What is the rank of $A^{n}$ if A is the zero ring? It's clearly not $n$ as many careless authors claim, since it's not even invariant. I don't think it's 0 either because it does have a linearly independent element(0, the only element).
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1$\begingroup$ The rank can be defined as $0$, $-\infty$, or $-7$... The only way to evaluate a definition is in the context of a specific body of results (how many exceptions will there be in the propositions proved about modules due to a particular choice? &c) $\endgroup$– Mariano Suárez-ÁlvarezCommented Jul 12, 2010 at 21:41
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4$\begingroup$ Or better, avoid defining ranks of modules over the zero ring. There is clearly nothing to gain from assigning a numerical invariant to a class of objects that are all isomorphic. $\endgroup$– Pete L. ClarkCommented Jul 12, 2010 at 21:43
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1$\begingroup$ "Not a real question?" or am I being misguidedly harsh here? $\endgroup$– Yemon ChoiCommented Jul 12, 2010 at 21:46
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1$\begingroup$ Many of the "many authors" that you claim to be mistaken probably don't consider a ring with 0=1 to be a ring at all. If it is an elementary textbook they might have been careful enough to announce this somewhere, but even if they didn't it's really not a big deal. $\endgroup$– Jack HuizengaCommented Jul 12, 2010 at 21:58
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4$\begingroup$ One might just say that the zero ring doesn't have the invariant dimension property (it's the unique such commutative ring, but there are interesting noncommutative examples). Or, one might take the K-theoretic perspective that rank of a projective module (or even a nonprojective one) should take values in some group dependent on the ring, or that there are many possible ranks associated to the primes of A. Asking what rank really means and how to generalize it does lead to interesting mathematical questions. $\endgroup$– Tyler LawsonCommented Jul 12, 2010 at 22:59
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1 Answer
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As far as I'm concerned, "free $A$-module of rank $n$" means "$A$-module isomorphic to $A^n$. You just have to remember that a free module of rank $n$ is sometimes also a free module of rank $m$ even when $n$ is different from $m$. This happens for all $m$ and $n$ in the case of the $0$ ring. It also happens for some positive values of $m$ and $n$ for certain noncommutative rings.
EDIT Oh, I see: You were thinking of "rank of a module", not "rank of a free module". So if "rank" refers to how many independent elements there are, and if you have the urge to be all Bourbaki-careful about it, then you just have to choose between two definitions in the first place.
Option 1, the rank is the supremum of the sizes of sets of independent elements.
Option 2, it is the supremum of the sizes of indexed collections of independent elements.
The distinction has no effect (because two elements cannot be independent if they are equal) except in the case of the (unique up to unique isomorphism) module for the (unique up to unique isomorphism) $0$ ring, in which case Option 1 gives $1$ and Option 2 gives $\infty$. I think I prefer Option 2, because $1$ doesn't seem right. On the other hand, that $\infty$ doesn't really exist: it's trying to be the largest possible cardinal number. What a choice!
answered Jul 12, 2010 at 21:56
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1$\begingroup$ Well «$\forall y\in\emptyset,\forall x\in\emptyset, f(x)=y$» is true, no? $\endgroup$ Commented Jul 12, 2010 at 22:00
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$\begingroup$ But you can't give me two different $y$'s that are both values of $f$, so in that sense how can $f$ be nonconstant? $\endgroup$– KConradCommented Jul 12, 2010 at 22:13
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$\begingroup$ Sorry, folks. When I edited my answer I took out this really self-indulgent second part of the answer, in which I pointed out that two constant functions from $X$ to $Y$ determined by two different elements of $Y$ are the same if $X$ is empty. I went on to assert that the identity map from the empty set to itself is not a constant function because there is no $y$ such that for every $x$ $f(x)=y$. $\endgroup$ Commented Jul 12, 2010 at 22:22
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1$\begingroup$ @Mariano: Your argument proves that every constant map from the empty set to the empty set is equal to the identity map. That does not contradict my statement that the identity map from the empty set to itself is not a constant map. $\endgroup$ Commented Jul 12, 2010 at 22:24
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$\begingroup$ @KConrad: All I am saying is that I prefer one of two possible definitions of "constant map from $X$ to $Y$ -- definitions which only disagree when both $X$ and $Y$ are empty. (1) there exists $y$ such that for every $x$ $f(x)=y$, (2) when $f(x_1)=f(x_2)$ always. $\endgroup$ Commented Jul 12, 2010 at 22:27