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 ith 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 ath and bth 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 bth column to the ath column of A.

It is clear that interchanging x_a and x_b in B merely interchanges the ath and bth 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.

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