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I have been trying to work some problems about determining a set of invariant factors given a relations matrix (in the sense of Jacobson) and vice versa. I am stuck and not sure if I am using the right tools. I have a couple questions but first I will state the relevant definitions and facts from Jacobson Algebra I.

Let $D$ be a principal ideal domain and let $M$ be a module over $D$ generated by a finite set $\{x_1, \ldots , x_n\}$. First let $D^{(n)}$ denote the free module of rank $n$ over $D$ with generators $\{e_1, \ldots, e_n \}$ and consider the epimorphism $ \eta : \sum_{1}^{n} a_i e_i \rightarrow \sum_{1}^{n} a_i x_i$, $a_i \in D$. It follows that $ M \cong D^{(n)}/ K$ where $K = \ker \eta$ and by theorem 3.7 of Jacobson $K$ has a base of $m \leq n$ elements denoted by $\{ f_1 , \ldots , f_m \}$ . We then express the generators of $K$ in terms of the base $\{e_1, \ldots, e_n \}$ in the form

$\begin{align*} f_1 &= a_{11} e_1 + a_{12} e_2 + \cdots + a_{1 n}e_n \\ f_2 &= a_{21} e_1 + a_{22}e_2 + \cdots + a_{2 n}e_n \\ \vdots \\ f_m &= a_{m1}e_1 + a_{m 2 } e_2 + \cdots + a_{mn} e_n \end{align*}$

Then the $m \times n$ matrix $A = (a_{ij})$ of these relations is called the relations matrix for the ordered set of generators $\{ f_1 , \ldots , f_m \}$ in terms of the ordered basis $\{e_1, \ldots, e_n \}$.

Fundamental Theorem: Let $D$ be a pid and let $M$ be a finitely generated $D$ module. Then $M$ is isomorphic to the direct sum of finitely many cyclic modules. More precisely, $M \cong D^r \oplus D/(a_1) \oplus D/(a_2) \oplus \ldots \oplus D/(a_m)$ for some integer $ r \geq 0$ and nonzero elements $a_1, a_2, \ldots, a_m$ called the invariant factors which are not units in $D$ and satisfy the divisibility relations $a_1 | a_2 | \ldots | a_m$.

I am trying to use the relations matrix to study the direct sum decomposition of $M$ in terms of the fundamental theorem for finitely generated modules over a p.i.d. I am trying to determine the quickest method for computing invariant factors given a relations matrix (because I will eventually take a qualifying exam on many of these type of problems). So far the only tool I have found in this regard is the following:

Jacobson Theorem 3.9: Let $A$ be an $m \times n$ matrix with entries in a p.i.d $D$ and suppose the rank of $A$ to be $r$. For each $i \leq r$ let $\Delta_i$ be a g.c.d of the $i$-rowed minors of $A$. Then any set of invariant factors for $A$ differ by unit multipliers from the elements $d_1 = \Delta_1, d_2 = \Delta_2 \Delta_{1}^{-1}, \ldots , d_r = \Delta_r \Delta_{r-1}^{-1}$.

Question 1: What is meant by $i$-th rowed minor in this context? The definition I am familiar with is for a minor usually involves $M_{i,j}$ specifying which row and column is "crossed out" before we take a determinant.

Question 2: Let $V$ be a finite-dimensional vector space over the field $F$ and let $T: V \rightarrow V$ be a linear operator. Give $V$ an $F[x]$ module structure by defining $x \alpha = T(\alpha)$ for each $\alpha \in V$. If $F = \mathbb{C}$ and let $ A = \left( \begin{array}{ccc} x^2(x-1)^2 & 0 & 0 \\ 0 & x(x-1)(x-2)^2 & 0 \\ 0 & 0 & x(x-2)^3 \end{array} \right) $ be a relation matrix for V with respect to $\{v_1, v_2, v_3\}$ generators of $V$. How do we use theorem 3.9 to determine the invariant factors of A with respect to $\{v_1, v_2, v_3\}$ or is the a better way than the theorem I listed? (I have struggled to find any examples in this sort of language I am using so any reference suggestions would be greatly appreciated as well).

Question 3: Given a set of invariant factors is it possible to determine a relations matrix corresponding to the invariant factors? Or how to we go to from a normal form to a relations matrix.

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    You get an $i$-rowed minor by picking any $i$ (distinct) rows and any $i$ columns, and computing the determinant of the thus created $i\times i$ matrix. In general there are ${n\choose i}$ ways of selecting the rows and ${n\choose i}$ way of selecting the columns, so ${n\choose i}^2\,$ $i$-rowed minors, so that's a lot of minors. In your exam cases the problem does appear to be manageable as there are many zeros.2011-07-22

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