0
$\begingroup$

Let $R$ be a ring and $M$ be an $R$-module.

Thm1. If $R$ is a division ring, then $M$ is free

Thm2. If $R$ is commutative, the rank of $M$ is unique.

First of all, are these statements true? (There's no reference, I just formulated by guessing)

Secondly, what is the generalized statement of the above theorems?

$\endgroup$
  • $\begingroup$ Thm1 is true, see e.g. math.stackexchange.com/q/75866/15416, Thm2 is in general false, en.wikipedia.org/wiki/Invariant_basis_number. What do you mean by "generalized statement"? Do you mean a common generalisation of those two theorems? $\endgroup$ – Julian Kuelshammer Mar 6 '15 at 20:12
  • $\begingroup$ @JulianKuelshammer I was asking generalization for each theorem and invariant basis number is what I was looking for. By the way, the link you gave says that a commutative ring is IBN, hence the rank is unique. Why Thm2 is false in general? $\endgroup$ – Rubertos Mar 6 '15 at 20:14
  • $\begingroup$ Sorry, my mistake. $\endgroup$ – Julian Kuelshammer Mar 6 '15 at 20:18
  • $\begingroup$ Artin-Wedderburn Theorem can be viewed as a generalisation of Thm1: en.wikipedia.org/wiki/Artin%E2%80%93Wedderburn_theorem $\endgroup$ – Julian Kuelshammer Mar 6 '15 at 20:19
  • $\begingroup$ @JulianKuelshammer Thank you :) Would you please write that as an answer? $\endgroup$ – Rubertos Mar 6 '15 at 20:37
1
$\begingroup$

Both Theorems are correct. For a reference of Theorem 1, see e.g. this math.SE question, discussing the theorem that a unital ring is a division ring if and only if every unital left module is free. For the second theorem, see e.g. this math.SE question discussing precisely this question.

As for generalisations. For Theorem 1, Artin-Wedderburn Theorem can be viewed as a generalisation of it. The class of free modules is replaced by the more general class of projective modules (which can e.g. be defined as direct summands of free modules). One formulation of this is then, that every module over a ring is projective if and only if the ring is isomorphic to a direct product of finitely many matrix rings over division rings. See e.g. the wikipedia article on this theorem and the references therein.

For Theorem 2, the property you are looking for is called the invariant basis number, meaning that all finitely generated free modules have a well-defined rank. This is satisfied for any commutative ring, any left noetherian ring and any semilocal ring. But there are also counterexamples for general rings. For references, see e.g. the wikipedia article or this mathoverflow question and the references therein.

$\endgroup$

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.