Finitely Generated Projective Modules are Locally Free

2026-01-22·22 min

Table of Contents

I have known for a long time that the book Commutative Algebra: Constructive Methods by Lombardi and Quitté contains (several) constructive proofs of the fact that finitely generated projective modules over a commutative ring are locally free (geometrically, this means that they are vector bundles on the specrum). But since the proofs are buried somewhere in the middle of this 1000 pages book and many prerequisites are scattered over a number of chapters, I never quite pushed through to understand the proofs. This is what I have done now and I'm writing down those things that seemed to me to contain the heart of the argument, while freely using those things that have been obvious or well-known to me.

It was a lot of fun to work out the details, and I learned a few new things or deepened my understanding about exterior algebra, $\lambda$-rings (I don't mention those in the article), the method of undetermined coefficients and I became an even bigger fan of the functorial view on algebraic geometry.

Before getting into the nitty-gritty, let me mention that there is also this proof by Ingo Blechschmidt, but since I don't speak topos language, I prefer an elementary external argument.

Determinants

First, the theory of determinants (or, more generally, of exterior products) must be developed for finitely generated free modules. We don't repeat this here.

A module $V$ over $k$ is finitely generated projective if and only if there exists a module $V'$ (which is then also finitely generated projective) such that $V \oplus V'$ is a finitely generated free module.

Lemma 1. Let $\alpha \from V \to V$ be an endomorphism of $V$. Let $V_1$ and $V_2$ be two modules such that $V \oplus V_1$ and $V \oplus V_2$ are finitely generated free modules. Then we have $\det(\alpha \oplus \id_{V_1}) = \det(\alpha \oplus \id_{V_2})$.

Proof. We have $\det(\alpha \oplus \id_{V_1}) = \det((\alpha \oplus \id_{V_1}) \oplus \id_{V \oplus V_2})$. However the endomorphism $(\alpha \oplus \id_{V_1}) \oplus \id_{V \oplus V_2}$ of the finitely generated free module $(V \oplus V_1) \oplus (V \oplus V_2)$ fits into a commutative square $$ \begin{CD} (V \oplus V_1) \oplus (V \oplus V_2) @>(\alpha \oplus \id_{V_1}) \oplus \id_{V \oplus V_2}>> (V \oplus V_1) \oplus (V \oplus V_2)\\ @VV\sigma V @VV\sigma V\\ (V \oplus V_2) \oplus (V \oplus V_1) @>(\alpha \oplus \id_{V_2}) \oplus \id_{V \oplus V_1}>> (V \oplus V_2) \oplus (V \oplus V_1) \end{CD} $$ where $\sigma$ is the isomorphism mapping $\sigma((v, v_1),(v', v_2)) = ((v, v_2), (v', v_1))$. It follows that $\det((\alpha \oplus \id_{V_1}) \oplus \id_{V \oplus V_2}) = \det((\alpha \oplus \id_{V_2}) \oplus \id_{V \oplus V_1})$ and this in turn is equal to $\det(\alpha \oplus \id_{V_2})$. ∎

It follows that if $V$ is a finitely generated projective module and $\alpha \from V \to V$ is an endomorphism of $V$, then we can define $\det(\alpha) := \det(\alpha \oplus \id_{V'})$ where $V'$ is any module such that $V \oplus V'$ is finitely generated free. Of course, if $V$ is already free, we can choose $V' := 0$ and recover the usual definition of the determinant for endomorphisms of free modules.

Lemma 2. Let $V$ be a finitely generated projective module.

We have $\det(\id_V) = 1$.
If $\alpha \from V \to V$, $\beta \from V \to V$ are two endomorphisms, then $\det(\beta \circ \alpha) = \det(\beta) \cdot \det(\alpha)$.

Proof.

Picking a complement such that $V \oplus V'$ is free, we have $\det(\id_V) = \det(\id_V \oplus \id_{V'}) = 1$.
Similarly, we have $\det(\beta \circ \alpha) = \det((\beta \circ \alpha) \oplus \id_{V'}) = \det((\beta \oplus \id_{V'}) \circ (\alpha \oplus \id_{V'})) = \det(\beta) \cdot \det(\alpha)$. ∎

Lemma 3. Let $V$ be a finitely generated projective module and let $\alpha \from V \to V$ be an endomorphism of $V$. Let $k \to k'$ be a ring extension and denote by $V_{k'} := V \otimes_k k'$ the scalar extension to $k'$. Then the determinant $\det(\alpha_{k'})$ of the scalar extension $\alpha_{k'} \from V_{k'} \to V_{k'}$ equals the image of $\det(\alpha)$ under the homomorphism $k \to k'$.

Proof. Reduce to the free case. ∎

The Rank Polynomial

If $V$ is a finitely generated free module, then the determinant of an endomorphism of the form $a \cdot \blank \from V \to V$ given by multiplication with an element $a \in k$ from the ground ring is not very interesting: It is just given by $a^n$ where $n$ is the rank of $V$. For a finitely generated projective module however, we don't have an a priori notion of a rank. We can use the above observation to define the rank of $V$. The question is just: Which element $a$ should we take, so that we can actually read off $n$ from the exponent $a^n$? The trick is to base-change to the polynomial ring $k[X]$ and use the indeterminate $X$.

Let $V$ be a finitely generated projective module. It is clear that if $k \to k'$ is any homomorphism of commutative rings, then the scalar extension $V \otimes_k k'$ is again finitely generated projective. In particular, the module $V[X]$ of polynomials with coefficients in $V$ is a finitely generated projective module over the polynomial algebra $k[X]$. We define the rank polynomial of $V$ to be $R_V(X) := \det(X \cdot \blank)$ where we regard $X$ as an endomorphism of $V[X]$ by multiplication.

By our remark above, if $V$ is free of rank $n$, then $R_V(X) = X^n$, so we can recover the traditional rank $n$ from the rank polynomial. More generally, we say that a finitely generated projective module has constant rank $n$ if $R_V(X) = X^n$.

Recall that a system $(e_i)_{i \in I}$ of elements $e_i \in k$, indexed by a finite set $I$ is called a fundamental system of idempotents if $\sum_{i \in I} e_i = 1$ and if $e_i \cdot e_j = 0$ for $i \neq j$. (Multiplying the first equation with $e_j$ then shows that $e_j ^ 2 = e_j$, i.e. the family indeed consists of idempotent elements.) This is equivalent to the geometric condition that $\Spec(k)$ is a disjoint open covering of the basic opens $D(e_i)$. (In particular, the spectrum of $k$ is connected if and only if every fundamental system of idempotents is trivial, in that there exists one $i \in I$ such that $e_i = 1$.)

Lemma 4. Let $V$ be a finitely generated projective module, let $k \to k'$ be a ring extension and let $a \in k'$ be an element. Let $V_{k'} := V \otimes_k k'$ be the base changed module and consider the operator $a \cdot \blank \from V_{k'} \to V_{k'}$ given by multiplication with $a$. Then we have $\det(a \cdot \blank) = R_V(a)$.

Proof. The element $a$ defines a homomorphism $\ev_a \from k[X] \to k'$ such that $k \to k'$ factors as $k \to k[X] \to k'$. By functoriality of base change, the module $V_{k'}$ is the base change of $V[X]$ along $\ev_a \from k[X] \to k'$. Moreover, the operator $a \cdot \blank \from V_{k'} \to V_{k'}$ is the base change of $X \cdot \blank \from V[X] \to V[X]$. Since determinants are compatible with base change, we get that $\det(a \cdot \blank)$ is the image of $\det(X \cdot \blank)$ under $\ev_a \from k[X] \to k'$. This is the claim. ∎

Proposition 5. Let $V$ be a finitely generated projective module. The coefficients of $R_V$ form a fundamental system of idempotents.

Proof. This proof is a bit magic. We add two variables $X, Y$ and consider the ring extension $k \to k[X, Y]$. We consider the operator $XY \cdot \blank \from V[X, Y] \to V[X, Y]$ on the base change $V[X, Y]$ of $V$ along $k \into k[X, Y]$. By Lemma 4, we have $\det(XY \cdot \blank) = R_V(XY)$. At the same time, by multiplicativity of the determinant map, we have $\det(XY \cdot \blank) = \det(X \cdot \blank) \det(Y \cdot \blank)$ where we regard the actions of $X$ and $Y$ on $V[X, Y]$. The determinants on the right are $R_V(X)$ and $R_V(Y)$, regarded as elements of $k[X, Y]$. So the polynomial $P := R_V \in k[X]$ satisfies $P(X Y) = P(X) P(Y)$ in $k[X, Y]$. Similarly, by Lemma 4, we have $P(1) = \det(\id_V) = 1$.

Writing $P = a_0 + a_1 X + \dots + a_n X^n$, a comparision of the coefficients shows that the formula $P(XY) = P(X) P(Y)$ means exactly that the coefficients form a fundamental system of idempotents. ∎

Perhaps we can illuminate the proof a little bit. As we have seen in the proof, the key point was to show that $P := R_V \in k[X]$ is a multiplicative polynomial, i.e. satisfies $P(X Y) = P(X) P(Y)$. In geometric terms, a polynomial $P \in k[X]$ corresponds to a morphism of affine schemes $\IA_k^1 \to \IA_k^1$. In the functorial view on algebraic geometry, $\IA_k^1$ is the functor $\Alg(k) \to \Set$ given by $\IA_k^1(k') = k'$ mapping an algebra to its underlying set. A polynomial $P \in k[X]$ describes the morphism of functors given by $\IA_k^1(k') \to \IA_k^1(k')$, $a \mapsto P(a)$. The polynomial $P$ is multiplicative if and only if the induced morphism $\IA_k^1 \to \IA_k^1$ is a morphism of monoid objects in the category of schemes, i.e. if for every algebra $k'$ and every pair of elements $a, b \in \IA_k^1(k')$ we have $P(ab) = P(a) P(b)$ and we have $P(1_{k'}) = 1_{k'}$. What is the morphism of monoid objects $\IA_k^1 \to \IA_k^1$ described by the rank polynomial $R_V$? Well, by Lemma 4 above, it is the map which maps an element $a \in k'$ to the determinant of the induced map $a \cdot \blank \from V_{k'} \to V_{k'}$.

We also could have started the other way round: Let $V$ be a finitely generated projective module. The fact that the determinant is compatible with base change implies that $V$ induces a morphism of functors $\IA_k^1 \to \IA_k^1$ given by $\IA_k^1(k') \to \IA_k^1(k')$, $a \mapsto \det(a \cdot \blank)$. The multiplicativity of the determinant translates into the fact that this morphism of functors is a morphism of monoid objects. In general, the polynomial describing a morphism of functors is obtained by substituting $k' = k[X]$ and $a = X$. This way we also obtain the rank polynomial $R_V = \det(X \cdot \blank)$ and the fact that its coefficients form a fundamental system of idempotents.

Corollary 6. Let $V$ be a finitely generated projective module. Then there exists a fundamental system of orthogonal idempotents $(e_i)_{i \in I}$ such that the localized module $V_{e_i}$ over the localized ring $k_{e_i}$ has constant rank for each $i \in I$.

Proof. We let $(e_i)_{i \in I}$ be the coefficients of the rank polynomial. Of course the rank polynomial is stable under base change, so after localizing at one of the coefficients, the localized module has a rank polynomial where one of the coefficients is a unit. The idempotency of this coefficient then implies that this unit is $1$ and the orthogonality relation implies that all other coefficients are $0$, so the rank polynomial is a monomial. ∎

Other Polynomials Defined Through Determinants

Recall that the rank polynomial has the property that for any element $a$ in any $k$-algebra $k'$, we have $R_V(a) = \det(a \cdot \blank)$. In particular, we have $R_V(X + 1) = \det((X + 1) \cdot \blank \from V[X] \to V[X])$. It turns out that the polynomial $R_V(X + 1)$ is a little bit easier to handle (for example we are going to compute its coefficients below) than the rank polynomial $R_V(X)$. Of course, it carries the same information as $R_V(X)$, because it arises from $R_V(X)$ under the automorphism of $k[X]$ given by $X \mapsto X + 1$. In particular, the module $V$ is of constant rank $n$ if and only if we have $R_V(X + 1) = (X + 1)^n$.

(Isn't the polynomial $X^n$ much nicer to work with than $(X + 1)^n$, you ask? $(X + 1)^n$ has all these annoying binomial coefficients, you say? Yes it does. Those are the ranks of all the exterior powers of $V$, but establishing this is actually not so easy).)

The polynomial $R_V(X + 1)$ is a special case of the following more general construction: Let $\alpha \from V \to V$ be an endomorphism. Then we define $$ B_\alpha(X, Y) \defby \det(X\id_{V[X, Y]} + Y \alpha \from V[X, Y] \to V[X, Y]). $$ I don't know if this polynomial has an established name (or an established symbol). I'm writing $B_\alpha$ because I found the computation of its coefficients in Bourbaki's Algebra book. It is not mentioned in Lombardi-Quitté, only the following to substitutions: The polynomial $\chi_\alpha(X) := B_\alpha(X, -1) = \det(\id_{V[X]} - X \alpha)$ is the well-known characteristic polynomial of $\alpha$. The polynomial $F_\alpha(X) := B_\alpha(1, X) = \det(\id_{V[X]} + X \alpha)$ is called the fundamental polynomial of $\alpha$. In particular, we recover the (shifted) rank polynomial as $F_{\id_V}(X) = \det(\id_{V[X]} + X \id_{V[X]}) = \det((X + 1) \cdot \blank) = R_V(X + 1)$.

In Bourbaki's Algebra book, Chapter 3, §8.5, Prop. 11 we find the computation of the coefficients of $B_\alpha(X, Y)$ (in the case of a free module):

Proposition 7. Let $V$ be a free module of rank $n$ and let $\alpha \from V \to V$ be an endomorphism. Then we have $$ B_\alpha(X, Y) = \sum_{k = 0}^n \tr(\wedge^k \alpha) X^{n - k} Y^k. $$ In particular, we have $$ \chi_\alpha(X) = \sum_{k = 0}^n (-1)^k \tr(\wedge^k \alpha) X^{n - k} $$ and $$ F_\alpha(X) = \sum_{k = 0}^n \tr(\wedge^k \alpha) X^k. $$

Proof. The proof in Bourbaki is a relatively straight-forward exterior algebra computation. ∎

To compute the fundamental polynomial, we can always reduce to the case of a free module by the following lemma:

Lemma 8. Let $V$ be a finitely generated projective $k$-module and let $\alpha \from V \to V$ be an endomorphism of $V$. Let $V'$ be another finitely generated projective $A$-module and consider the zero endomorphism $0 \from V' \to V'$. Then we have $F_\alpha(X) = F_{\alpha \oplus 0}(X)$.

Proof. The determinant of $\id_{V[X]} + \alpha X$ agrees with the determinant of $(\id_{V[X]} + \alpha X) \oplus \id_{V'[X]} = (\id_{(V \oplus V')[X]}) + X(\alpha \oplus 0)$. ∎

Let $V$ be a finitely generated projective $k$-module. Pick $V'$ such that $V \oplus V'$ is free. Then the map $\pi := \id_{V} \oplus 0 \from V \oplus V' \to V \oplus V'$ satisfies $\pi^2 = \pi$ and has $V$ as its image. Conversely, if $W$ is a finitely generated free module and $\pi \from W \to W$ satisfies $\pi^2 = \pi$, then $\Im(\pi) \oplus \Im(\id - \pi) \cong W$, so $V := \Im(\pi)$ is finitely generated projective. In this case, by Lemma 8 above, we can compute the (shifted) rank polynomial of $V$ through the fundamental polynomial of $\pi$ $$ R_V(X + 1) = F_{\id_V}(X) = F_{\id_V \oplus 0}(X) = F_\pi(X). $$

We will need one last proposition before we can stick everything together. In fact I want to state the proposition before defining the term $(\extpow^r F_\alpha)(X)$ appearing in it.

Proposition 9. Let $V$ be a finitely generated free module of rank $n$ and let $\alpha \from V \to V$ be an endomorphism. Then for the fundamental polynomial of the exterior power $\extpow^r \alpha \from \extpow^r V \to \extpow^r V$ we have $$ F_{\extpow^r \alpha}(X) = ({\textstyle\extpow^r} F_\alpha)(X). $$ In particular, if the fundamental polynomial of $\alpha$ splits as $$ F_{\alpha}(X) = (1 + \lambda_1 X) \cdots (1 + \lambda_n X), $$ then for we have $$ F_{\extpow^r \alpha}(X) = \prod_{1 \le i_1 < \dots < i_r \le n} (1 + \lambda_{i_1} \cdots \lambda_{i_r} \cdot X). $$

We still need to define the polynomial $(\extpow^r F_\alpha)(X)$. In fact, we can define $(\wedge^r f)(X)$ for every polynomial $f \in k[X]$ with $f(0) = 1$. I'm going to give a definition in terms of a functorial language which many people won't like but which fits with how I like to think about things. Let us denote by $\IP_n(k)$ the set of all polynomials of degree $\le n$ which satisfy $f(0) = 1$. Note that $\IP_n \from \Ring \to \Set$ is a functor. In fact, it is even an affine scheme; namely it is isomorphic to the affine space $\IA^n$ satisfying $\IA^n(k) = k^n$, by identifying the coefficients of $f \in \IP_n(k)$ in front of degree $X^1, \dots, X^n$ with the coordinates of $\IA^n$.

Lemma 10. There exists a unique morphism of functors $$ {\textstyle\extpow}^r \from \IP_n \to \IP_{\binom{n}{r}} $$ such that whenever $f \in \IP_n(k)$ factors as $f(X) = \prod_{i \in I} (1 + \lambda_i X)$, where $I$ is some index set of cardinality $n$ and $\lambda_i \in k$, then $$ (\wedge^r f)(X) = \prod_{J \in \binom{I}{r}} (1 + (\prod_{j \in J} \lambda_j) \cdot X). $$ Here, $\binom{I}{r}$ denotes the set of subset of cardinality $r$ of $I$.

Proof. We fix the set $I$ (for example we could take $I = \{1, \dots, n\}$). We want to show that there is a unique morphism of functors $\extpow^r \from \IP_n \to \IP_{\binom{n}{r}}$ such that the diagram $$ \begin{CD} \IA^I @>\extpow^r>> \IA^{\binom{I}{r}}\\ @VV\pi V @VV\pi V\\ \IP_n @>\extpow^r>> \IP_{\binom{n}{r}} \end{CD} $$ commutes. Here, $\IA^I \from \Ring \to \Set$ denotes the functor given by $\IA^I(k) = k^I$, $\pi \from \IA^I \to \IP_n$ denotes the morphism of functors given by $\pi((\lambda_i)_{i \in I}) = \prod_{i \in I}(1 + \lambda_i X)$, and similarly for the morphism $\pi \from \IA^{\binom{I}{r}} \to \IP_{\binom{n}{r}}$. Moreover, $\wedge^r \from \IA^I \to \IA^{\binom{I}{r}}$ maps the family $(\lambda_i)_{i \in I}$ to the family $(\prod_{j \in J} \lambda_j)_{J \in \binom{I}{r}}$.

Now note that the morphism $\pi \from \IA^I \to \IP_n$ kills the action of the symmetric group $\SymG(I)$ on $\IA^I$: For any $\sigma \in \SymG(I)$, precomposing $\pi \from \IA^I \to \IP_n$ with the induced morphism $\sigma \from \IA^I \to \IA^I$ again yields the map $\pi$. In fact, if $\varphi \from \IA^I \to X$ is any map to an affine scheme (a representbale functor) killing the action of $\SymG(I)$, then there is a unique induced map $\IP_n \to X$ compatible with $\pi$. This is because by multiplying out the product $\prod_{i \in I}(1 + \lambda_i X)$, we see that the components of the map $\IA^I \to \IP_n$ are actually the elementary symmetric functions in the coordinates of $\IA^I$, so by the fundamental theorem of symmetric functions, the induced map on the coordinate rings embed $\sheaf{O}(\IP_n)$ as the fixed ring $\IZ[(T_i)_{i \in I}]^{\SymG(I)}$. Hence, maps $\IP_n \to X$ correspond to maps $\sheaf{O}(X) \to \IZ[(T_i)_{i \in I}]$ mapping into the fixed ring, i.e. maps $\IA^I \to X$ compatible with the action.

So we need to show that the composition $\IA^I \to \IA^{\binom{I}{r}} \to \IP_{\binom{n}{r}}$ kills the action of $\SymG(I)$. But a permutation $\sigma \from I \to I$ induces a permutation $\widetilde{\sigma} \from \binom{I}{r} \to \binom{I}{r}$ and then the claim follows because the diagram $$ \begin{CD} \IA^I @>\extpow^r>> \IA^{\binom{I}{r}}\\ @VV\sigma V @VV\widetilde{\sigma} V\\ \IA^I @>\extpow^r>> \IA^{\binom{I}{r}}\\ \end{CD} $$ commutes and because we have $\pi \circ \widetilde{\sigma} = \pi$. ∎

Proof of Proposition 9. Let us pick a basis for the free module $V$, indexed by a finite set $I$ of cardinality $n$. This induces a canonical basis for $\extpow^r V$ indexed by $\binom{I}{r}$. Then we can identify the set of endomorphisms of $I$ with the set $M_I(k)$ of $I \times I$ square matrices and similarly for endomorphisms of $\extpow^r V$. What we want to show then amounts to the commutativity of the square of functors $$ \begin{CD} M_I @>\extpow^r>> M_{\binom{I}{r}}\\ @VV F V @VV F V\\ \IP_n @>\extpow^r>> \IP_{\binom{n}{r}}\\ \end{CD} $$ where the vertical maps send a matrix to its fundamental polynomial. This is a square of affine schemes (representable functors). In order to establish the commutativity, it is enough to evaluate both paths $M_I \to \IP_{\binom{n}{r}}$ on the universal point of $M_I$, i.e. the matrix $u := (T_{ij})_{i,j \in I} \in M_I(A)$ where $A = \IZ[(T_{ij})_{i,j \in I}]$ is the ring representing the functor $M_I$. Now by Proposition III.5.3 in Lombardi-Quitté, there exists an injective ring extension $A \into B$ such that the image of $u$ in $M_I(B)$ is diagonalizable. Since the map $\IP_{\binom{n}{r}}(A) \to \IP_{\binom{n}{r}}(B)$ is also injective, it is enough to show that $F_{\extpow^r u}(X) = (\extpow^r F_u)(X)$ in $\IP_\binom{n}{r}(B)$ where $u \in M_I(B)$ is now diagonalizable. However, if $u$ is similar to a diagonalizable matrix $\mathrm{Diag}((\lambda_i)_{i \in I})$, then $\extpow^r u$ is similar to $\mathrm{Diag}((\prod_{j \in J}\lambda_j)_{J \in \binom{I}{r}})$. For a diagonal matrix $\mathrm{Diag}((\lambda_i)_{i \in I})$ we compute easily that the fundamental polynomial is given by $\prod_{i \in I} (1 + X \lambda_i)$ and hence the desired equality of fundamental polynomails follows from the definition of the exterior power of polynomials in Lemma 10. ∎

Putting Everything Together

Corollary 11. Let $V$ be a finitely generated projective module of constant rank $n$. Then for each $r$, the exterior power $\extpow^r V$ is of constant rank $\binom{n}{r}$.

Proof. Pick $V'$ such that $V \oplus V'$ is finitely generated free and let $\pi = \id_V \oplus 0 \from V \oplus V' \to V \oplus V'$ be the idempotent map with image $V$. By Lemma 8, we have $R_V(X + 1) = F_\pi(X)$, so the fact that $V$ is of constant rank $n$ translates into the property $F_\pi(X) = (X + 1)^n$ of the morphism $\pi$.

Now the map $\extpow^r \pi$ is an idempotent map with image $\extpow^r V$. Hence we have $R_{\extpow^r V}(X + 1) = F_{\extpow^r \pi}(X) = \extpow^r ((X + 1) ^n)$. By the definition of the action of $\extpow^r$ on polynomials in Lemma 10 we have $ \extpow^r ((X + 1) ^n) = (X + 1)^{\binom{n}{r}}$ which proves the claim. ∎

Lemma 12. Let $V$ be a finitely generated projective $k$-module. If $V$ is of constant rank $0$, then $V = 0$.

Proof. If $V$ is of constant rank $0$, i.e. $R_V(X) = X^0 = 1$, then for every $a$ (in any algebra over $k$) we have $\det(a \cdot \blank) = 1$. In particular, the zero endomorphism of $V$ has determinant $1$. It is therefore enough to show that if an endomorphism $\alpha$ of a finitely generated projective $k$-module has an invertible determinant, then the endomorphism itself is invertible. Picking a complement $V'$ such that $V \oplus V'$ is free, we get that $\det(\alpha \oplus \id_{V'})$ is invertible. For free modules it is well-known by Cramer's rule that this implies $\alpha \oplus \id_{V'}$ is invertible and this in turn implies that $\alpha$ is invertible. ∎

Theorem 13. Let $V$ be a finitely generated projective $k$-module of constant rank $n$. Let $\pi$ be an idempotent endomorphism of a finitely generated free module with image $V$. Let $(f_i)$ denote the family of principal minors of order $n$ of $\pi$. Then we have $\sum f_i = 1$ and each of the localized modules $V_{f_i}$ is free of rank $n$ over the localized ring $k_{f_i}$.

Proof. First we claim that all $(n + 1) \times (n + 1)$ minors of $\pi$ vanish, or what is equivalent, that $\extpow^{n + 1} \pi = 0$. But $\extpow^{n + 1} \pi$ has image $\extpow^{n + 1} V$ which has constant rank $0$ by Corollary 11 and therefore vanishes by Lemma 12.

Furthermore, by Proposition 7, we have $\sum_{k} \tr(\extpow^k \pi) X^k = F_{\pi}(X) = R_V(X + 1) = (X + 1)^n$. In particular, comparing the coefficients of degree $n$, we see that $\tr(\extpow^n \pi) = 1$. But this trace is exactly the sum of the principal minors of order $n$ of $\pi$.

The localized module $V_{f_i}$ is the image of the base change of $\pi$ to $k_{f_i}$. This base-changed matrix still has the property that all $(n + 1) \times (n + 1)$ minors vanish but that one $n \times n$ minor is invertible. It is easy to see that this implies that the matrix is equivalent to a matrix of the form $\id_n \oplus 0$, so that its image $V_{f_i}$ is free. ∎