# Are finitely generated modules over a commutative ring always a direct sum of cyclic submodules?

Let's first motivate my question by looking at a finitely generated $$k$$-algebra $$A$$ over a field $$k$$.

Then $$A$$ in general does not have the form $$k[a_1,a_2,\ldots,a_n]$$ where $$\{a_1,a_2,\ldots,a_n\}$$ is a generating set for $$A$$. For example consider the two-dimensional irreducible representation $$V$$ of the quarternion group $$Q_8$$, then the ring of invariants is finitely generated by Hilbert's finiteness theorem, but the algebra of invariants, which is a subalgebra of a polynomial algebra in two variables holds the form $$\mathbb{C}[V]^{Q_8} = \dfrac{\mathbb{C}[f,g,h]}{(h^2-f^2g+4g^3)},$$ where $$f$$ and $$g$$ are invariant polynomials of degree 4, and $$h$$ is of degree 6. The reason is that the generating polynomials are not algebraically independent.

Now consider a commutative ring $$R$$, and $$M$$ a finitely generating $$R$$-module, and $$\{m_1,m_2,\ldots,m_n\}$$ a generating set for $$M$$, I want to know whether it is true that $$M$$ holds the form $$M = \bigoplus_{i=1}^n Rm_i$$

I think this is not true, but this is true if and only if $$M$$ is a finitely generated $$\textit{free}$$ module over $$R$$.

Can someone enlighten me?

You're right, it is not generally true. Rings for which finitely generated modules (and also simply just all modules) are direct sums of cyclic modules have pretty good coverage in literature.

Being free is sufficient for being that form, but it will not usually be necessary. For example, if the commutative ring has a nontrivial ideal, $$R/I$$ is clearly cyclic, and is certainly non-free (it has nonzero annihilator.) (Simplified thanks to metalspringpro)
• Just to add, as long as $R$ has a proper ideal $I$, then $R/I$ is cyclic but not free. So there are always nonfree modules having that form unless $R$ is a field. – metalspringpro Aug 7 at 14:36