My professor gave us this example of a module that is not free :
$R = k[x,y,z]$ where $k$ is a field. $I = xyR + yz R + xz R \subset R.$ Take $u = xy, v = yz, w = xz$ then $I = uR + vR + wR$, and so $$zu - xv = 0, xv - yw = 0, zu - yw = 0.$$
Claim: I is not a free $R$-module.
And she gave us the following
Proof:
" Assume by contradiction that $I$ is free. Let $S$ be an $R$-basis of $I.$ Choose $s_1, \dots , s_n \in S$ so that:
$$u = r_1s_1 + \dots + r_ns_n, v = r_1^{'}s_1 + \dots + r_n^{'}s_n.$$ Since $zu -xv = 0 = (zr_1 - xr_1^{'})s_1 + \dots + (zr_n - xr_n^{'})s_n$ then $zr_i - xr_i^{'} = 0\ \forall i$ which implies that $z \mid r_i^{'}, x\mid r_i$ (because $R$ is a UFD).
Now, write $r_i^{'} = a_i z, r_i = b_i x.$ Then $xy = u = \sum r_is_i = \sum b_i x s_i$ and so $y = \sum b_i s_i \in I$ which is a contradiction."
But I do not understand why she needed to use that $R$ is a UFD and I do not understand the contradiction. Is not $x,y,z$ are linearly dependent because we have $$\begin{bmatrix} z & -x & 0\\ 0 & x & -y\\ z & 0 & -y \end{bmatrix}\begin{bmatrix} u \\ v \\ w \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix}, $$
$\det A = 0.$
So what is wrong in writing $y$ as a linear combination of elements of the basis?
Could anyone clarify these points to me please?