Krdytkyu

Let $A subseteq B subseteq C$ be a tower of commutative rings, such that $B$ is free of rank $m$ as an $A$-module and $C$ is free of rank $n$ as a $B$-module.

For $b in B$, we can define the norm $N_{B/A}(b)$ as the determinant of multiplication by $b$ seen as a linear map $B to B$. If $A,B,C$ are fields, one can show that

$$N_{C/A} = N_{B/A} circ N_{C/B}$$

using properties of their automorphism groups. See for example Associativity of norms in inseparable extensions. A set of handwritten notes of a number theory course mentions (exercise) that this is true for general rings. I couldn't find any proof of this and haven't succeeded proving it:

Attempt. If we fix a basis $(b_i)$ of $B/A$, a basis $(c_j)$ of $C/B$ so that the $b_ic_j$ are a basis of $C/A$, then for the matrix representation of left multiplication by $c in C$ we have $M_{C/A}(c)_{(i,j), (k,l)} = M_{B/A}(M_{C/B}(c)_{(j,l)})_{(i,k)}$. Using this, the abuse of notation $m = {1, ldots, m }$ and defining the sign of a function $f : S to S$ to be zero when it is not a permutation, I get
$$DeclareMathOperator{sgn}{sgn}begin{align*}
N_{C/A}(c) &= sum_{sigma : m times n to m times n} sgn(sigma) prod_{i,j} M_{B/A}(M_{C/B}(c)_{j,pi_2sigma(i,j)})_{i,pi_1sigma(i,j)} \
&= sum_{tau : m times n to n} sum_{rho : m times n to m} sgn(tau, rho) prod_{i,j} M_{B/A}(M_{C/B}(c)_{j,tau(i,j)})_{i,rho(i,j)} tag{1}
end{align*}$$
where I write $pi_1, pi_2 : m times n to m, n$ for the projections. For the composition of norms, I find
$$begin{align*}
N_{B/A}left( N_{C/B}(c) right) &= sum_{sigma : m to m} sgn(sigma) prod_{i=1}^m M_{B/A} left( sum_{tau : n to n} sgn(tau) prod_{j=1}^n M_{C/B}(c)_{j, tau(j)} right)_{i, sigma(i)} \
&= sum_{tau : m times n to n} left( prod_{i=1}^m sgn left( tau|_{{i} times n} right) right) sum_{rho : m times n to m} sgn left( rho|_{m times {n}} right) \
&prod_{i=1}^m prod_{j=1}^n M_{B/A} left( M_{C/B}(c)_{j, tau(i, j)} right)_{rho(i,j-1), rho(i, j)} qquad(rho(i,0)=i) tag{2}
end{align*}$$
using the linearity of $M_{B/A}$, expanding its argument as a matrix product and expanding the product over $i$.

Observe that $(1)$ and $(2)$ look similar, but are still very different. I can get some cancellation among the terms in $(2)$, but not enough to obtain $(1)$.

How to prove that $(1)=(2)$?

Independent observations:

• When $A$ is a field, one can reduce to the case of fields as follows: take a maximal ideal $mathfrak m$ of $C$ to obtain an extension of fields $A = A/mathfrak m_A subseteq B/mathfrak m_B subseteq C/mathfrak m$, and use that the norm commutes with projections on quotients.


• When $A$ is an integral domain, localizing everything in $A^times$ reduces to the case where $A$ is a field.


• When $A$ is reduced, tensoring by $A/mathfrak p$ for all prime ideals of $A$ reduces to the case where $A$ is an integral domain.


• For general $A$, this gives that the difference $N_{C/A}(c) - N_{B/A}(N_{C/B}(c))$ is nilpotent.

In fact, the reference from the comments: Cassels, Local Fields, Appendix B Lemma 4 is for general rings, not just for fields.