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.

Start with

Behboodi, M., and G. Behboodi Eskandari. "On rings over which every finitely generated module is a direct sum of cyclic modules." Hacettepe Journal of Mathematics and Statistics 45.5 (2016): 1335-1342.

and follow the citations.

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

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)

| cite | improve this answer | |
  • $\begingroup$ Incidentally, what a great example of titling an article well! Makes it easy to find the answer to this problem right away, anyhow! $\endgroup$ – rschwieb Aug 7 at 14:19
  • 3
    $\begingroup$ 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. $\endgroup$ – metalspringpro Aug 7 at 14:36
  • 2
    $\begingroup$ @metalspringpro Good observation: originally I had been thinking about projectve modules that aren't free. But in hindsight that doesn't even matter. I hope you don't mind if I simplify that using your explanation. $\endgroup$ – rschwieb Aug 7 at 14:50
  • $\begingroup$ Sure, please do. $\endgroup$ – metalspringpro Aug 7 at 14:59

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.