I am trying to understand the construction of Artin and Mumford of a non-rational unirational threefold in ([1], p.90).

Assume $S$ is a smooth projective surface over $\mathbb{C}$ with a smooth curve $C\subset S$.

Let $R$ be the local ring of a point $p_0\in S$ and let $A$ be the $R$-algebra generated by elements $i,j$ with relations $i^2=a$, $j^2=bt$ and $ij=-ji$ where $a$ and $b$ are units in $R$ and $t=0$ is a local equation for $C$. So $A$ is some kind of generalized quaternion algebra over $R$.

Let $L\subset A\otimes O_{S'}$ be a left ideal, locally free of rank 2 such that $(A\otimes O_{S'})/L$ is also free of rank 2 for some $S'\rightarrow S$. All such ideals are principal.

Such an ideal contains, up to scalars, a unique element $x\in L$ with $x=p+qi+rj$. Then we have $\text{rank}((A\otimes O_{S'})x)\geq 2$ since $x$ and $ix$ are linearly independent.

The claim I am trying to understand is:

- The element $x$ generates a left ideal of rank two, i.e. ($\text{rank}((A\otimes O_{S'})x)=2$) if and only if $x$, $ix$ and $jx$ are linearly dependent.

The direction $\text{rank}((A\otimes O_{S'})x)=2 =>$ linearly dependent is clear to me.

I can see that $x, ix, jx$ and $ijx$ form a generating set of $(A\otimes O_{S'})x$ over $O_{S'}$.

But how to show that the rank of $(A\otimes O_{S'})x$ is two if these three vectors are linearly dependent? Because the linear relation reads like $b_1x+b_2ix+b_3jx=0$ but these $b_i$ don't need to be units, so I cannot just use $b_i^{-1}$ to replace them in the representaion of elements of $(A\otimes O_{S'})x$. How to go on here?

[1] Artin and Mumford's paper