Base field 4.4.7600.1
Generator \(w\), with minimal polynomial \(x^{4} - 9x^{2} + 19\); narrow class number \(2\) and class number \(1\).
Form
| Weight: | $[2, 2, 2, 2]$ |
| Level: | $[1, 1, 1]$ |
| Dimension: | $4$ |
| CM: | no |
| Base change: | yes |
| Newspace dimension: | $4$ |
Hecke eigenvalues ($q$-expansion)
The Hecke eigenvalue field is $\Q(e)$ where $e$ is a root of the defining polynomial:
| \(x^{4} - 19x^{2} + 72\) |
Show full eigenvalues Hide large eigenvalues
| Norm | Prime | Eigenvalue |
|---|---|---|
| 4 | $[4, 2, -w^{3} + w^{2} + 5w - 6]$ | $\phantom{-}e$ |
| 9 | $[9, 3, -w^{2} - w + 4]$ | $\phantom{-}\frac{1}{6}e^{3} - \frac{13}{6}e$ |
| 9 | $[9, 3, w^{2} - w - 4]$ | $\phantom{-}\frac{1}{6}e^{3} - \frac{13}{6}e$ |
| 11 | $[11, 11, w + 1]$ | $-\frac{1}{3}e^{3} + \frac{10}{3}e$ |
| 11 | $[11, 11, w - 1]$ | $-\frac{1}{3}e^{3} + \frac{10}{3}e$ |
| 19 | $[19, 19, -w]$ | $\phantom{-}\frac{1}{3}e^{3} - \frac{13}{3}e$ |
| 19 | $[19, 19, -w^{2} - w + 6]$ | $-e^{2} + 8$ |
| 19 | $[19, 19, -w^{2} + w + 6]$ | $-e^{2} + 8$ |
| 25 | $[25, 5, 2w^{2} - 9]$ | $-e^{2} + 14$ |
| 29 | $[29, 29, -w^{3} + 4w + 2]$ | $\phantom{-}2e^{2} - 20$ |
| 29 | $[29, 29, -w^{3} + 4w - 2]$ | $\phantom{-}2e^{2} - 20$ |
| 41 | $[41, 41, 2w^{2} - w - 7]$ | $\phantom{-}\frac{1}{6}e^{3} - \frac{19}{6}e$ |
| 41 | $[41, 41, w^{3} - w^{2} - 6w + 4]$ | $\phantom{-}\frac{1}{6}e^{3} - \frac{19}{6}e$ |
| 61 | $[61, 61, -w^{3} + 3w^{2} + 6w - 14]$ | $\phantom{-}\frac{1}{6}e^{3} + \frac{5}{6}e$ |
| 61 | $[61, 61, w^{3} + 2w^{2} - 5w - 8]$ | $\phantom{-}\frac{1}{2}e^{3} - \frac{15}{2}e$ |
| 61 | $[61, 61, -w^{3} + 2w^{2} + 5w - 8]$ | $\phantom{-}\frac{1}{2}e^{3} - \frac{15}{2}e$ |
| 61 | $[61, 61, w^{3} + 3w^{2} - 6w - 14]$ | $\phantom{-}\frac{1}{6}e^{3} + \frac{5}{6}e$ |
| 89 | $[89, 89, -w^{3} + w^{2} + 6w - 9]$ | $-e^{2} + 4$ |
| 89 | $[89, 89, 2w^{3} - w^{2} - 10w + 10]$ | $-e^{2} + 4$ |
| 109 | $[109, 109, -w^{3} + 5w^{2} + 7w - 23]$ | $-\frac{7}{6}e^{3} + \frac{85}{6}e$ |
Atkin-Lehner eigenvalues
This form has no Atkin-Lehner eigenvalues since the level is \((1)\).