Lemma 15.108.6. 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

as elements of $R$.

Lemma 15.108.6. 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 $(BA)^ k = (BA)^{k - 1} A B$. $\square$

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