Kernel of a surjective module morphism $f : R^2 \rightarrow R$ is a free submodule I have to prove this :

Let $R$ be a commutative ring and : $f : R^2 \rightarrow R$ a morphism of
  $R$-module. I want to show : $\ker f$ is a free module.

My first try is to say : $R^2/\ker(f) \cong R$, so : $R^2/\ker(f) = <\overline{\alpha}>$, with $\alpha \notin \ker(f)$ (otherwise, $f$ would not be surjective). But it brings me nowhere.
So, my second try (inspired of a question posted here) : let $u \in R^2$ such that : $f(u) = 1$ by the surjectivity of $f$, and we are considering $V = <u>$. We can show that :
$$ R^2 = V \bigoplus \ker(f)$$
We have aswell, by considering a base $(e_1, e_2)$ of $R^2$, that : $e_1 = z_1 + \lambda_1u$, and $e_2 = z_2 + \lambda_2u$, with $z_1, z_2 \in \ker(f)$. Then, let $x \in \ker(f)$. We have :
$x = x_1e_1 + x_2e_2 = x_1z_1 + x_2z_2 + x_1\lambda_1u + x_2\lambda_2u$, so :
$x- (x_1z_1 + x_2z_2) \in V \cap \ker(f) = {0}$, and then : $x = x_1z_1 + x_2z_2$, and then : $\ker f = <z_1, z_2>$.
The problem is even if $(z_1, z_2)$ is generating $\ker f$, I don't succeed to show it's a free family.
Someone could help me ? 
Thank you !
 A: It is false without hypotheses on the ring $R$. If $f$ is surjective, all you can say is that the short exact sequence
$$0\longrightarrow \ker f\longrightarrow R^2\xrightarrow{\; f\;\,} R\longrightarrow 0$$
splits, i.e. $\;R^2\simeq\ker f\oplus R$, which means $ker f$ is a direct summand of the free $R$-module $R^2$, i.e. it is a projective $R$-modulde (of finite type).
However, whereas the converse is true, projective  modules are not necessarily free, without hypotheses on $R$, for instance $R$ is a P.I.D or a polynomial ring over a field, or a local ring.
A: Let $R$ be commutative, and $K$ your kernel. If $f$ is surjective,
then the short exact sequence
$$0\to K\to R^2\to R\to0$$
splits, as $R$ is projective as an $R$-module. Therefore
$$R^2\cong R\oplus K.\tag{*}$$
As $R$ is commutative, we have exterior powers. Taking the exterior
square of (*) gives
$$R\cong K\oplus {\bigwedge}^2 K.$$
Taking the exterior cube gives
$$\{0\}\cong {\bigwedge}^2 K\oplus {\bigwedge}^3 K.$$
Therefore ${\bigwedge}^2 K=\{0\}$ and so $R\cong K$.
Again, this is assuming that $f$ is surjective.
I have used the formulae
$${\bigwedge}^m(A\oplus B)\cong\bigoplus_{k=0}^m\left(
{\bigwedge}^k A\otimes {\bigwedge}^{m-k} B\right)$$
and
$${\bigwedge}^m R^n\cong R^{\binom nm}.$$
