@Noam D. Elkies : Thanks! I am no expert in this, but what do you mean by: "there is no criterion known for the existence of a negative unit"? I was reading the article "On real quadratic fields containing units with norm -1" by Yokoi and it says $\mathbb{Q}(\sqrt{D})$ has a unit of negative norm if and only if $D-4$ is a square. Or did I misunderstand something? What do you mean by "even $\mathbb{Q}(\sqrt{t^2+1})$"? Does it mean all fields I get using $\sqrt{t^2+4}$ I can get using $\sqrt{t_0^2+1}$ for some $t_0\neq t$?

@Imeasy : Thank you. I knew that. But I don't know the behaviour of the 16 nodes, the tropes etc. with respect to the birational map of the singular Kummers $S_C$ and $S_{C'}$. So I can recover the curves, but what is their relation ship now? Must they be isomorphic?

@Jason Starr : Okay. I phrased the question in terms of isomorphims and the K3 surfaces. My observation for Picard number one still remains. Can something be done for higher Picard numbers?

@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 $a,b\in k^{\times}$.

@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 $j^2=0$?

Thank you! Is it possible, using your description, to write down an explicit rational map $X --> Y$ or $Y --> X$? I am interested in finding a resolution $Z \rightarrow X$ and $Z \rightarrow Y$ of such a map by blow ups, so I can compare cohomolgy of bundles given on $X$ and $Y$ by studying them on $Z$.

Thank you. Now I get it. Nice one-dimensional example. I should have tried that one myself.