Certain isometry overfinite ring is product of isometries over each local factor

Here is how you can write the argument:

You'll need to use the following that the question mentions:

You have $R = R_1 \times \cdots \times R_m$ as the decomposition of $R$ into finite local rings $R_i$.

Given a matrix $A \in \operatorname{Mat}_{m \times n}(R)$, as you mentioned, you can write $A = (A_1, \ldots, A_m)$ where $A_i \in \operatorname{Mat}_{m \times n}(R_i)$.

You also have $\phi: \operatorname{Mat}_{m \times n}(R) \to \operatorname{Mat}_{m \times n}(R)$ as a linear map such that:

$\operatorname{length}_R(\operatorname{rowsp}(\phi(A))) = \operatorname{length}_R(\operatorname{rowsp}(A))$ for every matrix $A$.


Now, to show that $\phi$ decomposes into $\phi_1 \times \cdots \times \phi_m$, where each $\phi_i:\operatorname{Mat}_{m \times n}(R_i) \to \operatorname{Mat}_{m \times n}(R_i)$ preserves the length of the rowspace as an $R_i$-module.

You'll need to consider the projections $\pi_i: R \to R_i$. These projections extend to matrices, giving $\pi_i:\operatorname{Mat}_{m \times n}(R) \to \operatorname{Mat}_{m \times n}(R_i)$.

Defining $\phi_i$ for each $i$ by:

$\phi_i(A_i) = \pi_i(\phi(A)) \quad \text{for } A = (A_1, \ldots, A_m).$

This $\phi_i$ is a well-defined map from $\operatorname{Mat}_{m \times n}(R_i)$ to itself.

You need to check that $\phi_i$ preserves the length of the rowspace as an $R_i$-module.

Consider any matrix $A \in \operatorname{Mat}_{m \times n}(R)$ with decomposition $A = (A_1, \ldots, A_m)$.

We have:

$\phi(A) = (\phi_1(A_1), \ldots, \phi_m(A_m)).$

Since $\phi$ preserves the length of the rowspace, we have:

$\lambda_R(\operatorname{rowsp}(\phi(A))) = \lambda_R(\operatorname{rowsp}(A)).$

Using the fact that $R = R_1 \times \cdots \times R_m$ and the properties of modules over product rings, the length of the rowspace of $A$ in $R$ can be expressed as:

$\lambda_R(\operatorname{rowsp}(A)) = \sum_{i=1}^m \lambda_{R_i}(\operatorname{rowsp}(A_i)).$

Similarly, for $\phi(A)$, we have:

$\lambda_R(\operatorname{rowsp}(\phi(A))) = \sum_{i=1}^m \lambda_{R_i}(\operatorname{rowsp}(\phi_i(A_i))).$

By the preservation property of $\phi$, you get:

$\sum_{i=1}^m \lambda_{R_i}(\operatorname{rowsp}(\phi_i(A_i))) = \sum_{i=1}^m \lambda_{R_i}(\operatorname{rowsp}(A_i)).$

Since this equality holds for all $A$, it must hold componentwise for each $i$. Therefore:

$\lambda_{R_i}(\operatorname{rowsp}(\phi_i(A_i))) = \lambda_{R_i}(\operatorname{rowsp}(A_i)) \quad \text{for each } i.$

So, each $\phi_i$ preserves the length of the rowspace as an $R_i$-module.

So, $\phi$ decomposes into $\phi_1 \times \cdots \times \phi_m$, where each $\phi_i$ is a map on $\operatorname{Mat}_{m \times n}(R_i)$ that preserves the length of the rowspace as an $R_i$-module.