Skip to main content
MathOverflow

Return to Answer

added 1 character in body
Source Link
GH from MO
GH from MO
  • 105.2k
  • 8
  • 292
  • 398

$\psi_N$ is a Größencharakter modulo $\mathfrak{m}=\mathcal{O}_K$$\mathfrak{m}:=\mathcal{O}_K$ in the following way. Let $\chi_f$ be the trivial character on the $1$-element group $(\mathcal{O}_K/\mathfrak{m})^\times$, and let $\chi_\infty(z,\overline{z}):=\overline{z}^N$. Then, for any nonzero $a\in\mathcal{O}_K$, we have $\psi_N((a))=\overline{a}^N=\chi_f(a)\chi_\infty(a)$, and we are done.

Note that the condition $w_K\mid N$ ensures that $\psi_N$ is well-defined: if $a,b\in\mathcal{O}_K$ are nonzero, and $(a)=(b)$, then $a=bu$ for some unit $u\in\mathcal{O}_K^\times$, and hence $a^N=(bu)^N=b^N$.

$\psi_N$ is a Größencharakter modulo $\mathfrak{m}=\mathcal{O}_K$ in the following way. Let $\chi_f$ be the trivial character on the $1$-element group $(\mathcal{O}_K/\mathfrak{m})^\times$, and let $\chi_\infty(z,\overline{z}):=\overline{z}^N$. Then, for any nonzero $a\in\mathcal{O}_K$, we have $\psi_N((a))=\overline{a}^N=\chi_f(a)\chi_\infty(a)$, and we are done.

Note that the condition $w_K\mid N$ ensures that $\psi_N$ is well-defined: if $a,b\in\mathcal{O}_K$ are nonzero, and $(a)=(b)$, then $a=bu$ for some unit $u\in\mathcal{O}_K^\times$, and hence $a^N=(bu)^N=b^N$.

$\psi_N$ is a Größencharakter modulo $\mathfrak{m}:=\mathcal{O}_K$ in the following way. Let $\chi_f$ be the trivial character on the $1$-element group $(\mathcal{O}_K/\mathfrak{m})^\times$, and let $\chi_\infty(z,\overline{z}):=\overline{z}^N$. Then, for any nonzero $a\in\mathcal{O}_K$, we have $\psi_N((a))=\overline{a}^N=\chi_f(a)\chi_\infty(a)$, and we are done.

Note that the condition $w_K\mid N$ ensures that $\psi_N$ is well-defined: if $a,b\in\mathcal{O}_K$ are nonzero, and $(a)=(b)$, then $a=bu$ for some unit $u\in\mathcal{O}_K^\times$, and hence $a^N=(bu)^N=b^N$.

deleted 34 characters in body
Source Link
GH from MO
GH from MO
  • 105.2k
  • 8
  • 292
  • 398

$\psi_N$ is a Größencharakter modulo $\mathfrak{m}=\mathcal{O}_K$ in the following way. Let $\chi_f$ be the trivial character on the $1$-element group $(\mathcal{O}_K/\mathfrak{m})^\times$, and let $\chi_\infty(z,\overline{z}):=\overline{z}^N$. Then, for any nonzero $a\in\mathcal{O}_K$, we have $\psi_N((a))=\overline{a}^N=\chi_f(a)\chi_\infty(a)$, and we are done.

Note that the condition $w_K\mid N$ ensures that $\psi_N$ is well defined-defined: if $a,b\in\mathcal{O}_K$ generate the sameare nonzero principal ideal in, and $\mathcal{O}_K$$(a)=(b)$, then $a=bu$ for some unit $u\in\mathcal{O}_K^\times$, and hence $a^N=(bu)^N=b^N$.

$\psi_N$ is a Größencharakter modulo $\mathfrak{m}=\mathcal{O}_K$ in the following way. Let $\chi_f$ be the trivial character on the $1$-element group $(\mathcal{O}_K/\mathfrak{m})^\times$, and let $\chi_\infty(z,\overline{z}):=\overline{z}^N$. Then, for any nonzero $a\in\mathcal{O}_K$, we have $\psi_N((a))=\overline{a}^N=\chi_f(a)\chi_\infty(a)$, and we are done.

Note that the condition $w_K\mid N$ ensures that $\psi_N$ is well defined: if $a,b\in\mathcal{O}_K$ generate the same nonzero principal ideal in $\mathcal{O}_K$, then $a=bu$ for some unit $u\in\mathcal{O}_K^\times$, and hence $a^N=(bu)^N=b^N$.

$\psi_N$ is a Größencharakter modulo $\mathfrak{m}=\mathcal{O}_K$ in the following way. Let $\chi_f$ be the trivial character on the $1$-element group $(\mathcal{O}_K/\mathfrak{m})^\times$, and let $\chi_\infty(z,\overline{z}):=\overline{z}^N$. Then, for any nonzero $a\in\mathcal{O}_K$, we have $\psi_N((a))=\overline{a}^N=\chi_f(a)\chi_\infty(a)$, and we are done.

Note that the condition $w_K\mid N$ ensures that $\psi_N$ is well-defined: if $a,b\in\mathcal{O}_K$ are nonzero, and $(a)=(b)$, then $a=bu$ for some unit $u\in\mathcal{O}_K^\times$, and hence $a^N=(bu)^N=b^N$.

Source Link
GH from MO
GH from MO
  • 105.2k
  • 8
  • 292
  • 398

$\psi_N$ is a Größencharakter modulo $\mathfrak{m}=\mathcal{O}_K$ in the following way. Let $\chi_f$ be the trivial character on the $1$-element group $(\mathcal{O}_K/\mathfrak{m})^\times$, and let $\chi_\infty(z,\overline{z}):=\overline{z}^N$. Then, for any nonzero $a\in\mathcal{O}_K$, we have $\psi_N((a))=\overline{a}^N=\chi_f(a)\chi_\infty(a)$, and we are done.

Note that the condition $w_K\mid N$ ensures that $\psi_N$ is well defined: if $a,b\in\mathcal{O}_K$ generate the same nonzero principal ideal in $\mathcal{O}_K$, then $a=bu$ for some unit $u\in\mathcal{O}_K^\times$, and hence $a^N=(bu)^N=b^N$.