Show $ sum_f lambda(f)t^{deg f} = prod_g big(1 - lambda(g)t^{deg g}big)^{-1} $
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So I want to show that $$ sum_f lambda(f)t^{deg f} = prod_g big(1 - lambda(g)t^{deg g}big)^{-1} $$ where the sum is over monic polynomials and the product is over all monic irreducible polynomials in $F[x]$, where $F$ is a finite field.
Proof : So I know that this identity is proved by expanding each term $big(1-lambda(g)t^{deg g }big)^{-1}$ in a geometric series and using the fact that every monic polynomial can be written as a product of monic irreducible polynomials in a unique way. As you can see the details are left. Can you give me a complete proof?
So first expanding $big(1-lambda(g)t^{deg g }big)^{-1} $. We get $displaystylesum_{r=0}^{infty} big(lambda(g)t^{deg g}big)^r$. Now I have
to show $$displaystyle sum_f lambda(f)t^{deg f} =prod_g left(sum_{r=0}^{infty} big(lambda(g)t^{deg g}big)^rright),.$$ I only have to use the fact above. But how exactly? Attention definition of $lambda$ : For a monic polynom $f(x) =
x^n - c_1x^{n-1} + cdots + (-1)^nc_n$ in $F[x]$ we define $lambda(f) = psi(c_1)chi(c_n)$, where $psi$ is a character with additive structure and $chi$ is a multiplicative character. Moreover you should know that $lambda$ is multiplicative. So $lambda(fg) = lambda(f) lambda(g)$.
calculus number-theory polynomials irreducible-polynomials geometric-series
asked Dec 13 '18 at 10:00
MugumbleMugumble
410213
$endgroup$
|
show 2 more comments
$begingroup$
So I want to show that $$ sum_f lambda(f)t^{deg f} = prod_g big(1 - lambda(g)t^{deg g}big)^{-1} $$ where the sum is over monic polynomials and the product is over all monic irreducible polynomials in $F[x]$, where $F$ is a finite field.
Proof : So I know that this identity is proved by expanding each term $big(1-lambda(g)t^{deg g }big)^{-1}$ in a geometric series and using the fact that every monic polynomial can be written as a product of monic irreducible polynomials in a unique way. As you can see the details are left. Can you give me a complete proof?
So first expanding $big(1-lambda(g)t^{deg g }big)^{-1} $. We get $displaystylesum_{r=0}^{infty} big(lambda(g)t^{deg g}big)^r$. Now I have
to show $$displaystyle sum_f lambda(f)t^{deg f} =prod_g left(sum_{r=0}^{infty} big(lambda(g)t^{deg g}big)^rright),.$$ I only have to use the fact above. But how exactly? Attention definition of $lambda$ : For a monic polynom $f(x) =
x^n - c_1x^{n-1} + cdots + (-1)^nc_n$ in $F[x]$ we define $lambda(f) = psi(c_1)chi(c_n)$, where $psi$ is a character with additive structure and $chi$ is a multiplicative character. Moreover you should know that $lambda$ is multiplicative. So $lambda(fg) = lambda(f) lambda(g)$.
calculus number-theory polynomials irreducible-polynomials geometric-series
asked Dec 13 '18 at 10:00
MugumbleMugumble
410213
$endgroup$
1
$begingroup$
What is $lambda(f)$?
$endgroup$
– Pedro Tamaroff♦
Dec 13 '18 at 12:32
$begingroup$
And what is $F$? An arbitrary field?
$endgroup$
– Paul Frost
Dec 13 '18 at 12:38
$begingroup$
It can only makle sense if $lambda(f) = 0$ for all but finitely many $f$ of a fixed degree.
$endgroup$
– Paul Frost
Dec 13 '18 at 12:38
$begingroup$
I am sorry that I haven't said what $F$ and $lambda(f)$ are. $F$ is a finite field. $lambda(f)$ is $psi(c_1)chi(c_d)$ for $f = x^d - c_1x^{d-1} + ... + (-1)^dc_d$. Moreover $psi$ and $chi $ are characters. I dont think that this is important but $psi$ has "additive" structure and $chi$ has "multiplicative" structure.
$endgroup$
– Mugumble
Dec 13 '18 at 13:37
$begingroup$
Your claim is true iff $lambda$ is completely multiplicative (on monic polynomials). This is exactly the same as the Euler product of the Riemann zeta function.
$endgroup$
– reuns
Dec 13 '18 at 21:10
|
show 2 more comments
$begingroup$
So I want to show that $$ sum_f lambda(f)t^{deg f} = prod_g big(1 - lambda(g)t^{deg g}big)^{-1} $$ where the sum is over monic polynomials and the product is over all monic irreducible polynomials in $F[x]$, where $F$ is a finite field.
Proof : So I know that this identity is proved by expanding each term $big(1-lambda(g)t^{deg g }big)^{-1}$ in a geometric series and using the fact that every monic polynomial can be written as a product of monic irreducible polynomials in a unique way. As you can see the details are left. Can you give me a complete proof?
So first expanding $big(1-lambda(g)t^{deg g }big)^{-1} $. We get $displaystylesum_{r=0}^{infty} big(lambda(g)t^{deg g}big)^r$. Now I have
to show $$displaystyle sum_f lambda(f)t^{deg f} =prod_g left(sum_{r=0}^{infty} big(lambda(g)t^{deg g}big)^rright),.$$ I only have to use the fact above. But how exactly? Attention definition of $lambda$ : For a monic polynom $f(x) =
x^n - c_1x^{n-1} + cdots + (-1)^nc_n$ in $F[x]$ we define $lambda(f) = psi(c_1)chi(c_n)$, where $psi$ is a character with additive structure and $chi$ is a multiplicative character. Moreover you should know that $lambda$ is multiplicative. So $lambda(fg) = lambda(f) lambda(g)$.
calculus number-theory polynomials irreducible-polynomials geometric-series
asked Dec 13 '18 at 10:00
MugumbleMugumble
410213
$endgroup$
So I want to show that $$ sum_f lambda(f)t^{deg f} = prod_g big(1 - lambda(g)t^{deg g}big)^{-1} $$ where the sum is over monic polynomials and the product is over all monic irreducible polynomials in $F[x]$, where $F$ is a finite field.
Proof : So I know that this identity is proved by expanding each term $big(1-lambda(g)t^{deg g }big)^{-1}$ in a geometric series and using the fact that every monic polynomial can be written as a product of monic irreducible polynomials in a unique way. As you can see the details are left. Can you give me a complete proof?
So first expanding $big(1-lambda(g)t^{deg g }big)^{-1} $. We get $displaystylesum_{r=0}^{infty} big(lambda(g)t^{deg g}big)^r$. Now I have
to show $$displaystyle sum_f lambda(f)t^{deg f} =prod_g left(sum_{r=0}^{infty} big(lambda(g)t^{deg g}big)^rright),.$$ I only have to use the fact above. But how exactly? Attention definition of $lambda$ : For a monic polynom $f(x) =
x^n - c_1x^{n-1} + cdots + (-1)^nc_n$ in $F[x]$ we define $lambda(f) = psi(c_1)chi(c_n)$, where $psi$ is a character with additive structure and $chi$ is a multiplicative character. Moreover you should know that $lambda$ is multiplicative. So $lambda(fg) = lambda(f) lambda(g)$.
calculus number-theory polynomials irreducible-polynomials geometric-series
calculus number-theory polynomials irreducible-polynomials geometric-series
asked Dec 13 '18 at 10:00
MugumbleMugumble
410213
asked Dec 13 '18 at 10:00
MugumbleMugumble
410213
edited Jan 7 at 18:57
Mugumble
asked Dec 13 '18 at 10:00
MugumbleMugumble
410213
asked Dec 13 '18 at 10:00
MugumbleMugumble
410213
asked Dec 13 '18 at 10:00
MugumbleMugumble
410213
410213
1
$begingroup$
What is $lambda(f)$?
$endgroup$
– Pedro Tamaroff♦
Dec 13 '18 at 12:32
$begingroup$
And what is $F$? An arbitrary field?
$endgroup$
– Paul Frost
Dec 13 '18 at 12:38
$begingroup$
It can only makle sense if $lambda(f) = 0$ for all but finitely many $f$ of a fixed degree.
$endgroup$
– Paul Frost
Dec 13 '18 at 12:38
$begingroup$
I am sorry that I haven't said what $F$ and $lambda(f)$ are. $F$ is a finite field. $lambda(f)$ is $psi(c_1)chi(c_d)$ for $f = x^d - c_1x^{d-1} + ... + (-1)^dc_d$. Moreover $psi$ and $chi $ are characters. I dont think that this is important but $psi$ has "additive" structure and $chi$ has "multiplicative" structure.
$endgroup$
– Mugumble
Dec 13 '18 at 13:37
$begingroup$
Your claim is true iff $lambda$ is completely multiplicative (on monic polynomials). This is exactly the same as the Euler product of the Riemann zeta function.
$endgroup$
– reuns
Dec 13 '18 at 21:10
|
show 2 more comments
1
$begingroup$
What is $lambda(f)$?
$endgroup$
– Pedro Tamaroff♦
Dec 13 '18 at 12:32
$begingroup$
And what is $F$? An arbitrary field?
$endgroup$
– Paul Frost
Dec 13 '18 at 12:38
$begingroup$
It can only makle sense if $lambda(f) = 0$ for all but finitely many $f$ of a fixed degree.
$endgroup$
– Paul Frost
Dec 13 '18 at 12:38
$begingroup$
I am sorry that I haven't said what $F$ and $lambda(f)$ are. $F$ is a finite field. $lambda(f)$ is $psi(c_1)chi(c_d)$ for $f = x^d - c_1x^{d-1} + ... + (-1)^dc_d$. Moreover $psi$ and $chi $ are characters. I dont think that this is important but $psi$ has "additive" structure and $chi$ has "multiplicative" structure.
$endgroup$
– Mugumble
Dec 13 '18 at 13:37
$begingroup$
Your claim is true iff $lambda$ is completely multiplicative (on monic polynomials). This is exactly the same as the Euler product of the Riemann zeta function.
$endgroup$
– reuns
Dec 13 '18 at 21:10
1
1
$begingroup$
What is $lambda(f)$?
$endgroup$
– Pedro Tamaroff♦
Dec 13 '18 at 12:32
$begingroup$
What is $lambda(f)$?
$endgroup$
– Pedro Tamaroff♦
Dec 13 '18 at 12:32
$begingroup$
And what is $F$? An arbitrary field?
$endgroup$
– Paul Frost
Dec 13 '18 at 12:38
$begingroup$
And what is $F$? An arbitrary field?
$endgroup$
– Paul Frost
Dec 13 '18 at 12:38
$begingroup$
It can only makle sense if $lambda(f) = 0$ for all but finitely many $f$ of a fixed degree.
$endgroup$
– Paul Frost
Dec 13 '18 at 12:38
$begingroup$
It can only makle sense if $lambda(f) = 0$ for all but finitely many $f$ of a fixed degree.
$endgroup$
– Paul Frost
Dec 13 '18 at 12:38
$begingroup$
I am sorry that I haven't said what $F$ and $lambda(f)$ are. $F$ is a finite field. $lambda(f)$ is $psi(c_1)chi(c_d)$ for $f = x^d - c_1x^{d-1} + ... + (-1)^dc_d$. Moreover $psi$ and $chi $ are characters. I dont think that this is important but $psi$ has "additive" structure and $chi$ has "multiplicative" structure.
$endgroup$
– Mugumble
Dec 13 '18 at 13:37
$begingroup$
I am sorry that I haven't said what $F$ and $lambda(f)$ are. $F$ is a finite field. $lambda(f)$ is $psi(c_1)chi(c_d)$ for $f = x^d - c_1x^{d-1} + ... + (-1)^dc_d$. Moreover $psi$ and $chi $ are characters. I dont think that this is important but $psi$ has "additive" structure and $chi$ has "multiplicative" structure.
$endgroup$
– Mugumble
Dec 13 '18 at 13:37
$begingroup$
Your claim is true iff $lambda$ is completely multiplicative (on monic polynomials). This is exactly the same as the Euler product of the Riemann zeta function.
$endgroup$
– reuns
Dec 13 '18 at 21:10
$begingroup$
Your claim is true iff $lambda$ is completely multiplicative (on monic polynomials). This is exactly the same as the Euler product of the Riemann zeta function.
$endgroup$
– reuns
Dec 13 '18 at 21:10
|
show 2 more comments
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1
$begingroup$
What is $lambda(f)$?
$endgroup$
– Pedro Tamaroff♦
Dec 13 '18 at 12:32
$begingroup$
And what is $F$? An arbitrary field?
$endgroup$
– Paul Frost
Dec 13 '18 at 12:38
$begingroup$
It can only makle sense if $lambda(f) = 0$ for all but finitely many $f$ of a fixed degree.
$endgroup$
– Paul Frost
Dec 13 '18 at 12:38
$begingroup$
I am sorry that I haven't said what $F$ and $lambda(f)$ are. $F$ is a finite field. $lambda(f)$ is $psi(c_1)chi(c_d)$ for $f = x^d - c_1x^{d-1} + ... + (-1)^dc_d$. Moreover $psi$ and $chi $ are characters. I dont think that this is important but $psi$ has "additive" structure and $chi$ has "multiplicative" structure.
$endgroup$
– Mugumble
Dec 13 '18 at 13:37
$begingroup$
Your claim is true iff $lambda$ is completely multiplicative (on monic polynomials). This is exactly the same as the Euler product of the Riemann zeta function.
$endgroup$
– reuns
Dec 13 '18 at 21:10