A question about Galois characters












1












$begingroup$


Let $F$ be a number field and $chi:mathrm{Gal}(overline{mathbb{Q}}/F)tooverline{mathbb{Q}_ell}^{times}$ ($ell$ a prime) a Galois character. My question is: Can we find a finite extension $K/F$ such that $chi_{|mathrm{Gal}(overline{mathbb{Q}}/K)}=1$






number-theory algebraic-number-theory galois-representations







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  • $begingroup$
    this might be what you're looking for
    $endgroup$
    – Ryan Keleti
    Jan 16 at 23:08
















1












$begingroup$


Let $F$ be a number field and $chi:mathrm{Gal}(overline{mathbb{Q}}/F)tooverline{mathbb{Q}_ell}^{times}$ ($ell$ a prime) a Galois character. My question is: Can we find a finite extension $K/F$ such that $chi_{|mathrm{Gal}(overline{mathbb{Q}}/K)}=1$






number-theory algebraic-number-theory galois-representations







share|cite|improve this question









$endgroup$












  • $begingroup$
    this might be what you're looking for
    $endgroup$
    – Ryan Keleti
    Jan 16 at 23:08














1












1








1





$begingroup$


Let $F$ be a number field and $chi:mathrm{Gal}(overline{mathbb{Q}}/F)tooverline{mathbb{Q}_ell}^{times}$ ($ell$ a prime) a Galois character. My question is: Can we find a finite extension $K/F$ such that $chi_{|mathrm{Gal}(overline{mathbb{Q}}/K)}=1$






number-theory algebraic-number-theory galois-representations







share|cite|improve this question









$endgroup$




Let $F$ be a number field and $chi:mathrm{Gal}(overline{mathbb{Q}}/F)tooverline{mathbb{Q}_ell}^{times}$ ($ell$ a prime) a Galois character. My question is: Can we find a finite extension $K/F$ such that $chi_{|mathrm{Gal}(overline{mathbb{Q}}/K)}=1$






number-theory algebraic-number-theory galois-representations




number-theory algebraic-number-theory galois-representations






share|cite|improve this question













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asked Jan 16 at 23:05









AZMEHAZMEH

133




133












  • $begingroup$
    this might be what you're looking for
    $endgroup$
    – Ryan Keleti
    Jan 16 at 23:08


















  • $begingroup$
    this might be what you're looking for
    $endgroup$
    – Ryan Keleti
    Jan 16 at 23:08
















$begingroup$
this might be what you're looking for
$endgroup$
– Ryan Keleti
Jan 16 at 23:08




$begingroup$
this might be what you're looking for
$endgroup$
– Ryan Keleti
Jan 16 at 23:08










1 Answer
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2












$begingroup$

By Galois theory, your question is equivalent to asking whether all $ell$-adic Galois characters have finite image. Unlike with complex-valued characters, there are plenty of infinite image $ell$-adic Galois characters.



The most important example is the $ell$-adic cyclotomic character. Take $F=mathbb Q$ and define $chi$ as follows:



$$
begin{align}
mathrm{Gal}(overline{mathbb Q}/mathbb Q) &to mathrm{Gal}(mathbb Q(zeta_{ell^infty}) / mathbb Q)\
&= varprojlim_{n} mathrm{Gal}(mathbb Q(zeta_{ell^n})/mathbb Q)\
&=varprojlim_{n} (mathbb Z/ell^nmathbb Z)^times\
&= mathbb Z_ell^timessubset mathbb Q_ell^times.
end{align}
$$



Here $zeta_{ell^n}$ is a primitive $ell^n$-th root of unity, and $mathbb Q(zeta_{ell^infty})$ is the field obtained by adjoining all $ell$-power roots of unity. This map is surjective (onto $mathbb Z_ell^times)$, so has infinite image. It only becomes trivial after restriction to $mathbb Q(zeta_{ell^infty})$, which is an infinite extension.






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  • $begingroup$
    Thank you very much for your answer.
    $endgroup$
    – AZMEH
    Jan 21 at 23:02











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$begingroup$

By Galois theory, your question is equivalent to asking whether all $ell$-adic Galois characters have finite image. Unlike with complex-valued characters, there are plenty of infinite image $ell$-adic Galois characters.



The most important example is the $ell$-adic cyclotomic character. Take $F=mathbb Q$ and define $chi$ as follows:



$$
begin{align}
mathrm{Gal}(overline{mathbb Q}/mathbb Q) &to mathrm{Gal}(mathbb Q(zeta_{ell^infty}) / mathbb Q)\
&= varprojlim_{n} mathrm{Gal}(mathbb Q(zeta_{ell^n})/mathbb Q)\
&=varprojlim_{n} (mathbb Z/ell^nmathbb Z)^times\
&= mathbb Z_ell^timessubset mathbb Q_ell^times.
end{align}
$$



Here $zeta_{ell^n}$ is a primitive $ell^n$-th root of unity, and $mathbb Q(zeta_{ell^infty})$ is the field obtained by adjoining all $ell$-power roots of unity. This map is surjective (onto $mathbb Z_ell^times)$, so has infinite image. It only becomes trivial after restriction to $mathbb Q(zeta_{ell^infty})$, which is an infinite extension.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you very much for your answer.
    $endgroup$
    – AZMEH
    Jan 21 at 23:02
















2












$begingroup$

By Galois theory, your question is equivalent to asking whether all $ell$-adic Galois characters have finite image. Unlike with complex-valued characters, there are plenty of infinite image $ell$-adic Galois characters.



The most important example is the $ell$-adic cyclotomic character. Take $F=mathbb Q$ and define $chi$ as follows:



$$
begin{align}
mathrm{Gal}(overline{mathbb Q}/mathbb Q) &to mathrm{Gal}(mathbb Q(zeta_{ell^infty}) / mathbb Q)\
&= varprojlim_{n} mathrm{Gal}(mathbb Q(zeta_{ell^n})/mathbb Q)\
&=varprojlim_{n} (mathbb Z/ell^nmathbb Z)^times\
&= mathbb Z_ell^timessubset mathbb Q_ell^times.
end{align}
$$



Here $zeta_{ell^n}$ is a primitive $ell^n$-th root of unity, and $mathbb Q(zeta_{ell^infty})$ is the field obtained by adjoining all $ell$-power roots of unity. This map is surjective (onto $mathbb Z_ell^times)$, so has infinite image. It only becomes trivial after restriction to $mathbb Q(zeta_{ell^infty})$, which is an infinite extension.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you very much for your answer.
    $endgroup$
    – AZMEH
    Jan 21 at 23:02














2












2








2





$begingroup$

By Galois theory, your question is equivalent to asking whether all $ell$-adic Galois characters have finite image. Unlike with complex-valued characters, there are plenty of infinite image $ell$-adic Galois characters.



The most important example is the $ell$-adic cyclotomic character. Take $F=mathbb Q$ and define $chi$ as follows:



$$
begin{align}
mathrm{Gal}(overline{mathbb Q}/mathbb Q) &to mathrm{Gal}(mathbb Q(zeta_{ell^infty}) / mathbb Q)\
&= varprojlim_{n} mathrm{Gal}(mathbb Q(zeta_{ell^n})/mathbb Q)\
&=varprojlim_{n} (mathbb Z/ell^nmathbb Z)^times\
&= mathbb Z_ell^timessubset mathbb Q_ell^times.
end{align}
$$



Here $zeta_{ell^n}$ is a primitive $ell^n$-th root of unity, and $mathbb Q(zeta_{ell^infty})$ is the field obtained by adjoining all $ell$-power roots of unity. This map is surjective (onto $mathbb Z_ell^times)$, so has infinite image. It only becomes trivial after restriction to $mathbb Q(zeta_{ell^infty})$, which is an infinite extension.






share|cite|improve this answer









$endgroup$



By Galois theory, your question is equivalent to asking whether all $ell$-adic Galois characters have finite image. Unlike with complex-valued characters, there are plenty of infinite image $ell$-adic Galois characters.



The most important example is the $ell$-adic cyclotomic character. Take $F=mathbb Q$ and define $chi$ as follows:



$$
begin{align}
mathrm{Gal}(overline{mathbb Q}/mathbb Q) &to mathrm{Gal}(mathbb Q(zeta_{ell^infty}) / mathbb Q)\
&= varprojlim_{n} mathrm{Gal}(mathbb Q(zeta_{ell^n})/mathbb Q)\
&=varprojlim_{n} (mathbb Z/ell^nmathbb Z)^times\
&= mathbb Z_ell^timessubset mathbb Q_ell^times.
end{align}
$$



Here $zeta_{ell^n}$ is a primitive $ell^n$-th root of unity, and $mathbb Q(zeta_{ell^infty})$ is the field obtained by adjoining all $ell$-power roots of unity. This map is surjective (onto $mathbb Z_ell^times)$, so has infinite image. It only becomes trivial after restriction to $mathbb Q(zeta_{ell^infty})$, which is an infinite extension.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 17 at 1:04









Mathmo123Mathmo123

17.9k33166




17.9k33166












  • $begingroup$
    Thank you very much for your answer.
    $endgroup$
    – AZMEH
    Jan 21 at 23:02


















  • $begingroup$
    Thank you very much for your answer.
    $endgroup$
    – AZMEH
    Jan 21 at 23:02
















$begingroup$
Thank you very much for your answer.
$endgroup$
– AZMEH
Jan 21 at 23:02




$begingroup$
Thank you very much for your answer.
$endgroup$
– AZMEH
Jan 21 at 23:02


















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