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Let $R$ be a commutative ring.

What are propositions about $R$-modules that do not generally hold but are true when $R$ is a field?

So-called "fundamental theorem of linear algebra"

$$(\mathrm{Im} f)^\perp=\mathrm{Ker} f^*$$

is not a feature of linear spaces, because for any $R$-homomorphism $f:M\to N$ between $R$-modules,

$$\begin{align} \varphi \in (\mathrm{Im} f)^\perp &\Leftrightarrow \forall n \in \mathrm{Im}f. \varphi(n)=0\\ &\Leftrightarrow \forall m\in M. \varphi(f(m))=0\\ &\Leftrightarrow f^*(\varphi)=0\\ &\Leftrightarrow \varphi\in \mathrm{Ker}f^*. \end{align}$$

I know only two examples.

  1. Basis theorem. Every linear space has a basis, but not all modules.
  2. If $a_1, \dots, a_m$ are linearly dependent, one of them is a linear combination of others.

(linear space theory) - (module theory) = (dimension theory)?

If $a_1, \dots, a_m\in V$ are linearly independent and $b_1, \dots, b_n \in V$ spans $V$, $m\leq n$ holds. However, this is also true for modules over a commutative ring(link).

Many properties of linear spaces can be generalized to $R$-modules. I want to know counterexamples.

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    $\begingroup$ modules aren't usually vector spaces. $\endgroup$
    – Rushabh Mehta
    Nov 3, 2019 at 3:04
  • $\begingroup$ A vector space is a module over a field. $\endgroup$
    – Ponta
    Nov 3, 2019 at 3:15
  • $\begingroup$ Indeed. That's why it's an example. Although depending on your definitions, that could be obvious, in which case literally any vector space property which doesn't hold in an arbitrary module, i.e., torsion free. $\endgroup$
    – Rushabh Mehta
    Nov 3, 2019 at 3:16
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    $\begingroup$ This is a broad question. There are many properties that only some modules have that all vector spaces have. $\endgroup$
    – PersonX
    Nov 4, 2019 at 20:13
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    $\begingroup$ Possibly the most relevant difference is that for $x\in M$, a module over a ring $R$, and $r\in R$, it does not hold that $rx=0$ implies either $r=0$ or $x=0$. All about linear independence and bases in vector spaces stems from this property: you lose the existence of linearly independent sets, for instance. $\endgroup$
    – egreg
    Nov 5, 2019 at 0:27

1 Answer 1

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The "most important" fact about vector spaces which fails for modules in general is exactly what you noted: every vector space has a basis, but not every module has a basis. Indeed, the following are equivalent (for the rest of this post, $R$ will be a commutative unital ring):

  • Every $R$-module is free (i.e. every $R$-module has a basis)
  • If $a_1, \dots, a_n$ are linearly dependent elements of an $R$-module $M$, then one of them is a linear combination of the others.
  • $R$ is a field (i.e. $R$-modules are vector spaces)

Another commonly-used fact in linear algebra is that every subspace has a complementary subspace. Rings for which this is true of all submodules need not be fields, but they are nicely classified. Precisely, the following are equivalent:

  • If $M$ is an $R$-module with a submodule $N$, then there is some submodule $N'$ of $M$ such that $M = N \oplus N'$.
  • $R$ is isomorphic (as a ring) to a direct product of finitely many fields.
  • $R$ (viewed as a module over itself) is a direct sum of simple $R$-modules.
  • Every $R$-module is a direct sum of simple $R$-modules.
  • $R$ is Artinian and reduced.

For many properties (in the language of modules) we have a good understanding of the rings $R$ for which the property holds for all $R$-modules. Do you have any specific properties you're curious about? It's a bit silly to ask for examples of properties of vector spaces and modules for which those properties don't hold: there are simply too many to list.

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