# A quaternion x generates a left ideal of rank 2 if and only if x, ix and jx are linearly dependent?

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

• For $S'$ the spectrum of a field, every left ideal in $A\otimes_{\mathcal{O}_S} \mathcal{O}_{S'}$ has rank equal to either $0$ (for the zero ideal), $2$ (for the ideals that you are considering), or $4$ (for the ideal that is the entire algebra). Thus, if there is any nontrivial linear relation among the four generators $x$, $ix$, $jx$, $ijx$ of this left ideal, then the rank is not $4$, so the ideal is either the zero ideal or a left ideal of rank $2$. – Jason Starr Jan 13 '17 at 15:09
• @Jason Starr : Okay. But why are the only possibilities for the rank 0, 2 and 4? I see that this is the case if $A$ was a "real" quaternion algebra, so that if we split it, we just have a 4-dim matrix algebra. But here the equation $t=0$ of $C$ comes into play. So we cannot get the whole matrix algebra I guess. Why are rank one and three not possible, for example if $S'$ is the residue field of a poit on $C$, where the relation $j^2=bt$ gets $j^2=0$? – Bernie Jan 14 '17 at 8:23
• I advise you to read this in a book about central simple algebras. The claim about the dimensions can be checked after base change from Spec of the field to Spec of the algebraic closure. However, already after adjoining the square roots of $a$, $b$ and $-1$, the base change of your quaternion algebra is isomorphic to the algebra of all $2\times 2$ matrices. – Jason Starr Jan 14 '17 at 8:56
• @Jason Starr : Don't we also need to adjoin a square root of $t$? My problem is not with cenral simple algebras but rather with with this special kind of quaternion-ish algebra. It degenerates over the curve $C$ to an algebra, which is not a quaternion algebra, because on $C$ it is given by the symbol $(a,0)$. But a quaternion algebra is given by a symbol $(a,b)$ with $a,b\in k^{\times}$. – Bernie Jan 16 '17 at 10:51
• Yes, please adjoin a square root of $bt$. I read the post too quickly. I assumed that $t$ had been absorbed into the element $b$. Anyway, the rank of a finitely generated module over a local ring equals the dimension of the base change to a fraction field. So the claim you state in your post can be checked after base change to the algebraic closure of the fraction field. – Jason Starr Jan 16 '17 at 11:45