# Finitely generated Module over Non-PID

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.

• What is the definition of $R^f$? – Robert Lewis Nov 18 '17 at 4:16
• @RobertLewis $R\oplus\dots\oplus R$ $f$ times – yoyostein Nov 18 '17 at 4:23

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.