1
$\begingroup$

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!

$\endgroup$
1
  • $\begingroup$ You want to add "nonzero" to discuss the converse; otherwise, a trivial example has $I=(0)=J$. $\endgroup$ – Arturo Magidin Mar 30 '12 at 3:16
4
$\begingroup$

(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 $IJ=(px) = I\cap J$).

$\endgroup$
1
  • $\begingroup$ Thanks so much, Arturo! Now I know I'm not just wasting my time. $\endgroup$ – Cameron Buie Apr 12 '12 at 3:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.