Projectives and the Devious Determinant

The determinant is certainly a fascinating beast. But what is the determinant? Is it a really just a number or a function on matrices? In this post I hope to convince you that the answer is 'no'. In fact, we will see that the determinant, suitably modified, can be used to classify certain types of projective modules over nice rings.

Determinants of Matrices

Let $ R$ be a commutative ring and $ n$ be a natural number. Just as in the case of vector spaces, an $ R$-module map $ f:R^n\to R^n$ can be given by an $ n\times n$ matrix with coefficients in $ R$. Moreover, we can compute the determinant of this matrix just as in linear algebra. In fact, various notions of "determinants" also exist when $ R$ is not commutative, but we will stick with the commutative case.

Now, consider the exterior power $ \wedge^n R^n$. There is an $ R$-module isomorphism $ \wedge^n R\cong R$. If for instance $ e_1,\dots,e_n$ is a basis of $ R^n$ as an $ R$-module, then $ e_1\wedge\cdots\wedge e_n$ is a basis for $ \wedge^n R^n$ as an $ R$-module. What about $ f$? Does it extend to a map $ f:\wedge^n R^n\to \wedge^n R^n$? It does! In order to see this, observe that the map $ R^n\to \wedge^n R^n$ given by

$ (r_1,\dots,r_n)\mapsto f(r_1)\wedge\cdots\wedge f(r_n)$

is an alternating $ R$-multilinear map (not morphism). Hence $ f$ indeed does extend. Let us write $ \mathrm{det}(R^n)$ for $ \wedge^n R^n$ and $ \mathrm{det}(f)$ for the induced map from $ f$. Let's do an example! Suppose $ f:R^2\to R^2$ is given by the matrix

$ \begin{pmatrix} a & b\\ c & d\end{pmatrix}$

where $ a,b,c,d\in R$ in the basis $ \{ e_1,e_2\}$. Then the induced map $ \mathrm{det}(f):\wedge^2 R^2\to\wedge^2 R^2$ sends $ e_1\wedge e_2$ to

$ (ae_1 + ce_2)\wedge (be_1 + de_2) = ae_1\wedge de_2 – be_1\wedge ce_2 = (ad-bc)e_1\wedge e_2$

Hence the term "determinant" is really justified! Of course, this should not be too surprising, since as a function of the rows, the determinant is the unique $ r$-multilinear alternating map that gives $ 1$ on the identity matrix, a result that is usually proven in an undergraduate linear algebra course. It is easy to see that the coefficient that appears in front of the basis element has these properties.

The Rank of a Projective Module

There is no reason that we can't define the determinant $ \mathrm{det}(M)$ for an arbitrary $ R$-module except for the fact that we used the $ n$-th exterior power of $ R^n$, but one way we could remove our choice of $ n$ is take the exterior algebra of $ M$; this is even functorial so if we do this then we can think of the determinant as a functor, and the number that is the determinant can then be recovered in the case of $ R^n$ as usual. However, in this post we will stick with the construction $ \wedge^n P$ for $ P$ a free module of rank $ n$ or a projective module of (constant) rank $ n$.

How can this be used? Let us return to the case where we think of the determinant as an exterior power. Something interesting happens in the case of finitely generated constant rank projective modules.

Commutative rings have the invariant basis property: if $ R^n\cong R^m$ where $ m,n$ are natural numbers then $ n = m$. Also, in this previous blog post we saw that projective modules over local rings are free. Hence for a prime ideal $ \mathfrak{p}$ of $ R$, the projective module $ P_{\mathfrak{p}}$ is free of some rank $ n$ over the local ring $ R_{\mathfrak{p}}$. We define $ \mathrm{rank}(P)(\mathfrak{p}) = n$ for this prime. This gives a function $ \mathrm{Spec}(R)\to \mathbb{Z}$, which is a continuous function if $ \mathbb{Z}$ is given the discrete topology.

We note that the rank of a module can be defined in more general cases ($ P$ does not have to be projective or finitely generated) but then we would need to use different definitions and the rank would not be as well-behaved.

Line Bundles

We will continue in the following situation: $ R$ is a commutative ring and $ P$ is a finitely generated projective module of constant rank. In this situation it is not difficult to prove that $ \mathrm{det}(P) := \wedge^n P$ is a finitely generated projective module of constant rank $ 1$ (exercise!). In fact, this follows from the neat and not difficult to prove isomorphism of $ R$-modules

$ \wedge^n(P\oplus Q) \cong \bigoplus_{i=1}^n [\wedge^i P\otimes \wedge^{n-i}Q]$

Finitely generated projective modules of rank $ 1$ are sometimes called line bundles, since they correspond to line bundles over the affine scheme $ \mathrm{Spec}(R)$. Similarly, finitely generated projective modules of rank $ n$ correspond to $ n$-dimensional vector bundles.

The following is a theorem of J.-P. Serre and forms part what is known as the Bass-Serre cancellation theorem:

Theorem. Suppose $ R$ is a commutative Noetherian ring of Krull dimension $ d$. If $ P$ is a projective module of constant rank $ n > d$ then there exists a projective module $ Q$ of constant rank $ d$ such that $ P\cong Q\oplus R^{n-d}$.

Now, if $ d=1$, then this theorem says that under the hypotheses on $ R$, every projective $ R$-module $ P$ of constant rank $ n > 1$ can be written as $ P = Q\oplus R^{n-1}$ where $ Q$ is a line bundle. Then it is a straightforward calculation to see that $ \mathrm{det}(P)\cong \mathrm{det}(Q)$. If $ P$ has rank $ 1$, then also we have $ \mathrm{det}(P)\cong P$.

In other words, we have shown the following corollary of Serre's theorem:

Theorem. Two constant (finite) rank projective modules over a commutative Noetherian ring of Krull dimension one are isomorphic if and only if they have the same rank and determinant.

Thus the determinant, suitably reinterpreted, is a convenient way to classify projective modules of finite constant rank over a commutative Noetherian ring of Krull dimension $ 1$, such as a Dedekind domain!


It often helps to think of various functions or even operations in a more categorical setting, and the determinant is just one such mathematical idea that falls under this paradigm. I learned most of this material from Lenstra's book "Galois Theory for Schemes" available here and from C. Weibel's book, "The K-Book: An Introduction to Algebraic K-Theory", and I recommend that the reader look into these books, both available for download from the above links.

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