Another thing your argument shows: $\rho_K\left(A,B\right)=\rho_K\left(B,A\right)$. Okay, this actually follows from $A$ being similar to $A^T$ and $B$ to $B^T$, but this is another consequence of your argument.

By the way, nice exercise for linear algebra classes (if someone reading this is holding any): Given a homomorphism between two chain complexes of vector spaces over a field $k$. (Both chain complexes are assumed to be bounded from below.) If this homomorphism is $0$ in homology, then show that it is a chain homotopy (i. e., $dw+wd$ for some $w$). But please check it first; I am not 100% sure about my proof.

I am not done thinking over this yet, but let me add that tensoring with $\mathbb{Z}\slash p$ and checking whether it's still homologically zero is not enough. For instance, if we alter David's second example such that the horizontal maps are multiplication by $p^2$ and the vertical map is multiplication by $p$. In this case, our map of chain complexes is homologically $0$ and remains so after tensoring with $\mathbb{Z}\slash p$ and $\mathbb{Z}\slash q$ for any other prime $q$, but it's not $du+vd$. So we should at least try $\mathbb{Z}\slash p^n$, if not even all $\mathbb{Z}\slash m$.

..., and if $a_i^{\ast}a_j=0$ for all $i\neq j$, then it also satisfies $v^{\ast}v=a_1^{\ast}a_1\cdot a_2^{\ast}a_2\cdot ...\cdot a_{n-1}^{\ast}a_{n-1}$ (due to the Cauchy-Binet formula), so that when $a_1$, $a_2$, ..., $a_{n-1}$ are $n-1$ normalized eigenvectors of $H$, then $v$ is a normalization of the $n$-th one.

I forgot to say: "The (generalized to $n$ dimensions) cross product" of $n-1$ vectors $a_1$, $a_2$, ..., $a_{n-1}$ in $\mathbb{C}^n$ means the vector $v=\left(v_1,v_2,...,v_n\right)$, where $v_i$ is the conjugate of ($\left(-1\right)^{i-1}\cdot$ the determinant of the matrix formed by the vectors $a_1$, $a_2$, ..., $a_{n-1}$ without their respective $i$-th coordinates) for every $i$. This vector $v$ satisfies $a_i^{\ast}v=0$ for every $i$.

So I assume that you are talking about the second example, and the question is whether any map between complexes of $\mathbb{Z}$-modules that is homologically $0$ even after tensoring with any arbitrary $\mathbb{Z}\slash p$ must be a map of the form $du+vd$. Well, this is an interesting question which I'll try to examie. But I fear that I don't really have the prerequisites for this, because I don't even know the definition of a spectral sequence.

"in David's examples": do you mean David's second example? As far as I understand, he made two examples: one of a $du+vd$ map which is not a $dw+wd$ map, and one of a homologically trivial map which is not a $du+vd$ map. The former can only become "better" (i. e., become $dw+wd$) after tensoring with $\mathbb{Z}\slash p$ (and it actually MUST become $dw+wd$, because after tensoring with $\mathbb{Z}\slash p$, every module becomes a vector space, and for vector spaces, $du+vd$ always rewrites as $dw+wd$ - at least, if our complex is bounded from one side).

Great! I knew about invariant factors, but I didn't make the observation that they are invariant under base change (i. e., that the $f_i$ are the same over $K$ and over $L$). Of course, this is trivial, but one doesn't think of it if one has always been thinking of primary decomposition instead of invariant factors.

"Componentwise" or "coordinatewise" sounds more like standard terminology than "piecewise", but I don't know a really common name for this.

Thanks! Though honestly I still have no idea when I have to use the grave accent and when I don't - seems perfectly random to me. But using it is a safe way, at least.

Still, even when the Bockstein homomorphisms are well-defined, what does "preserving" them mean? Aren't they canonical and therefore always preserved by a map of chain complexes?