## 15.118 Determinants

Let $R$ be a ring. Let $M$ be a finite projective $R$-module. There exists a product decomposition $R = R_0 \times \ldots \times R_ t$ such that in the corresponding decomposition $M = M_0 \times \ldots \times M_ t$ of $M$ we have that $M_ i$ is finite locally free of rank $i$ over $R_ i$. This follows from Algebra, Lemma 10.78.2 (to see that the rank is locally constant) and Algebra, Lemmas 10.21.3 and 10.24.3 (to decompose $R$ into a product). In this situation we define

$\det (M) = \wedge ^0_{R_0}(M_0) \times \ldots \times \wedge ^ t_{R_ t}(M_ t)$

as an $R$-module. This is a finite locally free module of rank $1$ as each term is finite locally free of rank $1$. If $\varphi : M \to N$ is an isomorphism of finite projective $R$-modules, then we obtain a canonical isomorphism

$\det (\varphi ) : \det (M) \longrightarrow \det (N)$

of locally free modules of rank $1$. More generally, if for all primes $\mathfrak p$ of $R$ the ranks of the free modules $M_\mathfrak p$ and $N_\mathfrak p$ are the same, then any $R$-module homomorphism $\varphi : M \to N$ induces an $R$-module map $\det (\varphi ) : \det (M) \to \det (N)$. Finally, if $M = N$ then $\det (\varphi ) : \det (M) \to \det (M)$ is an endomorphism of an invertible $R$-module. Since $R = \mathop{\mathrm{Hom}}\nolimits _ R(L, L)$ for an invertible $R$-module we may and do view $\det (\varphi )$ as an element of $R$. In this way we obtain the determinant

$\det : \mathop{\mathrm{Hom}}\nolimits _ R(M, M) \longrightarrow R$

which is a multiplicative map.

Remark 15.118.1. Let $R$ be a ring. Let $M$ be a finite projective $R$-module. Then we can consider the graded commutative $R$-algebra exterior algebra $\wedge ^*_ R(M)$ on $M$ over $R$. A formula for $\det (M)$ is that $\det (M) \subset \wedge ^*_ R(M)$ is the annihilator of $M \subset \wedge ^*_ R(M)$. This is sometimes useful as it does not refer to the decomposition of $R$ into a product. Of course, to prove this satisfies the desired properties one has to either decompose $R$ into a product (as above), or one has to look at the localizations at primes of $R$.

Next, we consider what happens to the determinant give a short exact sequence of finite projective modules.

Lemma 15.118.2. Let $R$ be a ring. Let

$0 \to M' \to M \to M'' \to 0$

be a short exact sequence of finite projective $R$-modules. Then there is a canonical isomorphism

$\gamma : \det (M') \otimes \det (M'') \longrightarrow \det (M)$

First proof. First proof. Decompose $R$ into a product of rings $R_{ij}$ such that $M' = \prod M'_{ij}$ and $M'' = \prod M''_{ij}$ where $M'_{ij}$ has rank $i$ and $M''_{ij}$ has rank $j$. Of course then $M = \prod M_{ij}$ and $M_{ij}$ has rank $i + j$. This reduces us to the case where $M'$ and $M''$ have constant rank say $i$ and $j$. In this case we have to construct a canonical map

$\wedge ^ i(M') \otimes \wedge ^ j(M'') \longrightarrow \wedge ^{i + j}(M)$

To do this choose $m'_1, \ldots , m'_ i$ in $M'$ and $m''_1, \ldots , m''_ j$ in $M''$. Denote $m_1, \ldots , m_ i \in M$ the images of $m'_1, \ldots , m'_ i$ and denote $m_{i + 1}, \ldots , m_{i + j} \in M$ elements mapping to $m''_1, \ldots , m''_ j$ in $M''$. Our rule will be that

$m'_1 \wedge \ldots \wedge m'_ i \otimes m''_1 \wedge \ldots \wedge m''_ j \longmapsto m_1 \wedge \ldots \wedge m_{i + j}$

We omit the detailed proof that this is well defined and an isomorphism. $\square$

Second proof. We will use the description of $\det (M)$, $\det (M')$, and $\det (M'')$ given in Remark 15.118.1. Consider the $R$-algebra maps $\wedge ^*_ R(M') \to \wedge ^*_ R(M)$ and $\wedge ^*_ R(M) \to \wedge ^*_ R(M'')$. The first is injective and the second is surjective. Take an element $x' \in \det (M') \subset \wedge ^*_ R(M')$ and an element $x'' \in \det (M'') \subset \wedge ^*_ R(M'')$. Choose an element $y'' \in \wedge ^*(M)$ mapping to $x''$ and set

$\gamma (x' \otimes x'') = x' \wedge y'' \in \det (M) \subset \wedge ^*_ R(M)$

The reader verifies easily by looking at localizations at primes that this well defined and an isomorphism. Moreover, this construction gives the same map as the construction given in the first proof. $\square$

Lemma 15.118.3. Let $R$ be a ring. Let

$\xymatrix{ 0 \ar[r] & M' \ar[r] \ar[d]^ u & M \ar[r] \ar[d]^ v & M'' \ar[r] \ar[d]^ w & 0 \\ 0 \ar[r] & K' \ar[r] & K \ar[r] & K'' \ar[r] & 0 }$

be a commutative diagram of finite projective $R$-modules whose vertical arrows are isomorphisms. Then we get a commutative diagram of isomorphisms

$\xymatrix{ \det (M') \otimes \det (M'') \ar[r]_-\gamma \ar[d]_{\det (u) \otimes \det (w)} & \det (M) \ar[d]^{\det (v)} \\ \det (K') \otimes \det (K'') \ar[r]^-\gamma & \det (K) }$

where the horizontal arrows are the ones constructed in Lemma 15.118.2.

Proof. Omitted. Hint: use the second construction of the maps $\gamma$ in Lemma 15.118.2. $\square$

Lemma 15.118.4. Let $R$ be a ring. Let

$K \subset L \subset M$

be $R$-modules such that $K$, $L/K$, and $M/L$ are finite projective $R$-modules. Then the diagram

$\xymatrix{ \det (K) \otimes \det (L/K) \otimes \det (M/L) \ar[r] \ar[d] & \det (L) \otimes \det (M/L) \ar[d] \\ \det (K) \otimes \det (M/K) \ar[r] & \det (M) }$

commutes where the maps are those of Lemma 15.118.2.

Proof. Omitted. Hint: after localizing at a prime of $R$ we can assume $K \subset L \subset M$ is isomorphic to $R^{\oplus a} \subset R^{\oplus a + b} \subset R^{\oplus a + b + c}$ and in this case the result is an evident computation. $\square$

Lemma 15.118.5. Let $R$ be a ring. Let $M'$ and $M''$ be two finite projective $R$-modules. Then the diagram

$\xymatrix{ \det (M') \otimes \det (M'') \ar[r] \ar[d]_{\epsilon \cdot (\text{switch tensors})} & \det (M' \oplus M'') \ar[d]^{\det (\text{switch summands})} \\ \det (M'') \otimes \det (M') \ar[r] & \det (M'' \oplus M') }$

commutes where $\epsilon = \det ( -\text{id}_{M' \otimes M''}) \in R^*$ and the horizontal arrows are those of Lemma 15.118.2.

Proof. Omitted. $\square$

Lemma 15.118.6. Let $R$ be a ring. Let $M$, $N$ be finite projective $R$-modules. Let $a : M \to N$ and $b : N \to M$ be $R$-linear maps. Then

$\det (\text{id} + a \circ b) = \det (\text{id} + b \circ a)$

as elements of $R$.

Proof. It suffices to prove the assertion after replacing $R$ by a localization at a prime ideal. Thus we may assume $R$ is local and $M$ and $N$ are finite free. In this case we have to prove the equality

$\det (I_ n + AB) = \det (I_ m + BA)$

of usual determinants of matrices where $A$ has size $n \times m$ and $B$ has size $m \times n$. This reduces to the case of the ring $R = \mathbf{Z}[a_{ij}, b_{ji}; 1 \leq i \leq n, 1 \leq j \leq m]$ where $a_{ij}$ and $b_{ij}$ are variables and the entries of the matrices $A$ and $B$. Taking the fraction field, this reduces to the case of a field of characteristic zero. In characteristic zero there is a universal polynomial expressing the determinant of a matrix of size $\leq N$ in the traces of the powers of said matrix. Hence it suffices to prove

$\text{Trace}((I_ n + AB)^ k) = \text{Trace}((I_ m + BA)^ k)$

for all $k \geq 1$. Expanding we see that it suffices to prove $\text{Trace}((AB)^ k) = \text{Trace}((BA)^ k)$ for all $k \geq 0$. For $k = 1$ this is the well known fact that $\text{Trace}(AB) = \text{Trace}(BA)$. For $k > 1$ it follows from this by writing $(AB)^ k = A(BA)^{k - 1}B$ and $(BA)^ k = (BA)^{k - 1} A B$. $\square$

Recall that we have defined in Algebra, Section 10.55 a group $K_0(R)$ as the free group on isomorphism classes of finite projective $R$-modules modulo the relations $[M'] + [M''] = [M' \oplus M'']$.

Lemma 15.118.7. Let $R$ be a ring. There is a map

$\det : K_0(R) \longrightarrow \mathop{\mathrm{Pic}}\nolimits (R)$

which maps $[M]$ to the class of the invertible module $\wedge ^ n(M)$ if $M$ is a finite locally free module of rank $n$.

Proof. This follows immediately from the constructions above and in particular Lemma 15.118.2 to see that the relations are mapped to $0$. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).