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 5, 2013 at 14:16 | comment | added | Will Sawin | This is correct. The product is a polynomial in the $\alpha_i$ and $\beta_i$, with the coefficient of $\beta_{i_j}$ a degree $j-1$ monomial in the $\alpha_i$. Thus from the polynomial we can determine the sequence $i_j$, so it is injective. | |
| Aug 5, 2013 at 14:11 | comment | added | Derrick Stolee | Thanks! I'm pretty sure you are right with the free semigroup, which is how I developed the construction (that is, until I realized I needed inverses as well). | |
| Aug 5, 2013 at 14:10 | vote | accept | Derrick Stolee | ||
| Aug 3, 2013 at 0:21 | comment | added | Michael Zieve | Incidentally, while the $\phi_i$ cannot generate a free group when $k>1$, one can probably show that they generate a free semigroup. I would guess that the answer to the original question would be "yes" if one required all the $e_j$ to equal $1$. | |
| Aug 2, 2013 at 23:57 | history | answered | Michael Zieve | CC BY-SA 3.0 |