Show $ sum_f lambda(f)t^{deg f} = prod_g big(1 - lambda(g)t^{deg g}big)^{-1} $












2












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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


















2












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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
















2












2








2


2



$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 7 at 18:57







Mugumble

















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
















  • 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












0






active

oldest

votes











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3037828%2fshow-sum-f-lambdaft-deg-f-prod-g-big1-lambdagt-deg-g-big%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























0






active

oldest

votes








0






active

oldest

votes









active

oldest

votes






active

oldest

votes
















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3037828%2fshow-sum-f-lambdaft-deg-f-prod-g-big1-lambdagt-deg-g-big%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Plaza Victoria

Image
Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l...
Read more

Puebla de Zaragoza

Image
Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc...
Read more

Musa

Image
Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
Read more