Show existence of $sigma$












1












$begingroup$


Let $K$ be a normal extension of $F$ and $fin F[x]$ be irreducible over $F$.



Let $g_1, g_2$ be irreducible factors of $f$ in the ring $K[x]$.



How could we show that there exists $sigma in G(K/F)$ such that $g_2=sigma (g_1)$ ?



Could you give me a hint for that?






abstract-algebra galois-theory







share|cite|improve this question









$endgroup$












  • $begingroup$
    IMHO the same question has appeared here. No answers, though. I'll keep looking...
    $endgroup$
    – Jyrki Lahtonen
    Oct 30 '18 at 6:39










  • $begingroup$
    A more promising match. Further checking required to decide whether this is a dupe. May be this?
    $endgroup$
    – Jyrki Lahtonen
    Oct 30 '18 at 6:44
















1












$begingroup$


Let $K$ be a normal extension of $F$ and $fin F[x]$ be irreducible over $F$.



Let $g_1, g_2$ be irreducible factors of $f$ in the ring $K[x]$.



How could we show that there exists $sigma in G(K/F)$ such that $g_2=sigma (g_1)$ ?



Could you give me a hint for that?






abstract-algebra galois-theory







share|cite|improve this question









$endgroup$












  • $begingroup$
    IMHO the same question has appeared here. No answers, though. I'll keep looking...
    $endgroup$
    – Jyrki Lahtonen
    Oct 30 '18 at 6:39










  • $begingroup$
    A more promising match. Further checking required to decide whether this is a dupe. May be this?
    $endgroup$
    – Jyrki Lahtonen
    Oct 30 '18 at 6:44














1












1








1


0



$begingroup$


Let $K$ be a normal extension of $F$ and $fin F[x]$ be irreducible over $F$.



Let $g_1, g_2$ be irreducible factors of $f$ in the ring $K[x]$.



How could we show that there exists $sigma in G(K/F)$ such that $g_2=sigma (g_1)$ ?



Could you give me a hint for that?






abstract-algebra galois-theory







share|cite|improve this question









$endgroup$




Let $K$ be a normal extension of $F$ and $fin F[x]$ be irreducible over $F$.



Let $g_1, g_2$ be irreducible factors of $f$ in the ring $K[x]$.



How could we show that there exists $sigma in G(K/F)$ such that $g_2=sigma (g_1)$ ?



Could you give me a hint for that?






abstract-algebra galois-theory




abstract-algebra galois-theory






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Oct 29 '18 at 19:42









Mary StarMary Star

3,10282470




3,10282470












  • $begingroup$
    IMHO the same question has appeared here. No answers, though. I'll keep looking...
    $endgroup$
    – Jyrki Lahtonen
    Oct 30 '18 at 6:39










  • $begingroup$
    A more promising match. Further checking required to decide whether this is a dupe. May be this?
    $endgroup$
    – Jyrki Lahtonen
    Oct 30 '18 at 6:44


















  • $begingroup$
    IMHO the same question has appeared here. No answers, though. I'll keep looking...
    $endgroup$
    – Jyrki Lahtonen
    Oct 30 '18 at 6:39










  • $begingroup$
    A more promising match. Further checking required to decide whether this is a dupe. May be this?
    $endgroup$
    – Jyrki Lahtonen
    Oct 30 '18 at 6:44
















$begingroup$
IMHO the same question has appeared here. No answers, though. I'll keep looking...
$endgroup$
– Jyrki Lahtonen
Oct 30 '18 at 6:39




$begingroup$
IMHO the same question has appeared here. No answers, though. I'll keep looking...
$endgroup$
– Jyrki Lahtonen
Oct 30 '18 at 6:39












$begingroup$
A more promising match. Further checking required to decide whether this is a dupe. May be this?
$endgroup$
– Jyrki Lahtonen
Oct 30 '18 at 6:44




$begingroup$
A more promising match. Further checking required to decide whether this is a dupe. May be this?
$endgroup$
– Jyrki Lahtonen
Oct 30 '18 at 6:44










1 Answer
1






active

oldest

votes


















1












$begingroup$

Let $L/K$ be the splitting field of $f$. Let $alpha_1$ be a root of $g_1$ and $alpha_2$ be a root of $g_2$. From field theory, since $alpha_1$ and $alpha_2$ have the same minimal polynomial over $F$ and $L/F$ is normal, we get $sigma in operatorname{Aut}(L/F)$ such that $sigma(alpha_1) = alpha_2$. Since $K/F$ is normal, $sigma|_K in operatorname{Aut}(K/F)$, still with $sigma|_K(alpha_1) = alpha_2$.



Now $g_2$ is irreducible in $K$ and $g_2(alpha_2) = 0 in L$, so $g_2$ is the minimal poolynomial of $alpha_2$ over $K$. However, $sigma|_K(g_1)$ is also irreducible in $K$, and $(sigma|_K(g_1))(alpha_2) = (sigma|_K(g_1))(sigma(alpha_1)) = sigma|_K(g_1(alpha_1)) = 0$, so $sigma|_K(g_1)$ is also the minimal polynomial of $alpha_2$ over $K$.



Since minimal polynomial is unique, we conclude $sigma|_K(g_1) = g_2$.




A counter-example when $K/F$ is not normal is $F=Bbb Q$ and $K=Bbb Q(sqrt[3]2)$ and $f = X^3-2 in F[X]$, with $f = (X-sqrt[3]2)(X^2+sqrt[3]2X+sqrt[3]4) in K[X]$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Sorry but could you explain why this counter-example works? Namely why there is no such automorphism?
    $endgroup$
    – ZFR
    Feb 26 at 18:02










  • $begingroup$
    Am I right it is because your irreducible factors of $f$, namely $X-sqrt[3]2$ and $X^2+sqrt[3]2X+sqrt[3]4$ have different degrees?
    $endgroup$
    – ZFR
    Feb 26 at 18:10






  • 1




    $begingroup$
    Yes, that is indeed the reason.
    $endgroup$
    – Kenny Lau
    Feb 26 at 18:33











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%2f2976583%2fshow-existence-of-sigma%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

Let $L/K$ be the splitting field of $f$. Let $alpha_1$ be a root of $g_1$ and $alpha_2$ be a root of $g_2$. From field theory, since $alpha_1$ and $alpha_2$ have the same minimal polynomial over $F$ and $L/F$ is normal, we get $sigma in operatorname{Aut}(L/F)$ such that $sigma(alpha_1) = alpha_2$. Since $K/F$ is normal, $sigma|_K in operatorname{Aut}(K/F)$, still with $sigma|_K(alpha_1) = alpha_2$.



Now $g_2$ is irreducible in $K$ and $g_2(alpha_2) = 0 in L$, so $g_2$ is the minimal poolynomial of $alpha_2$ over $K$. However, $sigma|_K(g_1)$ is also irreducible in $K$, and $(sigma|_K(g_1))(alpha_2) = (sigma|_K(g_1))(sigma(alpha_1)) = sigma|_K(g_1(alpha_1)) = 0$, so $sigma|_K(g_1)$ is also the minimal polynomial of $alpha_2$ over $K$.



Since minimal polynomial is unique, we conclude $sigma|_K(g_1) = g_2$.




A counter-example when $K/F$ is not normal is $F=Bbb Q$ and $K=Bbb Q(sqrt[3]2)$ and $f = X^3-2 in F[X]$, with $f = (X-sqrt[3]2)(X^2+sqrt[3]2X+sqrt[3]4) in K[X]$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Sorry but could you explain why this counter-example works? Namely why there is no such automorphism?
    $endgroup$
    – ZFR
    Feb 26 at 18:02










  • $begingroup$
    Am I right it is because your irreducible factors of $f$, namely $X-sqrt[3]2$ and $X^2+sqrt[3]2X+sqrt[3]4$ have different degrees?
    $endgroup$
    – ZFR
    Feb 26 at 18:10






  • 1




    $begingroup$
    Yes, that is indeed the reason.
    $endgroup$
    – Kenny Lau
    Feb 26 at 18:33
















1












$begingroup$

Let $L/K$ be the splitting field of $f$. Let $alpha_1$ be a root of $g_1$ and $alpha_2$ be a root of $g_2$. From field theory, since $alpha_1$ and $alpha_2$ have the same minimal polynomial over $F$ and $L/F$ is normal, we get $sigma in operatorname{Aut}(L/F)$ such that $sigma(alpha_1) = alpha_2$. Since $K/F$ is normal, $sigma|_K in operatorname{Aut}(K/F)$, still with $sigma|_K(alpha_1) = alpha_2$.



Now $g_2$ is irreducible in $K$ and $g_2(alpha_2) = 0 in L$, so $g_2$ is the minimal poolynomial of $alpha_2$ over $K$. However, $sigma|_K(g_1)$ is also irreducible in $K$, and $(sigma|_K(g_1))(alpha_2) = (sigma|_K(g_1))(sigma(alpha_1)) = sigma|_K(g_1(alpha_1)) = 0$, so $sigma|_K(g_1)$ is also the minimal polynomial of $alpha_2$ over $K$.



Since minimal polynomial is unique, we conclude $sigma|_K(g_1) = g_2$.




A counter-example when $K/F$ is not normal is $F=Bbb Q$ and $K=Bbb Q(sqrt[3]2)$ and $f = X^3-2 in F[X]$, with $f = (X-sqrt[3]2)(X^2+sqrt[3]2X+sqrt[3]4) in K[X]$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Sorry but could you explain why this counter-example works? Namely why there is no such automorphism?
    $endgroup$
    – ZFR
    Feb 26 at 18:02










  • $begingroup$
    Am I right it is because your irreducible factors of $f$, namely $X-sqrt[3]2$ and $X^2+sqrt[3]2X+sqrt[3]4$ have different degrees?
    $endgroup$
    – ZFR
    Feb 26 at 18:10






  • 1




    $begingroup$
    Yes, that is indeed the reason.
    $endgroup$
    – Kenny Lau
    Feb 26 at 18:33














1












1








1





$begingroup$

Let $L/K$ be the splitting field of $f$. Let $alpha_1$ be a root of $g_1$ and $alpha_2$ be a root of $g_2$. From field theory, since $alpha_1$ and $alpha_2$ have the same minimal polynomial over $F$ and $L/F$ is normal, we get $sigma in operatorname{Aut}(L/F)$ such that $sigma(alpha_1) = alpha_2$. Since $K/F$ is normal, $sigma|_K in operatorname{Aut}(K/F)$, still with $sigma|_K(alpha_1) = alpha_2$.



Now $g_2$ is irreducible in $K$ and $g_2(alpha_2) = 0 in L$, so $g_2$ is the minimal poolynomial of $alpha_2$ over $K$. However, $sigma|_K(g_1)$ is also irreducible in $K$, and $(sigma|_K(g_1))(alpha_2) = (sigma|_K(g_1))(sigma(alpha_1)) = sigma|_K(g_1(alpha_1)) = 0$, so $sigma|_K(g_1)$ is also the minimal polynomial of $alpha_2$ over $K$.



Since minimal polynomial is unique, we conclude $sigma|_K(g_1) = g_2$.




A counter-example when $K/F$ is not normal is $F=Bbb Q$ and $K=Bbb Q(sqrt[3]2)$ and $f = X^3-2 in F[X]$, with $f = (X-sqrt[3]2)(X^2+sqrt[3]2X+sqrt[3]4) in K[X]$.






share|cite|improve this answer









$endgroup$



Let $L/K$ be the splitting field of $f$. Let $alpha_1$ be a root of $g_1$ and $alpha_2$ be a root of $g_2$. From field theory, since $alpha_1$ and $alpha_2$ have the same minimal polynomial over $F$ and $L/F$ is normal, we get $sigma in operatorname{Aut}(L/F)$ such that $sigma(alpha_1) = alpha_2$. Since $K/F$ is normal, $sigma|_K in operatorname{Aut}(K/F)$, still with $sigma|_K(alpha_1) = alpha_2$.



Now $g_2$ is irreducible in $K$ and $g_2(alpha_2) = 0 in L$, so $g_2$ is the minimal poolynomial of $alpha_2$ over $K$. However, $sigma|_K(g_1)$ is also irreducible in $K$, and $(sigma|_K(g_1))(alpha_2) = (sigma|_K(g_1))(sigma(alpha_1)) = sigma|_K(g_1(alpha_1)) = 0$, so $sigma|_K(g_1)$ is also the minimal polynomial of $alpha_2$ over $K$.



Since minimal polynomial is unique, we conclude $sigma|_K(g_1) = g_2$.




A counter-example when $K/F$ is not normal is $F=Bbb Q$ and $K=Bbb Q(sqrt[3]2)$ and $f = X^3-2 in F[X]$, with $f = (X-sqrt[3]2)(X^2+sqrt[3]2X+sqrt[3]4) in K[X]$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 1 at 23:53









Kenny LauKenny Lau

20k2160




20k2160












  • $begingroup$
    Sorry but could you explain why this counter-example works? Namely why there is no such automorphism?
    $endgroup$
    – ZFR
    Feb 26 at 18:02










  • $begingroup$
    Am I right it is because your irreducible factors of $f$, namely $X-sqrt[3]2$ and $X^2+sqrt[3]2X+sqrt[3]4$ have different degrees?
    $endgroup$
    – ZFR
    Feb 26 at 18:10






  • 1




    $begingroup$
    Yes, that is indeed the reason.
    $endgroup$
    – Kenny Lau
    Feb 26 at 18:33


















  • $begingroup$
    Sorry but could you explain why this counter-example works? Namely why there is no such automorphism?
    $endgroup$
    – ZFR
    Feb 26 at 18:02










  • $begingroup$
    Am I right it is because your irreducible factors of $f$, namely $X-sqrt[3]2$ and $X^2+sqrt[3]2X+sqrt[3]4$ have different degrees?
    $endgroup$
    – ZFR
    Feb 26 at 18:10






  • 1




    $begingroup$
    Yes, that is indeed the reason.
    $endgroup$
    – Kenny Lau
    Feb 26 at 18:33
















$begingroup$
Sorry but could you explain why this counter-example works? Namely why there is no such automorphism?
$endgroup$
– ZFR
Feb 26 at 18:02




$begingroup$
Sorry but could you explain why this counter-example works? Namely why there is no such automorphism?
$endgroup$
– ZFR
Feb 26 at 18:02












$begingroup$
Am I right it is because your irreducible factors of $f$, namely $X-sqrt[3]2$ and $X^2+sqrt[3]2X+sqrt[3]4$ have different degrees?
$endgroup$
– ZFR
Feb 26 at 18:10




$begingroup$
Am I right it is because your irreducible factors of $f$, namely $X-sqrt[3]2$ and $X^2+sqrt[3]2X+sqrt[3]4$ have different degrees?
$endgroup$
– ZFR
Feb 26 at 18:10




1




1




$begingroup$
Yes, that is indeed the reason.
$endgroup$
– Kenny Lau
Feb 26 at 18:33




$begingroup$
Yes, that is indeed the reason.
$endgroup$
– Kenny Lau
Feb 26 at 18:33


















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%2f2976583%2fshow-existence-of-sigma%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

Måne

Image
För jordens måne, se Månen. Några månar i solsystemet i skala till vår måne En måne , naturlig satellit eller drabant är en himlakropp som kretsar kring en planet eller en asteroid i ett solsystem. Ordet i bestämd form – månen – avser planeten jordens måne. Det finns inte i något fall belagt att någon måne har egna månar. De flesta stora månar i solsystemet är bundna i synkron rotation med en och samma sida vänd mot planeten. Ett exempel på en måne som inte beter sig på detta vis, är Hyperion som kaotiskt roterar kring Saturnus. En asteroidmåne är en måne som kretsar kring en asteroid. En måne kan ha bildats samtidigt som moderplaneten eller senare infångats av planetens gravitationskraft. I vårt eget solsystem känner vi i nuläget till 173 månar runt sex av de åtta planeterna [ 1 ] : Jorden – månen Mars – Phobos och Deimos Jupiter – Ganymedes (som är solsystemets största måne), Io, Europa, Callisto och 59 andra kända månar Saturnus – Titan (den enda måne...
Read more

Storängen

Image
För stadsdelen i Mariehamn, se Storängen, Åland. För ett naturreservat i Norrtälje kommun, se Storängens naturreservat. Storängen från luften i september 1967, vy mot väst. Till vänster syns Järlasjön och Saltsjöbanans spår. Till höger löper Värmdövägen och längst ut till höger syns anläggningsarbetena för Värmdöleden. Storängen är ett villaområde i centrala Nacka i Stockholms län. Området är beläget strax sydost om Nacka Forum vid Saltsjöbanan mellan Lillängen och Saltsjö-Duvnäs. Storängens villasamhälle grundades 1904 av "Tjänstemännens Egnahemsförening vid Storängen" och de första villorna började byggas samma år. Redan 1909 var 107 tomter bebyggda av totalt runt 160. [ 1 ] År 2013 fanns drygt 150 villor med lika många familjer i området. Egnahemsföreningen existerar fortfarande och fungerar idag som intresseförening för de boende i Storängen. Storängen anses vara ett av Sveriges bäst bevarade villasamhällen från 1900-talets början, tack vare sina tidstypis...
Read more

VLT Carioca

Image
VLT Carioca Informações Proprietário Prefeitura do Rio de Janeiro Local [[Rio de Janeiro, RJ]] País {{country data Brasil| country flag2| name = Brasil| variant = | size = }} Tipo de transporte Bonde (Veículo leve sobre trilhos) Número de linhas 2 (+1 em construção) Número de estações 26 (+3 em construção) Tráfego 80 mil passageiros/dia [ 1 ] (dezembro/2018) Website www.vltrio.com.br Funcionamento Início de funcionamento 5 de junho de 2016 (2 anos) Operadora(s) Concessionária do VLT Carioca S.A. Número de veículos 32 Alstom Citadis Comprimento dos veículos 44 m (144 ft) Dados técnicos Extensão do sistema 28 km (17,4 mi) [ 2 ] Frequência 5–10 minutos Bitola Bitola padrão/internacional 1 435 mm (4,71 ft) Eletrificação APS, 750V DC Velocidade média 15 km/h (9,32 mph) [ 2 ] Velocidade máxima 50 km/h (31,1 mph) ...
Read more