I'm having trouble with the following proposition (4.2.15) in Springer's Linear Algebraic Groups.
Let $R$ be an integral domain with quotient field $K$, and let $A \in \textrm{Mat}_{m \times n}(R)$. Let $E_R(A)$ be the $R$-submodule of $R^n$ generated by $(a_{i1}, ... , a_{in}) : 1 \leq i \leq m$, and let $\mathcal M_R(A) = R^n/E_R(A)$. Notice for $S \subseteq R$ multiplicatively closed, $S^{-1}\mathcal M_R(A) = S^{-1} (R^n/E_R(A))$ can be identified as $S^{-1}R$-modules with $(S^{-1}R)^n/E_{S^{-1}R}(A) = \mathcal M_{S^{-1}R}(A)$.
Here's the proposition:
(i) There is an $f \in R$ such that $\mathcal M_{R_f}(A)$ is free of rank $n-r$, where $r$ is the rank of $A$.
(ii) There is an $f \in R$ satisfying (i) and also the following property: if $e_1, ... , e_n$ is the standard basis of $R_f^n$, then there is a subset of $n-r$ elements of $\{e_1, ... , e_n\}$ whose image in $\mathcal M_{R_f}(A)$ is an $R_f$-basis for $\mathcal M_{R_f}(A)$.
(i) is easy: you can find $B \in \textrm{GL}_m(K), C \in \textrm{GL}_n(K)$ such that $BAC$ is in the form $\begin{pmatrix} I_r & 0 \\ 0 & 0 \end{pmatrix}$ Taking a common denominator $f$ for all the entries of $B, C$ and their inverses, everything is in $R_f$ and of course $\mathcal M_{R_f}(A) \cong \mathcal M_{R_f}(BAC)$ as $R_f$-modules, and $\mathcal M_{R_f}(A)$ is clearly free of rank $n - r$.
I'm confused about the proof of (ii). Since localization is transitive, let's assume without loss of generality that $\mathcal M_R(A)$ is already free over $R$ of rank $n - r$. The proof says to take a basis $f_1, ... , f_{n-r}$ of $\mathcal M_R(A)$. Springer indicates that the images $e_i'$ in $R^n/E_R(A) = \mathcal M_R(A)$ of some subset $e_{r+1}, ... , e_n$ of the standard basis for $R^n$ are linearly independent over $K$.
I'm confused first of all about what this means. The $A$-module $R^n/E_R(A)$ is not an $K$-module. Is he saying that this sits inside $K^n/E_K(A)$?
Okay I figured it out. Since $M_R(A)$ is a free $R$-module of rank $n-r$, the point is that it naturally sits inside its localization $M_R(A) \otimes_R K = M_K(A)$, and we have a commutative diagram of $R$-modules $$\begin{matrix} R^n & \hookrightarrow & K^n \\ \downarrow & & \downarrow \\ \mathcal M_R(A) & \hookrightarrow & \mathcal M_K(A) \end{matrix}$$ where the vertical arrows are surjective. Now letting $e_i$ be the standard basis for $R^n$, the point is that the images $\overline{e_i}$ in $\mathcal M_K(A)$ span this vector space over $K$. Hence they can be reduced to a basis in some way, say (after relabelling) $\overline{e_{r+1}}, ... , \overline{e_n}$. But all these $\overline{e_i}$ are really in $\mathcal M_R(A)$, and they are linearly independent over $K$, hence over $R$. They might not span anymore, but that's okay, that's what this whole proof is for.