Timeline for Algebraically Independent Numbers and Affine Linear Maps
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 6, 2013 at 21:25 | comment | added | Michael Zieve | @DerrickStolee: Will Sawin isn't saying that the $\phi_i$ themselves are translations. He's arguing that, if $G$ is the group generated by your finitely many maps $\phi_i:x\mapsto \alpha_i x+\beta_i$, then the map $f\mapsto f'(0)$ defines a group homomorphism $G\to\mathbb{R}^*$. The kernel of this homomorphism consists of the translations in $G$, and hence is abelian. The image of the homomorphism is a subgroup of $\mathbb{R}^*$, so it too is abelian. Thus $G$ is solvable, so if $k>1$ then $G$ is not free. | |
| Aug 5, 2013 at 14:10 | comment | added | Derrick Stolee | The case k=1 is not important to me. Also, the group of translations are not within this group, since that would require $\alpha_1 = 1$, which is not transcendental over the rationals. | |
| Aug 3, 2013 at 0:00 | comment | added | Will Sawin | Even if it were trivial, say if $k=1$, then the group would still be solvable. | |
| Aug 2, 2013 at 20:27 | comment | added | Felipe Voloch | The way he is defining his group, the subgroup of translations may be trivial. | |
| Aug 2, 2013 at 19:29 | history | answered | Will Sawin | CC BY-SA 3.0 |