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 $$\DeclareMathOperator{\sgn}{sgn}\begin{align*} N_{C/A}(c) &= \sum_{\sigma : m \times n \to m \times n} \sgn(\sigma) \prod_{i,j} M_{B/A}(M_{C/B}(c)_{j,\pi_2\sigma(i,j)})_{i,\pi_1\sigma(i,j)} \\ &= \sum_{\tau : m \times n \to n} \sum_{\rho : m \times n \to m} \sgn(\tau, \rho) \prod_{i,j} M_{B/A}(M_{C/B}(c)_{j,\tau(i,j)})_{i,\rho(i,j)} \tag{1} \end{align*}$$ 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.