## Properties of the columns of a relations matrix

Let $R$ be a Euclidean domain and let $M$ be a finitely generated (left, unital) $R$-module. By this previous exercise, there is a surjective module homomorphism $\varphi : R^n \rightarrow M$ for some natural number $n$. By this previous exercise, $\mathsf{ker}\ \varphi \subseteq R^n$ is finitely generated as an $R$-module. If we let $B = \{x_1,\ldots,x_n\}$ be an (ordered) basis for $R^n$ and let $S = \{y_1, \ldots, y_m\}$ be an (ordered) generating set for $\mathsf{ker}\ \varphi$, then for each $y_i$, we have unique elements $a_{i,j} \in R$ such that $y_i = \sum_{j=1}^n a_{i,j} x_j$.

We now construct the $m \times n$ matrix $A = [a_{i,j}]_{i=1,j=1}^{m,n}$ whose $i$th row is simply the coefficients of $y_i$. This $A = \mathsf{Mat}^B_S(\varphi)$ is called a relations matrix with respect to the homomorphism $\varphi$, the basis $B$, and the generating set $S$.

1. Show that interchanging $x_a$ and $x_b$ in the basis $B$ has the effect of interchanging the $a$th and $b$th columns of $A$.
2. Show that, for any $r \in R$, replacing the element $x_b$ in $B$ by $x_b - rx_a$ ($a \neq b$) results in another basis. Show further that this has the effect of adding $r$ times the $b$th column to the $a$th column of $A$.

It is clear that interchanging $x_a$ and $x_b$ in $B$ merely interchanges the $a$th and $b$th columns of $A$, since the rows of $A$ are precisely the (ordered) expansions of the $y_i$ in terms of the $x_j$.

Now we will show that replacing $x_b$ by $x_b - rx_a$ results in a new basis. Recall that a basis is a linearly independent generating set. To see that $B^\prime = \{x_1, \ldots, x_a, \ldots \ldots, x_b - rx_a, \ldots, x_n\}$ is linearly independent, note that if $0 = \sum_{j \neq a,b} s_jx_j + s_ax_a + s_b(x_b - rx_a) = \sum_{j \neq a,b} s_jx_j + (s_a - rs_b)x_a + s_bx_b$, then $s_j = 0$ for all $j \neq a$ since $B$ is linearly independent, and then $s_a = 0$. Now if $\alpha \in R^n$, then since $B$ is a generating set, we have $\alpha = \sum s_jx_j$ for some $s_j \in R$. Certainly then $\alpha = \sum_{j \neq a,b} s_jx_j + (s_a + rs_b)x_a + s_b(x_b - rx_a)$, so that $B^\prime$ is a generating set, and thus a basis, for $R^n$.

Our argument that $B^\prime$ is a generating set now shows that the matrix for $B^\prime$ is obtained from that for $B$ by adding $r$ times column $b$ to column $a$.