Finitely generated Module over Non-PID

Question

It is known that finitely generated modules over a PID can be written in this form:

$$R^f \oplus R/(d_1)\oplus\dots\oplus R/(d_n)$$

Generalizing this, can we say that a finitely generated module $M$ over an arbitrary commutative ring $R$ can be written in this form:

$$M\cong R^f\oplus R/I_1\oplus\dots\oplus R/I_n$$

where $I_j$ are ideals of $R$?

Notation: $R^f=R\oplus\dots\oplus R$ means the f-fold direct sum of $R$ with itself.

Is there such a result?

Thanks.

Answer 1

No, even for Dedekind domains, which are in some sense the "nearest" generalization of a PID, this is no longer true. The problem is that, while over a PID a module is projective iff it is free, over a Dedekind domain (and many other types of rings), there are projective modules that are not free.

For a concrete example, consider the ideal $I = \left(2, 1 + \sqrt{-5}\right)$ of $R = \mathbb{Z}\left[\sqrt{-5}\right]$. One can show that $I$ is projective (it is finitely generated and principal in the localization $R_P$ for each prime ideal $P$) and has rank $1$, but it is not principal, hence not free. So you cannot find a decomposition for $I$ as in your question.

However, there is a generalization of the Fundamental Theorem of Finitely Generated Modules over a PID to Dedekind domains. See Theorem $22$ of $\S16.3$ in Dummit and Foote:

Theorem 22: Suppose $M$ is a finitely generated module over a Dedekind domain $R$. Let $n \geq 0$ be the rank of $M$ and let $M_\text{tors}$ be its torsion submodule. Then $$ M \cong \overbrace{R \oplus \cdots \oplus R \oplus I}^{n \text{ factors}} \oplus M_\text{tors} $$ for some ideal $I$ of $R$, and $$ M_\text{tors} \cong R/P_1^{e_1} \times \cdots \times R/P_s^{e_s} $$ for some $s \geq 0$ and powers of prime ideals $P_i$. The ideals $P_i^{e_i}$ are unique, and the ideal $I$ is unique up to multiplication by a principal ideal.