# comaximality of ideals in a commutative ring with unit

Suppose we have a commutative ring $R$ with unit. I'm curious about what condition(s) on $R$ would be sufficient (without Axiom of Choice) to give a converse to the following familiar result:

(#) If $I,J$ are comaximal ideals of $R$ (i.e.: $I+J=R$), then $IJ=I\cap J$.

I discovered that the converse of (#) holds if $R$ is a PID, so that's sufficient, but I was wondering if that can be weakened at all, perhaps to Dedekind domain, UFD, or even integral domain. I suspect it doesn't hold in general.

In particular, if the converse of (#) holds when $R$ is a Dedekind domain (every non-unit ideal of $R$ can be uniquely factored as a product of prime ideals), can anyone give me a (sketch of a) proof?

Thanks!

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You want to add "nonzero" to discuss the converse; otherwise, a trivial example has $I=(0)=J$. – Arturo Magidin Mar 30 '12 at 3:16

(I'm assuming $I$ and $J$ nonzero; otherwise, $I=J=(0)$ is a counterexample, with $IJ=(0)=I\cap J$, but $I+J\neq R$ in any ring.)

Yes, the result holds for Dedekind domains because $I\supset J = I\text{ divides }J$.

Thus, $I\cap J$ is the largest ideal that is a multiple of both $I$ and $J$, which is the lcm of $I$ and $J$. And $I+J$ is the smallest ideal that divides both $I$ and $J$, that is, their gcd.

So for Dedekind domains, the biconditional statement is equivalent to "$\gcd(I,J)=1$ if and only if $\mathrm{lcm}(I,J)=IJ$", which is true. Thus, both implications hold in Dedekind domains.

It does not hold for arbitrary UFDs; take $R=\mathbb{R}[x,y]$, $I=(x)$, $J=(y)$. Then $IJ=(xy)$. Let $p(x,y)\in I\cap J$. Then we can write $p(x,y) = x(q_0(x) + q_1(x)y + \cdots + q_m(x)y^m)$ for some polynomials $q_i$; since it is also a multiple of $y$, $y$ must divides $q_0(x)$, hence $q_0(x) = 0$. Thus, $xy$ divides $p(x,y)$, so $p(x,y)\in (xy)$. That is, $IJ=I\cap J$. However, $I+J = (x,y)\neq R$.

(Or $I=(2)$, $J=(x)$ in $R=\mathbb{Z}[x]$; more generally, for any UFD $D$ that is not a field, let $p$ be a prime of $D$ and take $I=(p)$, $J=(x)$ in $D[x]$; then $I+J=(p,x)\neq D$, but $IJ=(px) = I\cap J$).

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Thanks so much, Arturo! Now I know I'm not just wasting my time. – Cameron Buie Apr 12 '12 at 3:22