A small gap in Emil Artin's “Galois Theory”.












3












$begingroup$


I am reading Emil Artin's "Galois Theory" now.

There is a small gap in this book:




We shall now try to imitate the set $E_0$ without having an extension field $E$ and an element $a$ at our disposal. We shall assume only an irreducible polynomial $$f(x) = x^n + a_{n-1} x^{n-1} + cdots + a_0$$ as given. We select a symbol $xi$ and let $E_1$ be the set of all formal polynomials $$g(xi) = c_0 + c_1 xi + cdots + c_{n-1} xi^{n-1}$$ of a degree lower than $n$. This set forms a group under addition. We now introduce besides the ordinary multiplication a new kind of multiplication of two elements $g(xi)$ and $h(xi)$ of $E_1$ denoted by $g(xi) times h(xi)$. It is defined as the remainder $r(xi)$ of the ordinary product $g(xi) h(xi)$ under division by $f(xi)$. We first remark that any product of $m$ terms $g_1(xi), g_2(xi), cdots, g_m(xi)$ is again the remainder of the ordinary product $g_1(xi) g_2(xi) cdots g_m(xi)$. This is true by definition for $m = 2$ and follows for every $m$ by induction if we just prove the easy lemma: The remainder of the product of two remainders (of two polynomials) is the remainder of the product of these two polynomials.




And I tried to fill the gap.

Is the following proof ok or not?
$a, b, f$ is a polynomial.
$textrm{Mod}[a, f]$ is the remainder on division of $a$ by $f$.
$textrm{Quotient}[a, f]$ is the polynomial quotient of $a$ and $f$.




$a times b = a times textrm{Mod}[b, f]$




Proof:



$a b = textrm{Quotient}[a b,f] f + textrm{Mod}[a b,f]$



and



$b = textrm{Quotient}[b,f] f + textrm{Mod}[b,f]$.



$a b = a textrm{Quotient}[b,f] f + a textrm{Mod}[b,f]$.
$a textrm{Mod}[b,f] = textrm{Quotient}[a textrm{Mod}[b,f],f] f + textrm{Mod}[a textrm{Mod}[b,f],f]$.



$a b = a textrm{Quotient}[b,f] f + textrm{Quotient}[a textrm{Mod}[b,f],f] f + textrm{Mod}[a textrm{Mod}[b,f],f] = (a textrm{Quotient}[b,f] + textrm{Quotient}[a textrm{Mod}[b,f],f]) f + textrm{Mod}[a textrm{Mod}[b,f],f].$



$therefore textrm{Mod}[a b,f] = textrm{Mod}[a textrm{Mod}[b,f],f].$



$therefore a times b = a times textrm{Mod}[b, f]$.




$a times b = textrm{Mod}[a, f] times textrm{Mod}[b, f]$




Proof:



$a times b = a times textrm{Mod}[b, f] = textrm{Mod}[b, f] times a = textrm{Mod}[b, f] times textrm{Mod}[a, f] = textrm{Mod}[a, f] times textrm{Mod}[b, f]$.




$(g_1(xi) times g_2(xi)) times g_3(xi) = g_1(xi) times (g_2(xi) times g_3(xi)) = textrm{Mod}[g_1(xi) g_2(xi) g_3(xi), f]$




Proof:



$textrm{Mod}[g_1(xi) g_2(xi) g_3(xi), f] = (g_1(xi) g_2(xi)) times g_3(xi) = textrm{Mod}[g_1(xi) g_2(xi), f] times g_3(xi) = (g_1(xi) times g_2(xi)) times g_3(xi).$



$textrm{Mod}[g_1(xi) g_2(xi) g_3(xi), f] = g_1(xi) times (g_2(xi) g_3(xi)) = g_1(xi) times textrm{Mod}[g_2(xi) g_3(xi), f] = g_1(xi) times (g_2(xi) times g_3(xi)).$




$g_1(xi) times g_2(xi) times cdots times g_m(xi) times g_{m+1}(xi) = textrm{Mod}[g_1(xi) g_2(xi) cdots g_m(xi) g_{m+1}(xi), f]$




Proof:



I prove by induction.



$g_1(xi) times g_2(xi) times cdots times g_m(xi) times g_{m+1}(xi) = textrm{Mod}[g_1(xi) g_2(xi) cdots g_m(xi), f] times g_{m+1}(xi) = (g_1(xi) g_2(xi) cdots g_m(xi)) times g_{m+1}(xi) = textrm{Mod}[(g_1(xi) g_2(xi) cdots g_m(xi)) g_{m+1}(xi), f] = textrm{Mod}[g_1(xi) g_2(xi) cdots g_m(xi) g_{m+1}(xi), f].$






proof-verification polynomials modular-arithmetic







share|cite|improve this question











$endgroup$












  • $begingroup$
    The right way to do all this is to consider the quotient ring $K[X]/left<f(X)right>$.
    $endgroup$
    – Lord Shark the Unknown
    Jan 2 at 9:51










  • $begingroup$
    But Artin did not write a quotient ring in his book.
    $endgroup$
    – tchappy ha
    Jan 2 at 9:55












  • $begingroup$
    And I want to know the proof that Artin expects.
    $endgroup$
    – tchappy ha
    Jan 2 at 9:58
















3












$begingroup$


I am reading Emil Artin's "Galois Theory" now.

There is a small gap in this book:




We shall now try to imitate the set $E_0$ without having an extension field $E$ and an element $a$ at our disposal. We shall assume only an irreducible polynomial $$f(x) = x^n + a_{n-1} x^{n-1} + cdots + a_0$$ as given. We select a symbol $xi$ and let $E_1$ be the set of all formal polynomials $$g(xi) = c_0 + c_1 xi + cdots + c_{n-1} xi^{n-1}$$ of a degree lower than $n$. This set forms a group under addition. We now introduce besides the ordinary multiplication a new kind of multiplication of two elements $g(xi)$ and $h(xi)$ of $E_1$ denoted by $g(xi) times h(xi)$. It is defined as the remainder $r(xi)$ of the ordinary product $g(xi) h(xi)$ under division by $f(xi)$. We first remark that any product of $m$ terms $g_1(xi), g_2(xi), cdots, g_m(xi)$ is again the remainder of the ordinary product $g_1(xi) g_2(xi) cdots g_m(xi)$. This is true by definition for $m = 2$ and follows for every $m$ by induction if we just prove the easy lemma: The remainder of the product of two remainders (of two polynomials) is the remainder of the product of these two polynomials.




And I tried to fill the gap.

Is the following proof ok or not?
$a, b, f$ is a polynomial.
$textrm{Mod}[a, f]$ is the remainder on division of $a$ by $f$.
$textrm{Quotient}[a, f]$ is the polynomial quotient of $a$ and $f$.




$a times b = a times textrm{Mod}[b, f]$




Proof:



$a b = textrm{Quotient}[a b,f] f + textrm{Mod}[a b,f]$



and



$b = textrm{Quotient}[b,f] f + textrm{Mod}[b,f]$.



$a b = a textrm{Quotient}[b,f] f + a textrm{Mod}[b,f]$.
$a textrm{Mod}[b,f] = textrm{Quotient}[a textrm{Mod}[b,f],f] f + textrm{Mod}[a textrm{Mod}[b,f],f]$.



$a b = a textrm{Quotient}[b,f] f + textrm{Quotient}[a textrm{Mod}[b,f],f] f + textrm{Mod}[a textrm{Mod}[b,f],f] = (a textrm{Quotient}[b,f] + textrm{Quotient}[a textrm{Mod}[b,f],f]) f + textrm{Mod}[a textrm{Mod}[b,f],f].$



$therefore textrm{Mod}[a b,f] = textrm{Mod}[a textrm{Mod}[b,f],f].$



$therefore a times b = a times textrm{Mod}[b, f]$.




$a times b = textrm{Mod}[a, f] times textrm{Mod}[b, f]$




Proof:



$a times b = a times textrm{Mod}[b, f] = textrm{Mod}[b, f] times a = textrm{Mod}[b, f] times textrm{Mod}[a, f] = textrm{Mod}[a, f] times textrm{Mod}[b, f]$.




$(g_1(xi) times g_2(xi)) times g_3(xi) = g_1(xi) times (g_2(xi) times g_3(xi)) = textrm{Mod}[g_1(xi) g_2(xi) g_3(xi), f]$




Proof:



$textrm{Mod}[g_1(xi) g_2(xi) g_3(xi), f] = (g_1(xi) g_2(xi)) times g_3(xi) = textrm{Mod}[g_1(xi) g_2(xi), f] times g_3(xi) = (g_1(xi) times g_2(xi)) times g_3(xi).$



$textrm{Mod}[g_1(xi) g_2(xi) g_3(xi), f] = g_1(xi) times (g_2(xi) g_3(xi)) = g_1(xi) times textrm{Mod}[g_2(xi) g_3(xi), f] = g_1(xi) times (g_2(xi) times g_3(xi)).$




$g_1(xi) times g_2(xi) times cdots times g_m(xi) times g_{m+1}(xi) = textrm{Mod}[g_1(xi) g_2(xi) cdots g_m(xi) g_{m+1}(xi), f]$




Proof:



I prove by induction.



$g_1(xi) times g_2(xi) times cdots times g_m(xi) times g_{m+1}(xi) = textrm{Mod}[g_1(xi) g_2(xi) cdots g_m(xi), f] times g_{m+1}(xi) = (g_1(xi) g_2(xi) cdots g_m(xi)) times g_{m+1}(xi) = textrm{Mod}[(g_1(xi) g_2(xi) cdots g_m(xi)) g_{m+1}(xi), f] = textrm{Mod}[g_1(xi) g_2(xi) cdots g_m(xi) g_{m+1}(xi), f].$






proof-verification polynomials modular-arithmetic







share|cite|improve this question











$endgroup$












  • $begingroup$
    The right way to do all this is to consider the quotient ring $K[X]/left<f(X)right>$.
    $endgroup$
    – Lord Shark the Unknown
    Jan 2 at 9:51










  • $begingroup$
    But Artin did not write a quotient ring in his book.
    $endgroup$
    – tchappy ha
    Jan 2 at 9:55












  • $begingroup$
    And I want to know the proof that Artin expects.
    $endgroup$
    – tchappy ha
    Jan 2 at 9:58














3












3








3


0



$begingroup$


I am reading Emil Artin's "Galois Theory" now.

There is a small gap in this book:




We shall now try to imitate the set $E_0$ without having an extension field $E$ and an element $a$ at our disposal. We shall assume only an irreducible polynomial $$f(x) = x^n + a_{n-1} x^{n-1} + cdots + a_0$$ as given. We select a symbol $xi$ and let $E_1$ be the set of all formal polynomials $$g(xi) = c_0 + c_1 xi + cdots + c_{n-1} xi^{n-1}$$ of a degree lower than $n$. This set forms a group under addition. We now introduce besides the ordinary multiplication a new kind of multiplication of two elements $g(xi)$ and $h(xi)$ of $E_1$ denoted by $g(xi) times h(xi)$. It is defined as the remainder $r(xi)$ of the ordinary product $g(xi) h(xi)$ under division by $f(xi)$. We first remark that any product of $m$ terms $g_1(xi), g_2(xi), cdots, g_m(xi)$ is again the remainder of the ordinary product $g_1(xi) g_2(xi) cdots g_m(xi)$. This is true by definition for $m = 2$ and follows for every $m$ by induction if we just prove the easy lemma: The remainder of the product of two remainders (of two polynomials) is the remainder of the product of these two polynomials.




And I tried to fill the gap.

Is the following proof ok or not?
$a, b, f$ is a polynomial.
$textrm{Mod}[a, f]$ is the remainder on division of $a$ by $f$.
$textrm{Quotient}[a, f]$ is the polynomial quotient of $a$ and $f$.




$a times b = a times textrm{Mod}[b, f]$




Proof:



$a b = textrm{Quotient}[a b,f] f + textrm{Mod}[a b,f]$



and



$b = textrm{Quotient}[b,f] f + textrm{Mod}[b,f]$.



$a b = a textrm{Quotient}[b,f] f + a textrm{Mod}[b,f]$.
$a textrm{Mod}[b,f] = textrm{Quotient}[a textrm{Mod}[b,f],f] f + textrm{Mod}[a textrm{Mod}[b,f],f]$.



$a b = a textrm{Quotient}[b,f] f + textrm{Quotient}[a textrm{Mod}[b,f],f] f + textrm{Mod}[a textrm{Mod}[b,f],f] = (a textrm{Quotient}[b,f] + textrm{Quotient}[a textrm{Mod}[b,f],f]) f + textrm{Mod}[a textrm{Mod}[b,f],f].$



$therefore textrm{Mod}[a b,f] = textrm{Mod}[a textrm{Mod}[b,f],f].$



$therefore a times b = a times textrm{Mod}[b, f]$.




$a times b = textrm{Mod}[a, f] times textrm{Mod}[b, f]$




Proof:



$a times b = a times textrm{Mod}[b, f] = textrm{Mod}[b, f] times a = textrm{Mod}[b, f] times textrm{Mod}[a, f] = textrm{Mod}[a, f] times textrm{Mod}[b, f]$.




$(g_1(xi) times g_2(xi)) times g_3(xi) = g_1(xi) times (g_2(xi) times g_3(xi)) = textrm{Mod}[g_1(xi) g_2(xi) g_3(xi), f]$




Proof:



$textrm{Mod}[g_1(xi) g_2(xi) g_3(xi), f] = (g_1(xi) g_2(xi)) times g_3(xi) = textrm{Mod}[g_1(xi) g_2(xi), f] times g_3(xi) = (g_1(xi) times g_2(xi)) times g_3(xi).$



$textrm{Mod}[g_1(xi) g_2(xi) g_3(xi), f] = g_1(xi) times (g_2(xi) g_3(xi)) = g_1(xi) times textrm{Mod}[g_2(xi) g_3(xi), f] = g_1(xi) times (g_2(xi) times g_3(xi)).$




$g_1(xi) times g_2(xi) times cdots times g_m(xi) times g_{m+1}(xi) = textrm{Mod}[g_1(xi) g_2(xi) cdots g_m(xi) g_{m+1}(xi), f]$




Proof:



I prove by induction.



$g_1(xi) times g_2(xi) times cdots times g_m(xi) times g_{m+1}(xi) = textrm{Mod}[g_1(xi) g_2(xi) cdots g_m(xi), f] times g_{m+1}(xi) = (g_1(xi) g_2(xi) cdots g_m(xi)) times g_{m+1}(xi) = textrm{Mod}[(g_1(xi) g_2(xi) cdots g_m(xi)) g_{m+1}(xi), f] = textrm{Mod}[g_1(xi) g_2(xi) cdots g_m(xi) g_{m+1}(xi), f].$






proof-verification polynomials modular-arithmetic







share|cite|improve this question











$endgroup$




I am reading Emil Artin's "Galois Theory" now.

There is a small gap in this book:




We shall now try to imitate the set $E_0$ without having an extension field $E$ and an element $a$ at our disposal. We shall assume only an irreducible polynomial $$f(x) = x^n + a_{n-1} x^{n-1} + cdots + a_0$$ as given. We select a symbol $xi$ and let $E_1$ be the set of all formal polynomials $$g(xi) = c_0 + c_1 xi + cdots + c_{n-1} xi^{n-1}$$ of a degree lower than $n$. This set forms a group under addition. We now introduce besides the ordinary multiplication a new kind of multiplication of two elements $g(xi)$ and $h(xi)$ of $E_1$ denoted by $g(xi) times h(xi)$. It is defined as the remainder $r(xi)$ of the ordinary product $g(xi) h(xi)$ under division by $f(xi)$. We first remark that any product of $m$ terms $g_1(xi), g_2(xi), cdots, g_m(xi)$ is again the remainder of the ordinary product $g_1(xi) g_2(xi) cdots g_m(xi)$. This is true by definition for $m = 2$ and follows for every $m$ by induction if we just prove the easy lemma: The remainder of the product of two remainders (of two polynomials) is the remainder of the product of these two polynomials.




And I tried to fill the gap.

Is the following proof ok or not?
$a, b, f$ is a polynomial.
$textrm{Mod}[a, f]$ is the remainder on division of $a$ by $f$.
$textrm{Quotient}[a, f]$ is the polynomial quotient of $a$ and $f$.




$a times b = a times textrm{Mod}[b, f]$




Proof:



$a b = textrm{Quotient}[a b,f] f + textrm{Mod}[a b,f]$



and



$b = textrm{Quotient}[b,f] f + textrm{Mod}[b,f]$.



$a b = a textrm{Quotient}[b,f] f + a textrm{Mod}[b,f]$.
$a textrm{Mod}[b,f] = textrm{Quotient}[a textrm{Mod}[b,f],f] f + textrm{Mod}[a textrm{Mod}[b,f],f]$.



$a b = a textrm{Quotient}[b,f] f + textrm{Quotient}[a textrm{Mod}[b,f],f] f + textrm{Mod}[a textrm{Mod}[b,f],f] = (a textrm{Quotient}[b,f] + textrm{Quotient}[a textrm{Mod}[b,f],f]) f + textrm{Mod}[a textrm{Mod}[b,f],f].$



$therefore textrm{Mod}[a b,f] = textrm{Mod}[a textrm{Mod}[b,f],f].$



$therefore a times b = a times textrm{Mod}[b, f]$.




$a times b = textrm{Mod}[a, f] times textrm{Mod}[b, f]$




Proof:



$a times b = a times textrm{Mod}[b, f] = textrm{Mod}[b, f] times a = textrm{Mod}[b, f] times textrm{Mod}[a, f] = textrm{Mod}[a, f] times textrm{Mod}[b, f]$.




$(g_1(xi) times g_2(xi)) times g_3(xi) = g_1(xi) times (g_2(xi) times g_3(xi)) = textrm{Mod}[g_1(xi) g_2(xi) g_3(xi), f]$




Proof:



$textrm{Mod}[g_1(xi) g_2(xi) g_3(xi), f] = (g_1(xi) g_2(xi)) times g_3(xi) = textrm{Mod}[g_1(xi) g_2(xi), f] times g_3(xi) = (g_1(xi) times g_2(xi)) times g_3(xi).$



$textrm{Mod}[g_1(xi) g_2(xi) g_3(xi), f] = g_1(xi) times (g_2(xi) g_3(xi)) = g_1(xi) times textrm{Mod}[g_2(xi) g_3(xi), f] = g_1(xi) times (g_2(xi) times g_3(xi)).$




$g_1(xi) times g_2(xi) times cdots times g_m(xi) times g_{m+1}(xi) = textrm{Mod}[g_1(xi) g_2(xi) cdots g_m(xi) g_{m+1}(xi), f]$




Proof:



I prove by induction.



$g_1(xi) times g_2(xi) times cdots times g_m(xi) times g_{m+1}(xi) = textrm{Mod}[g_1(xi) g_2(xi) cdots g_m(xi), f] times g_{m+1}(xi) = (g_1(xi) g_2(xi) cdots g_m(xi)) times g_{m+1}(xi) = textrm{Mod}[(g_1(xi) g_2(xi) cdots g_m(xi)) g_{m+1}(xi), f] = textrm{Mod}[g_1(xi) g_2(xi) cdots g_m(xi) g_{m+1}(xi), f].$






proof-verification polynomials modular-arithmetic




proof-verification polynomials modular-arithmetic






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 2 at 9:47







tchappy ha

















asked Jan 2 at 9:32









tchappy hatchappy ha

766412




766412












  • $begingroup$
    The right way to do all this is to consider the quotient ring $K[X]/left<f(X)right>$.
    $endgroup$
    – Lord Shark the Unknown
    Jan 2 at 9:51










  • $begingroup$
    But Artin did not write a quotient ring in his book.
    $endgroup$
    – tchappy ha
    Jan 2 at 9:55












  • $begingroup$
    And I want to know the proof that Artin expects.
    $endgroup$
    – tchappy ha
    Jan 2 at 9:58


















  • $begingroup$
    The right way to do all this is to consider the quotient ring $K[X]/left<f(X)right>$.
    $endgroup$
    – Lord Shark the Unknown
    Jan 2 at 9:51










  • $begingroup$
    But Artin did not write a quotient ring in his book.
    $endgroup$
    – tchappy ha
    Jan 2 at 9:55












  • $begingroup$
    And I want to know the proof that Artin expects.
    $endgroup$
    – tchappy ha
    Jan 2 at 9:58
















$begingroup$
The right way to do all this is to consider the quotient ring $K[X]/left<f(X)right>$.
$endgroup$
– Lord Shark the Unknown
Jan 2 at 9:51




$begingroup$
The right way to do all this is to consider the quotient ring $K[X]/left<f(X)right>$.
$endgroup$
– Lord Shark the Unknown
Jan 2 at 9:51












$begingroup$
But Artin did not write a quotient ring in his book.
$endgroup$
– tchappy ha
Jan 2 at 9:55






$begingroup$
But Artin did not write a quotient ring in his book.
$endgroup$
– tchappy ha
Jan 2 at 9:55














$begingroup$
And I want to know the proof that Artin expects.
$endgroup$
– tchappy ha
Jan 2 at 9:58




$begingroup$
And I want to know the proof that Artin expects.
$endgroup$
– tchappy ha
Jan 2 at 9:58










0






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Province de Neuquén

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Province de Neuquén Provincia del Neuquén .mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Héraldique Drapeau Localisation de la province de Neuquén Administration Pays   Argentine Capitale Neuquén Gouverneur Omar Gutiérrez ISO 3166-2 AR-Q Démographie Gentilé Neuquino/a Population 550 344  hab. (2010) Densité 5,8  hab./km 2 Géographie Superficie 94 078  km 2 Liens Site web http://www.neuquen.gov.ar modifier   La province de Neuquén est une province d'Argentine, qui se trouve à l'extrémité nord-ouest de la Patagonie argentine. Avec une superficie de 94 378  km 2 , elle est la plus petite province des régions australes du pays. Sa géographie s'abîme sur l'imposante Cordillère des Andes. Sommaire 1 Histoire 1.1 Avant la découverte par les Espagnols 1.2 ...
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Journaliste

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Pour les articles homonymes, voir Journaliste (homonymie). Journaliste Journalistes Appellation Journaliste Secteur d'activité Média - Communication Professions voisines Journaliste sportif - Reporter-photographe - Éditorialiste Codes CITP 2451, 2642 IDEO (France) 10197 ROME (France) E1106 modifier - modifier le code - modifier Wikidata Un journaliste est une personne qui recueille, écrit ou distribue des informations. L'activité d'un journaliste est le journalisme. Un journaliste peut travailler sur des sujets généraux ou se spécialiser dans certaines questions. La plupart des journalistes ont tendance à se spécialiser et, en collaborant avec d'autres journalistes, rédigent un article ou mettent en forme un reportage qui couvre l'actualité. Par exemple, un journaliste sportif couvre les nouvelles dans le monde du sport, mais ce journaliste peut aussi faire partie d'un journal qui couvre d'autres su...
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