$L(s,chi)=1$ for non trivial Dirichlet character modulo a polynomial of degree $1$
0
$begingroup$
Let $m(x)in Bbb F_q[X]$ be a monic polynomial of degree $1$.
Show that for every non-trivial Dirichlet character modulo $m$ we have:
$L(s,chi)=1$.
I have seen a theorem that states that if $chi$ is non trivial, then $L(s,chi)$ is a polynomial in $q^{-s}$ of degree at most $deg(m)-1$. In my case, it means the degree is $0$, so $L(s,chi) in Bbb F_q$. I still can't understand why it equals $1$.
number-theory polynomials finite-fields dirichlet-series
asked Dec 25 '18 at 15:29
user401516user401516
92039
$endgroup$
add a comment |
0
$begingroup$
Let $m(x)in Bbb F_q[X]$ be a monic polynomial of degree $1$.
Show that for every non-trivial Dirichlet character modulo $m$ we have:
$L(s,chi)=1$.
I have seen a theorem that states that if $chi$ is non trivial, then $L(s,chi)$ is a polynomial in $q^{-s}$ of degree at most $deg(m)-1$. In my case, it means the degree is $0$, so $L(s,chi) in Bbb F_q$. I still can't understand why it equals $1$.
number-theory polynomials finite-fields dirichlet-series
asked Dec 25 '18 at 15:29
user401516user401516
92039
$endgroup$
add a comment |
0
0
0
$begingroup$
Let $m(x)in Bbb F_q[X]$ be a monic polynomial of degree $1$.
Show that for every non-trivial Dirichlet character modulo $m$ we have:
$L(s,chi)=1$.
I have seen a theorem that states that if $chi$ is non trivial, then $L(s,chi)$ is a polynomial in $q^{-s}$ of degree at most $deg(m)-1$. In my case, it means the degree is $0$, so $L(s,chi) in Bbb F_q$. I still can't understand why it equals $1$.
number-theory polynomials finite-fields dirichlet-series
asked Dec 25 '18 at 15:29
user401516user401516
92039
$endgroup$
Let $m(x)in Bbb F_q[X]$ be a monic polynomial of degree $1$.
Show that for every non-trivial Dirichlet character modulo $m$ we have:
$L(s,chi)=1$.
I have seen a theorem that states that if $chi$ is non trivial, then $L(s,chi)$ is a polynomial in $q^{-s}$ of degree at most $deg(m)-1$. In my case, it means the degree is $0$, so $L(s,chi) in Bbb F_q$. I still can't understand why it equals $1$.
number-theory polynomials finite-fields dirichlet-series
number-theory polynomials finite-fields dirichlet-series
asked Dec 25 '18 at 15:29
user401516user401516
92039
asked Dec 25 '18 at 15:29
user401516user401516
92039
edited Dec 25 '18 at 17:29
user401516
asked Dec 25 '18 at 15:29
user401516user401516
92039
asked Dec 25 '18 at 15:29
user401516user401516
92039
asked Dec 25 '18 at 15:29
user401516user401516
92039
92039
add a comment |
add a comment |
1 Answer
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$$L(T,chi) = sum_{f in F_q[X]_{monic}} chi(f) T^{deg(f)} = \
T^{deg(1)} + sum_{a in F_q[X]/(m)} chi(a) sum_{f in F_q[X]_{monic}} T^{deg(f m +a)}$$ Can you finish from there ?
answered Dec 25 '18 at 16:51
reunsreuns
20.1k21148
$endgroup$
$begingroup$
Why is $L(T,chi)=sum_{f ; monic}chi(f)T^{deg f}$ ? according to the definition I know: $L(T,chi)=sum_{f ; monic}frac{chi(f)}{|f|^T}$
$endgroup$
– user401516
Dec 25 '18 at 17:01
1
$begingroup$
@user401516 With $|f| = q^{deg(f)}$ then $|f|^{-s} = T^{deg(f)}$ where $T =q^{-s}$. We use the $q^{-s}$ notation only in the theory of global zeta functions. For local zeta functions we prefer to make clear they are power series in $T =q^{-s}$.
$endgroup$
– reuns
Dec 25 '18 at 17:04
$begingroup$
Thank you, I understand. Why is $T^{deg(a)}=sum_{f ; monic}T^{deg(fm+a)}$ ? sorry for being slow, pretty new to this subject.
$endgroup$
– user401516
Dec 25 '18 at 17:15
$begingroup$
$m$ is monic so any $h$ monic of degree $ge m$ is of the form $h = fm+a$ with $deg(a) < deg(m)$ and $f$ monic. That leaves us with $L(T,chi)= sum_{h in F_q[X]_{monic}, deg(h) < deg(m)} chi(h) T^{deg(h)}+ sum_{f in F_q[X]_{monic}} sum_{a in F_q[X]/(m)} chi(fm+a) T^{deg(f m +a)}$. Here $deg(m) = 1$ so the first sum is easy. The second sum simplifies as $chi(fm+a) = chi(a)$. @user401516
$endgroup$
– reuns
Dec 25 '18 at 17:47
add a comment |
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1 Answer
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1 Answer
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$begingroup$
$$L(T,chi) = sum_{f in F_q[X]_{monic}} chi(f) T^{deg(f)} = \
T^{deg(1)} + sum_{a in F_q[X]/(m)} chi(a) sum_{f in F_q[X]_{monic}} T^{deg(f m +a)}$$ Can you finish from there ?
answered Dec 25 '18 at 16:51
reunsreuns
20.1k21148
$endgroup$
$begingroup$
Why is $L(T,chi)=sum_{f ; monic}chi(f)T^{deg f}$ ? according to the definition I know: $L(T,chi)=sum_{f ; monic}frac{chi(f)}{|f|^T}$
$endgroup$
– user401516
Dec 25 '18 at 17:01
1
$begingroup$
@user401516 With $|f| = q^{deg(f)}$ then $|f|^{-s} = T^{deg(f)}$ where $T =q^{-s}$. We use the $q^{-s}$ notation only in the theory of global zeta functions. For local zeta functions we prefer to make clear they are power series in $T =q^{-s}$.
$endgroup$
– reuns
Dec 25 '18 at 17:04
$begingroup$
Thank you, I understand. Why is $T^{deg(a)}=sum_{f ; monic}T^{deg(fm+a)}$ ? sorry for being slow, pretty new to this subject.
$endgroup$
– user401516
Dec 25 '18 at 17:15
$begingroup$
$m$ is monic so any $h$ monic of degree $ge m$ is of the form $h = fm+a$ with $deg(a) < deg(m)$ and $f$ monic. That leaves us with $L(T,chi)= sum_{h in F_q[X]_{monic}, deg(h) < deg(m)} chi(h) T^{deg(h)}+ sum_{f in F_q[X]_{monic}} sum_{a in F_q[X]/(m)} chi(fm+a) T^{deg(f m +a)}$. Here $deg(m) = 1$ so the first sum is easy. The second sum simplifies as $chi(fm+a) = chi(a)$. @user401516
$endgroup$
– reuns
Dec 25 '18 at 17:47
add a comment |
2
$begingroup$
$$L(T,chi) = sum_{f in F_q[X]_{monic}} chi(f) T^{deg(f)} = \
T^{deg(1)} + sum_{a in F_q[X]/(m)} chi(a) sum_{f in F_q[X]_{monic}} T^{deg(f m +a)}$$ Can you finish from there ?
answered Dec 25 '18 at 16:51
reunsreuns
20.1k21148
$endgroup$
$begingroup$
Why is $L(T,chi)=sum_{f ; monic}chi(f)T^{deg f}$ ? according to the definition I know: $L(T,chi)=sum_{f ; monic}frac{chi(f)}{|f|^T}$
$endgroup$
– user401516
Dec 25 '18 at 17:01
1
$begingroup$
@user401516 With $|f| = q^{deg(f)}$ then $|f|^{-s} = T^{deg(f)}$ where $T =q^{-s}$. We use the $q^{-s}$ notation only in the theory of global zeta functions. For local zeta functions we prefer to make clear they are power series in $T =q^{-s}$.
$endgroup$
– reuns
Dec 25 '18 at 17:04
$begingroup$
Thank you, I understand. Why is $T^{deg(a)}=sum_{f ; monic}T^{deg(fm+a)}$ ? sorry for being slow, pretty new to this subject.
$endgroup$
– user401516
Dec 25 '18 at 17:15
$begingroup$
$m$ is monic so any $h$ monic of degree $ge m$ is of the form $h = fm+a$ with $deg(a) < deg(m)$ and $f$ monic. That leaves us with $L(T,chi)= sum_{h in F_q[X]_{monic}, deg(h) < deg(m)} chi(h) T^{deg(h)}+ sum_{f in F_q[X]_{monic}} sum_{a in F_q[X]/(m)} chi(fm+a) T^{deg(f m +a)}$. Here $deg(m) = 1$ so the first sum is easy. The second sum simplifies as $chi(fm+a) = chi(a)$. @user401516
$endgroup$
– reuns
Dec 25 '18 at 17:47
add a comment |
2
2
2
$begingroup$
$$L(T,chi) = sum_{f in F_q[X]_{monic}} chi(f) T^{deg(f)} = \
T^{deg(1)} + sum_{a in F_q[X]/(m)} chi(a) sum_{f in F_q[X]_{monic}} T^{deg(f m +a)}$$ Can you finish from there ?
answered Dec 25 '18 at 16:51
reunsreuns
20.1k21148
$endgroup$
$$L(T,chi) = sum_{f in F_q[X]_{monic}} chi(f) T^{deg(f)} = \
T^{deg(1)} + sum_{a in F_q[X]/(m)} chi(a) sum_{f in F_q[X]_{monic}} T^{deg(f m +a)}$$ Can you finish from there ?
answered Dec 25 '18 at 16:51
reunsreuns
20.1k21148
edited Dec 25 '18 at 17:07
answered Dec 25 '18 at 16:51
reunsreuns
20.1k21148
answered Dec 25 '18 at 16:51
reunsreuns
20.1k21148
answered Dec 25 '18 at 16:51
reunsreuns
20.1k21148
20.1k21148
$begingroup$
Why is $L(T,chi)=sum_{f ; monic}chi(f)T^{deg f}$ ? according to the definition I know: $L(T,chi)=sum_{f ; monic}frac{chi(f)}{|f|^T}$
$endgroup$
– user401516
Dec 25 '18 at 17:01
1
$begingroup$
@user401516 With $|f| = q^{deg(f)}$ then $|f|^{-s} = T^{deg(f)}$ where $T =q^{-s}$. We use the $q^{-s}$ notation only in the theory of global zeta functions. For local zeta functions we prefer to make clear they are power series in $T =q^{-s}$.
$endgroup$
– reuns
Dec 25 '18 at 17:04
$begingroup$
Thank you, I understand. Why is $T^{deg(a)}=sum_{f ; monic}T^{deg(fm+a)}$ ? sorry for being slow, pretty new to this subject.
$endgroup$
– user401516
Dec 25 '18 at 17:15
$begingroup$
$m$ is monic so any $h$ monic of degree $ge m$ is of the form $h = fm+a$ with $deg(a) < deg(m)$ and $f$ monic. That leaves us with $L(T,chi)= sum_{h in F_q[X]_{monic}, deg(h) < deg(m)} chi(h) T^{deg(h)}+ sum_{f in F_q[X]_{monic}} sum_{a in F_q[X]/(m)} chi(fm+a) T^{deg(f m +a)}$. Here $deg(m) = 1$ so the first sum is easy. The second sum simplifies as $chi(fm+a) = chi(a)$. @user401516
$endgroup$
– reuns
Dec 25 '18 at 17:47
add a comment |
$begingroup$
Why is $L(T,chi)=sum_{f ; monic}chi(f)T^{deg f}$ ? according to the definition I know: $L(T,chi)=sum_{f ; monic}frac{chi(f)}{|f|^T}$
$endgroup$
– user401516
Dec 25 '18 at 17:01
1
$begingroup$
@user401516 With $|f| = q^{deg(f)}$ then $|f|^{-s} = T^{deg(f)}$ where $T =q^{-s}$. We use the $q^{-s}$ notation only in the theory of global zeta functions. For local zeta functions we prefer to make clear they are power series in $T =q^{-s}$.
$endgroup$
– reuns
Dec 25 '18 at 17:04
$begingroup$
Thank you, I understand. Why is $T^{deg(a)}=sum_{f ; monic}T^{deg(fm+a)}$ ? sorry for being slow, pretty new to this subject.
$endgroup$
– user401516
Dec 25 '18 at 17:15
$begingroup$
$m$ is monic so any $h$ monic of degree $ge m$ is of the form $h = fm+a$ with $deg(a) < deg(m)$ and $f$ monic. That leaves us with $L(T,chi)= sum_{h in F_q[X]_{monic}, deg(h) < deg(m)} chi(h) T^{deg(h)}+ sum_{f in F_q[X]_{monic}} sum_{a in F_q[X]/(m)} chi(fm+a) T^{deg(f m +a)}$. Here $deg(m) = 1$ so the first sum is easy. The second sum simplifies as $chi(fm+a) = chi(a)$. @user401516
$endgroup$
– reuns
Dec 25 '18 at 17:47
$begingroup$
Why is $L(T,chi)=sum_{f ; monic}chi(f)T^{deg f}$ ? according to the definition I know: $L(T,chi)=sum_{f ; monic}frac{chi(f)}{|f|^T}$
$endgroup$
– user401516
Dec 25 '18 at 17:01
$begingroup$
Why is $L(T,chi)=sum_{f ; monic}chi(f)T^{deg f}$ ? according to the definition I know: $L(T,chi)=sum_{f ; monic}frac{chi(f)}{|f|^T}$
$endgroup$
– user401516
Dec 25 '18 at 17:01
1
1
$begingroup$
@user401516 With $|f| = q^{deg(f)}$ then $|f|^{-s} = T^{deg(f)}$ where $T =q^{-s}$. We use the $q^{-s}$ notation only in the theory of global zeta functions. For local zeta functions we prefer to make clear they are power series in $T =q^{-s}$.
$endgroup$
– reuns
Dec 25 '18 at 17:04
$begingroup$
@user401516 With $|f| = q^{deg(f)}$ then $|f|^{-s} = T^{deg(f)}$ where $T =q^{-s}$. We use the $q^{-s}$ notation only in the theory of global zeta functions. For local zeta functions we prefer to make clear they are power series in $T =q^{-s}$.
$endgroup$
– reuns
Dec 25 '18 at 17:04
$begingroup$
Thank you, I understand. Why is $T^{deg(a)}=sum_{f ; monic}T^{deg(fm+a)}$ ? sorry for being slow, pretty new to this subject.
$endgroup$
– user401516
Dec 25 '18 at 17:15
$begingroup$
Thank you, I understand. Why is $T^{deg(a)}=sum_{f ; monic}T^{deg(fm+a)}$ ? sorry for being slow, pretty new to this subject.
$endgroup$
– user401516
Dec 25 '18 at 17:15
$begingroup$
$m$ is monic so any $h$ monic of degree $ge m$ is of the form $h = fm+a$ with $deg(a) < deg(m)$ and $f$ monic. That leaves us with $L(T,chi)= sum_{h in F_q[X]_{monic}, deg(h) < deg(m)} chi(h) T^{deg(h)}+ sum_{f in F_q[X]_{monic}} sum_{a in F_q[X]/(m)} chi(fm+a) T^{deg(f m +a)}$. Here $deg(m) = 1$ so the first sum is easy. The second sum simplifies as $chi(fm+a) = chi(a)$. @user401516
$endgroup$
– reuns
Dec 25 '18 at 17:47
$begingroup$
$m$ is monic so any $h$ monic of degree $ge m$ is of the form $h = fm+a$ with $deg(a) < deg(m)$ and $f$ monic. That leaves us with $L(T,chi)= sum_{h in F_q[X]_{monic}, deg(h) < deg(m)} chi(h) T^{deg(h)}+ sum_{f in F_q[X]_{monic}} sum_{a in F_q[X]/(m)} chi(fm+a) T^{deg(f m +a)}$. Here $deg(m) = 1$ so the first sum is easy. The second sum simplifies as $chi(fm+a) = chi(a)$. @user401516
$endgroup$
– reuns
Dec 25 '18 at 17:47
add a comment |
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