseteq A$, we clearly have $M_N = A_N$. By property 4. of the theorem, we have that $M$ is pure.

Answer 2

Take $\alpha=\{0\}$ and $\beta$ any nonzero proper ideal in $A$. Then $\alpha\cap\beta=\alpha\beta$ but $\alpha+\beta\neq(1)$.