Timeline for On a special type of subring of $\mathbb C[x_0,...,x_{q-1}]$
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 26, 2019 at 6:35 | vote | accept | user521337 | ||
| Dec 19, 2018 at 13:35 | comment | added | Julian Rosen | Using the multinomial theorem, $y_k^p=\sum_{j_i} \frac{p!}{j_0!\cdots j_{q-1}!}x_0^{q_0}\cdots x_{q-1}^{j_{q-1}}\zeta^{k(0j_0+1j_1+\ldots (q-1)j_{q-1})}$. Now if $0j_0+1j_1+\ldots (q-1)j_{q-1}\equiv m\mod p$, then $\zeta^{k(0j_0+1j_1+\ldots (q-1)j_{q-1})}= \zeta^{km}$. So $y_k^p=\sum_m \zeta^{km}\sigma_m$. | |
| Dec 18, 2018 at 3:10 | comment | added | user521337 | ok, yeah I got that, thanks ... could you please explain why $y_k^p=\omega_k$ ? | |
| Dec 17, 2018 at 17:04 | comment | added | Julian Rosen | Sorry, I should have said this better. The matrix taking $x_0,\ldots,x_{q-1}$ to $y_0,\ldots,y_{q-1}$ is square and Vandermonde, so the span of the $x$'s equals the span of $y_0,\ldots,y_{q-1}$. The extra $y$'s ($y_q,\ldots,y_{p-1}$) are by definition in the span of the $x$'s. | |
| Dec 17, 2018 at 5:57 | comment | added | user521337 | Why is $\mathbb C$-span of $\{y_i\}$ the same as the $\mathbb C$-span of $\{x_i\}$ ? the change matrix is not even a square matrix ... | |
| Dec 16, 2018 at 14:25 | history | answered | Julian Rosen | CC BY-SA 4.0 |