Kernel of surjective map from rank two module is free and rank 1. I am having a bit of trouble with the following question 11(a) (section 2) in Waterhouse affine group schemes. It's really just a module theory question. It asks:
Let $M$ be a free rank $2$ $R$-module. Let $f: M \rightarrow R$ be linear and surjective. Show that $I = \ker (\epsilon)$ is free of rank $1$.
The idea is to show that $f(n)m - f(m)n$ is a basis of $I$, where $m,n$ is a basis of $M$.
Note: This question is the same, but one answer suggests this is false (I think this is the case without commutativity assumptions) and the other gives a complicated proof using exterior algebras - is there is simpler one?
 A: Let $(e_1,e_2)$ be a basis of $M$. Let $a_i =f(e_i)$. Since $f$ is surjective, there exists $y=y_1\cdot e_1+y_2\cdot e_2\in M$ such that $f(y)=1$.
Then we have $a_1y_1+a_2y_2=1$.  Now, $I=\{x=x_1\cdot e_1+x_2\cdot e_2\mid a_1x_1+a_2x_2=0\}$.
Claim. $z=a_2\cdot e_1-a_1\cdot e_2$ is a basis of $I$ (so $I$ is free of rank $1$).
Let $x=x_1\cdot e_1+x_2\cdot e_2\in I$. We have $a_1x_1+a_2x_2=0$, so $a_1y_1x_1+a_2y_1x_2=0$, that is $(1-a_2y_2)x_1+a_2y_1x_2=0$, and $x_1=a_2(y_2x_1-y_1x_2)$. Similarly, multiplying by $y_2$, we get $x_2=-a_1(y_2x_1-y_1x_2)$.
In other words, $x=\alpha\cdot  z$, where $\alpha=y_2x_1-y_1x_2\in R$.
Consequently, $I\subset R\cdot z$. Conversely, $z\in I$ , so  $R \cdot z \subset I$. Finally $I=R\cdot z$.
Hence $z$ generates $I$. It remains to shows that for all $\alpha\in R$, $\alpha\cdot z=0\Rightarrow z=0$ . If $\alpha\cdot  z=0$, since $(e_1,e_2)$ is a basis of $M$, we get $\alpha a_2=0$
and $-\alpha a_1=0$. Then $y_2(\alpha a_2)+y_1(\alpha a_1)=0+0=0=\alpha(a_1y_1+a_2y_2)=\alpha$. Hence $z$ is a basis of $I$, and $I$ is free of rank $1$.
