Skip to main content
MathOverflow

Return to Question

Inserted missing "unit" notation on the target.
Source Link
BCnrd
BCnrd
  • 7.1k
  • 2
  • 66
  • 74

Suppose $K$ is a quadratic imaginary field, and $\phi:G_K\rightarrow \overline{\mathbb{Q}}$$\phi:G_K\rightarrow \overline{\mathbb{Q}}^{\times}$ is a finite order Galois character satisfying $\phi\phi^c=1$ (where $c$ is a complex conjugation). Can we necessarily write $\phi=\psi^c/\psi$ for some finite order character $\psi:G_K\rightarrow \overline{\mathbb{Q}}$$\psi:G_K\rightarrow \overline{\mathbb{Q}}^{\times}$? What about for a CM field?

Suppose $K$ is a quadratic imaginary field, and $\phi:G_K\rightarrow \overline{\mathbb{Q}}$ is a finite order Galois character satisfying $\phi\phi^c=1$ (where $c$ is a complex conjugation). Can we necessarily write $\phi=\psi^c/\psi$ for some finite order character $\psi:G_K\rightarrow \overline{\mathbb{Q}}$? What about for a CM field?

Suppose $K$ is a quadratic imaginary field, and $\phi:G_K\rightarrow \overline{\mathbb{Q}}^{\times}$ is a finite order Galois character satisfying $\phi\phi^c=1$ (where $c$ is a complex conjugation). Can we necessarily write $\phi=\psi^c/\psi$ for some finite order character $\psi:G_K\rightarrow \overline{\mathbb{Q}}^{\times}$? What about for a CM field?

Source Link
blt
blt
  • 1.2k
  • 1
  • 10
  • 13

Can every Galois character phi with phi phi^c = 1 be written as psi/psi^c?

Suppose $K$ is a quadratic imaginary field, and $\phi:G_K\rightarrow \overline{\mathbb{Q}}$ is a finite order Galois character satisfying $\phi\phi^c=1$ (where $c$ is a complex conjugation). Can we necessarily write $\phi=\psi^c/\psi$ for some finite order character $\psi:G_K\rightarrow \overline{\mathbb{Q}}$? What about for a CM field?