This is a concept from Lectures on Modules and Rings of T. Y. Lam. It's on page 34-35 of the book.

Throughout this post, $R$ will be used to denote a commutative ring.

Let $P$ be a f.g $R-$projective module. For any prime ideal $\mathfrak{p} \subset R$, the localization $P_{\mathfrak{p}} := P \otimes_R R_{\mathfrak{p}}$ is also a f.g projective $R_{\mathfrak{p}}-$module. $\color{green}{\textbf{I get this part. Yay!!!}}$ Since $R_{\mathfrak{p}}$ is a local ring, $P_{\mathfrak{p}}$ must actually be free (by FC 19.29). $\color{green}{\textbf{Sounds good!!!}}$, say $P_{\mathfrak{p}} \cong R_{\mathfrak{p}}^{n_{\mathfrak{p}}}$ (for some $n_\mathfrak{p} \in \mathbb{N}$), so we have a function $f: \text{Spec}(R) \to \mathbb{Z}$, sending each $\mathfrak{p}$ to $n_\mathfrak{p}$. If that function $f$ is constant, i.e $n_\mathfrak{p} = n, \forall \mathfrak{p} \in \text{Spec}{R}$, we shall say that $P$ has rank $n$. And denote it as: $\text{rk}(P) = n$. $\color{red}{\textbf{What I don't really get is the idea behind this definition.}}$.

And then, the author leaves a fact unproved:

Fact. Let $P$, $Q$ be two f.g projective $R-$modules of rank $n$, and $m$ respectively. Show that $\text{rk}(P^*) = n$, and $\text{rk}(P\otimes_RQ) = nm$, where $P^* = \text{Hom}_R(P; R)$.

I think it should be very easy, since the author doesn't prove it. So here's my try on the problem.

Since $P$, $Q$ are both f.g $R-$projective, there exists $R-$modules $P'$, and $Q'$, such that $R^i = P\oplus P'$, and $R^j = Q\oplus Q'$. So we'll have:

  • $R^i = \text{Hom}(R^i,R) = \text{Hom}(P \oplus P',R) = \text{Hom}(P,R) \oplus \text{Hom}(P',R)$
  • $R^j = \text{Hom}(R^j,R) = \text{Hom}(Q \oplus Q',R) = \text{Hom}(Q,R) \oplus \text{Hom}(Q',R)$

So basically, we'll have $P \oplus P' \cong \text{Hom}(P,R) \oplus \text{Hom}(P',R)$, and $Q \oplus Q' \cong \text{Hom}(Q,R) \oplus \text{Hom}(Q',R)$.

How can I proceed from here? Is this the correct way to start?

I just wanna ask 2 things:

  • Firstly, can you guys please give me the idea (or some motivation) behind this definition for the rank of a projective modules?

  • Secondly, can you guys check my work (are there any subtle errors, or something), and give me a little push on the problem? :(

Thank you so much for your help,

And have a good day,


You know the rank of a free module, right? It is the cardinality of a basis (well-defined since $R$ is commutative and $R \neq 0$). This is already well-known from linear algebra ($R$ is a field), where it is called the dimension (but as you see, this is really the same concept). Now it turns out that finitely generated projective $R$-modules are "not far" from being free: They are locally free. This means that there is a Zariski covering of $\mathrm{Spec}(R)$ (the spectrum, consisting of prime ideals of $R$) such that the restrictions are free. More elementary: There are elements $f_1,\dotsc,f_n$ generating the unit ideal such that each localization at $f_i$ is a finitely generated free module over $R_{f_i}$. In particular, the localization at each prime ideal $\mathfrak{p}$ is a free module over $R_{\mathfrak{p}}$, so that we may consider its rank. It is a locally constant function on $\mathrm{Spec}(R)$. I don't know why Lam only considers the case that this is a constant function (which is automatic when $R$ has non nontrivial idempotents for example), since this is not necessary at all.

The formulas $\mathrm{rank}(M^*) = \mathrm{rank}(M)$, $\mathrm{rank}(M \oplus N) = \mathrm{rank}(M) + \mathrm{rank}(N)$ and $\mathrm{rank}(M \otimes N) = \mathrm{rank}(M) \cdot \mathrm{rank}(N)$ are clear if $M,N$ are finitely generated free modules. By localization (this process preserves duals, direct sums and tensor products) it follows more generally when $M,N$ are finitely generated projective modules.

A typical example is the ideal $I = (2,1+\sqrt{-5})$ of $R=\mathbb{Z}[\sqrt{-5}]$. One can prove $I \oplus I \cong R^2$, so that $I$ is a finitely generated projective $R$-module of (constant) rank $1$. But $I$ is not a principal ideal, and hence not free.

  • $\begingroup$ Thank you so much for your help, and your great explanation. I think I need some time to digest this. :) $\endgroup$ – user49685 Dec 16 '13 at 16:47
  • $\begingroup$ I'm sorry, but I still fail to see it. I know that $M \cong M^*$ (for f.g free module). And I know, what I should prove is that $M_{\mathfrak{p}} \cong M^*_{\mathfrak{p}}$. But is it correct that: $\text{Hom}(M, R)_{\mathfrak{p}} \cong \text{Hom}(M_{\mathfrak{p}}, R_{\mathfrak{p}})$? It seems so, but I just cannot find a reference for this. Can you confirm this for me? And also, can you recommend some book that has a quite comprehensive section on localization? I think I'm having some big troubles with it. Mac Donald's book doesn't seem enough. :( $\endgroup$ – user49685 Dec 16 '13 at 17:36
  • $\begingroup$ Ok, on thinking again, I think that congruence is exactly what you mean by 'localization preserves duality', right? But can you give me a sketch of that proof? Or am I misinterpreting your idea? @@ $\endgroup$ – user49685 Dec 16 '13 at 17:45
  • $\begingroup$ The canonical homomorphism $S^{-1} \hom_R(M,N) \to \hom_{S^{-1} R}(S^{-1} M,S^{-1} N)$ is an isomorphism when $M$ is finitely generated projective over $R$. The proof is a good exercise, but you can also find it somewhere in Bourbaki's Algèbre (which is another good exercise to find the correct chapter, section, paragraph, number, ...). Of course this book also covers everything else you have to know about the basics of module theory. $\endgroup$ – Martin Brandenburg Dec 16 '13 at 18:29
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    $\begingroup$ This reminds me of xkcd.com/230 $\endgroup$ – Martin Brandenburg Dec 31 '13 at 18:26

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