2
$\begingroup$

The theory of commutative absolutely flat rings (a.k.a. commutative von Neumann regular rings) is algebraic and furthermore it is the smallest variety containing all fields. Being a variety the category of (set-theoretic) models is arguably much better behaved than the category of fields.

How does linear algebra in modules over commutative absolutely flat rings compare to plain linear algebra, i.e. working with vector spaces? Is is it basically the same or are there some major differences?

Let's ask some more concrete questions:

Let $R$ be an commutative absolutely flat ring: Are finitely generated $R$-modules always free? Assuming the axiom of choice: Are all $R$-modules free?

$\endgroup$
5
  • $\begingroup$ The smallest variety containing all fields is the variety of all commutative rings. To see this it is enough to note that any free commutative ring is a domain, hence is embeddable in a field. $\endgroup$
    – Keith Kearnes
    Aug 15, 2016 at 18:28
  • $\begingroup$ @KeithKearnes Huh. I may have a serious misconception. I was basing that on this mathoverflow.net/a/91969/78650. $\endgroup$
    – Stefan Perko
    Aug 15, 2016 at 19:48
  • 1
    $\begingroup$ In that mathoverflow question they were expanding the language of rings to include a multiplicative inverse operation as a new fundamental operation. The objects obtained were no longer just rings, but rather objects they referred to as inverse rings. Analogously, the class of semigroups that are embeddable in groups is not a variety, but becomes one if you add the inverse as a new operation. $\endgroup$
    – Keith Kearnes
    Aug 15, 2016 at 20:11
  • 1
    $\begingroup$ Let me correct my previous comment. Where I say "the class of semigroups embeddable in groups" I should have said "the class of all reducts of groups to the language of semigroups". $\endgroup$
    – Keith Kearnes
    Aug 15, 2016 at 20:59
  • $\begingroup$ @KeithKearnes Yes, I didn't mean "variety with the same language", I meant variety in general. $\endgroup$
    – Stefan Perko
    Aug 16, 2016 at 6:52

2 Answers 2

Reset to default
2
$\begingroup$

[OP]

"Let's ask some more concrete questions:

Let R be an commutative absolutely flat ring: Are finitely generated R-modules always free? Assuming the axiom of choice: Are all R-modules free?"

[Answer]

Not necessarily.

Every nontrivial (unital) ring has a simple module, which is finitely generated, in fact 1-generated. R has a free simple left module iff R has no proper nontrivial left ideals iff R is a division ring. [The least obvious part of this claim is that if R has a free simple left module, then it has no proper nontrivial left ideals. You see this by observing that the only free module that could possibly be simple is the rank 1 free module, ${}_R R$. Then notice that this module is simple iff $R$ has no proper nontrivial left ideals.]

In the commutative case this shows that all (f.g.) $R$-modules are free iff $R$ is a field.

$\endgroup$
1
$\begingroup$

No. Simple example: let $k$ be field. The ring $R=k\times k$ is absolutely flat, and the first factor $k$, as an $R$-module, is the cyclic $R$-module generated by the idempotent $e=(1,0)$. It's not free, since $e$ is killed by the orthogonal idempotent $1-e=(0,1)$.

Actually one proves a commutative ring $R$ is absolutely flat if and only if its Krull dimension is $0$, and its localisations at maximal ideals are fields.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .