Fix a dimension $n \geq 1$ and a number $k \geq 1$ and suppose that $a\_1,...,a\_k \in \mathbb{R}^n$ are points in $n$-dimensional space such that among all the $nk$ coordinates there is a subset of size at least $nk - \binom{n+1}{2}$ that is algebraically independent (over $\mathbb{Q}$).


Is there an isometry $T$ of $\mathbb{R}^n$ such that the set of the coordinates of $Ta\_1,...,Ta\_k$ is algebraically independent? In other words, can we "erase" the algebraic dependency using isometries? Note that the group of isometries acting on $\mathbb{R}^n$ has $\binom{n+1}{2}$ degrees of freedom, that is, we can prescribe the position of $\binom{n+1}{2}$ coordinates, so intuitively the conditions are the weakest such that one can still hope that the answer is positive. 


This is easy for $n=1$, and in general, it is easy to see how translations affect algebraic dependency. [This paper](https://link.springer.com/article/10.1007/s00454-005-1225-8) has an explicit construction for $n=2$ (Lemma 3.5), but it seems difficult to generalize this to arbitrary dimension (or maybe it's just that my knowledge of the Euclidean group is lacking). I wonder if there's a high-level approach to this, or indeed, if the interplay between the 'algebraic' and 'geometric' structures on $\mathbb{R}^n$, given by algebraic independence and isometries, respectively, has been studied before.



Algebraic independence of points under isometry
The answer is no.


A rational edge length is an algebraic dependence between the coordinates that you cannot get rid of by an isometry.
One can find $k$ points in $\mathbb{R}^n$ with all but one coordinate algebraically independent and $\sum\_{i=1}^n (a\_1^i - a\_2^i)^2 = 1$.
(For this, take $k$ generic points with $a\_1$ and $a\_2$ at distance less than $1$, and then change one of the coordinates of $a\_2$ so that the distance from $a\_1$ becomes $1$.)


