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stub for rank
I think it is wrong. There is a nontrivial ring $R$ for which there is a possibility of having two different numbers $m$ and $n$ such that the free modules $R^n$ and $R^m$ are isomorphic. In other words, the rank of a free module over an arbitrary ring is not necessarily unique. This is related to the number of issues including those important in the business of Serre’s conjecture (every finitely generated projective module over a polynomial ring is free). As wikipedia puts it in article module (mathematics) “However, modules can be quite a bit more complicated than vector spaces; for instance, not all modules have a basis, and even those that do, free modules, need not have a unique rank if the underlying ring does not satisfy the invariant basis number condition, unlike vector spaces which always have a basis whose cardinality is then unique (assuming the axiom of choice).”
dang it. Could you fix that? I need to rush off now.
Well fix it later (or I will even later). I do not fully understand what you wanted to say there, it would require to think too thoroughly for such a crazy moment before the trip for which I am not yet even starting packing.
There is a whole large part of ring Theory very much influenced by P. M.Cohn and his book ‘Free rings and their relations’. I handles invariant basis number rings (It is over 30 years since I worked on that stuff but I remember it was nice… and very well written.)
But it should be rewritten. It is clear for example that many proofs there hide almost apparent tricks with quasideterminants (which were introduced around 1990). Also one should take into account huge shift in understanding Cohn localization by Vogel and then by Ranicki and Neeman in terms of Bousfield localization for unbounded derived categories.
Urs, it looked like you set this page up to be a page on all of the meanings of ‘rank’, which I like. So I formatted it a bit to go with that, and added another meaning.
I also fixed the bit about ranks’ being unique.
Yes, thanks!
A finitely generated free module is dualizable, and therefore we can consider the trace of its identity map, which can be defined in a basis-invariant way. If the rank of the module is not well-defined, then which of its possible ranks does the trace of the identity give you?
In the case of a commutative ring, I think your argument works, Mike (since left modules form a monoidal category and “dualizable” makes obvious sense there). It looks to me more problematic in the noncommutative ring case. In any case, Wikipedia gives an example here of what Zoran is referring to.
A finitely generated free (or even projective) module over a noncommutative ring is still dualizable in the bicategory of rings and bimodules. In that bicategory there is a notion of trace, called the Hattori-Stallings trace, which is a special case of a general notion of trace in a bicategory defined by Kate Ponto, cf. for instance arxiv:0910.1306. In general, the trace is not necessarily an integer, but rather an element of the ground ring (in the commutative case) or of a suitable quotient of the ring (in the noncommutative case). If I have a moment, maybe I’ll think about what the trace of the identity would be for those modules over $\mathcal{M}_{\mathbb{N}}(R)$.
Ah, of course, the answer is that $1=2$ in $\langle\langle \mathcal{M}_{\mathbb{N}}(R) \rangle\rangle$, which is where traces of endomorphisms of dualizable $\mathcal{M}_{\mathbb{N}}(R)$-modules live. By definition, $\langle\langle \mathcal{M}_{\mathbb{N}}(R) \rangle\rangle$ is the abelian group obtained by quotienting $\mathcal{M}_{\mathbb{N}}(R)$ by the additive subgroup generated by $x y - y x$, for all $x$ and $y$. In this case, if $1$ denotes the identity matrix, $A$ the odd columns of the identity matrix, $B$ the even columns of the identity matrix, and $A^\top$ and $B^\top$ their transposes, then we have $A A^\top + B B^\top = 1$, but $A^\top A = B^\top B = 1$, hence $1 = 1 + 1$ in the quotient $\langle\langle \mathcal{M}_{\mathbb{N}}(R) \rangle\rangle$.
Of course that implies $0=1$ in $\langle\langle \mathcal{M}_{\mathbb{N}}(R) \rangle\rangle$ as well, and so the “rank” of any finite-dimensional free module is zero according to this definition. Seeing that, I wouldn’t be surprised if $\langle\langle \mathcal{M}_{\mathbb{N}}(R) \rangle\rangle = 0$, but I don’t see how to prove it at this time of night.
(Note that the trace-theoretic notion of “rank” has other issues arising from where it lives, even in the commutative case. For instance, this “rank” of a $p$-dimensional $\mathbb{F}_p$-vector-space is zero, since it is $p$ regarded as an element of $\mathbb{F}_p$.)
I put in Wikipedia’s examples of three broad classes of rings with IBN, as well as its example of a nontrivial ring without IBN (but explained a bit more abstractly and concisely).
Rings with IBN have other nice properties…. but forget what!!!
Added to rank the definition of the rank of a sheaf of modules on a locally ringed space $X$ and its characterization in the internal language of the sheaf topos $\mathrm{Sh}(X)$.
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