# Can a product of non-principal ideals be principal, in a local ring?

In a commutative ring $R$, we can define the ideal class monoid $\mathrm{ICl}(R)$ of $R$ as follows. The elements of $\mathrm{ICl}(R)$ are equivalence classes of invertible ideals (where an ideal $I$ is invertible if there exists an ideal $J$ with $IJ$ being principal). Two ideals $I$ and $J$ are equivalent if there exist nonzero $a,b \in R$ with $aI=bJ$. The resulting quotient is a (commutative) monoid under ideal product, with identity the equivalence class of any principal ideal.[1]

## Q: Suppose we have two ideals $I$, $J$ in a local ring $R$, with their product $IJ = (a)$ being principal, and $a$ is not a zerodivisor.[2] Then are $I$ and $J$ themselves principal?

In particular: Is the ideal class monoid of a local integral domain always trivial?

This is probably answerable with only elementary commutative algebra, but I couldn't see it. A few related examples that motivated the question are:

• The class group $\mathrm{Cl}(R)$ of a Dedekind domain $R$. In a Dedekind domain, all ideals are invertible so $\mathrm{ICl}(R)$ recovers the usual definition of $\mathrm{Cl}(R)$. This group is often nontrivial, but for local Dedekind domains (i.e. discrete valuation rings), it is always trivial: it is easy to see that all ideals in a discrete valuation ring are principal.
• The (Weil) divisor class group of a scheme. For simplicity, consider the case of $\mathrm{Spec}(R)$, with $R$ an integrally closed Noetherian domain. We can then define the Weil class group $\mathrm{Cl}(R)$ as the group of Weil divisors modulo principal divisors, i.e. $\mathbb{Z}$-linear combinations of height $1$ prime ideals of $R$ modulo principal divisors. This also can be nontrivial, even when $R$ is local. The typical example is $R = \mathbb{C}[[x,y,z]]/(xy-z^2)$, where the divisor given by $I = (x,z)$ is non-principal, yet $2 I$ is principal, being the divisor of the rational function $x$.

I would appreciate any geometrical intuition towards $\mathrm{ICl}(R)$, for any $R$. Does it relate in any meaningful way to the geometry of $\mathrm{Spec}(R)$, e.g. to its Weil divisor class group?

[1]: Note that this differs from some definitions, which consider all ideals instead of only invertible ideals.

[2]: The nonzerodivisor assumption is to rule out easy counterexamples along the lines of $\mathbb{C}[x,y]/(x^2,xy,y^2)$, $I = J = (x,y)$, $IJ=(0)$.

Let $(R,\mathfrak m)$ be a local (not necessarily noetherian) ring, $a\in R$ a non-zero divisor, and $I,J\subset R$ ideals such that $IJ=(a)$. Then $I$ and $J$ are principal.

Write $a=\sum_{i}a_ib_i$ with $a_i\in I$ and $b_i\in J$. Then $I(a^{-1}J)=R$. (Note that $a^{-1}J$ is an $R$-submodule of $Q(R)$, the total ring of fractions of $R$, and this is the frame where the last equation holds.) From $1=\sum_{i}a_i(a^{-1}b_i)$ it follows that $a_i(a^{-1}b_i)\notin\mathfrak m$ for some $i$. Now show that $I=(a_i)$.

• This is an extension of Kaplansky's proof to local rings which are not necessarily integral domains. Commented Nov 9, 2015 at 10:25

In a local domain, every invertible ideal is principal. The proof of this fact can be found in Kaplansky's book "Commutative rings", page 37, Theorem 59.

I needed this statement recently in a form which was slightly more general than those presented in the other answers (in order to avoid dealing with going to stalks in a slightly messy situation). Luckily I found the statement on the Stacks Project at tag 09ME. I'll reproduce the statement and proof here for completeness:

Lemma. Let $$A$$ be a ring. Let $$I,J\subset A$$ be nonzero ideals of $$A$$ such that $$IJ=(f)$$ for some non-zerodivisor $$f\in A$$. Then $$I$$ and $$J$$ are finitely generated ideals and finite locally free of rank 1 as $$A$$-modules.

Proof. It suffices to show that $$I$$ and $$J$$ are finite locally free $$A$$-modules of rank 1 by tag 00NX. To do this, write $$f=\sum_{i=1}^n x_iy_i$$ with $$x_i\in I$$ and $$y_i\in J$$. We can also write $$x_iy_i=a_if$$ for some $$a_i\in A$$. Since $$f$$ is a non-zerodivisor, we see that $$\sum_{i=1}^n a_i=1$$. Thus it suffices to show that $$I_{a_i}$$ and $$J_{a_i}$$ are free of rank 1 over $$A_{a_i}$$. After replacing $$A$$ by $$A_{a_i}$$, we conclude that $$f=xy$$ for some $$x\in I$$ and $$y\in J$$. Note that both $$x$$ and $$y$$ are non-zerodivisors. We claim that $$I=(x)$$ and $$J=(y)$$ which finishes the proof. Namely, if $$x'\in I$$, then $$x'y=af=axy$$ for some $$a\in A$$. Hence $$x'=ax$$ and we win.