Galois group of $f(x)=x^5+2x+1inmathbb{Z}_3[x]$












2












$begingroup$


consider $f(x)=x^5+2x+1inmathbb{Z}_3[x]$,what is the splitting field of $f$ and its Galois group?



I know it is a Galois extension. But i do not know the degree of the extension for i have no idea of the root of $f$,
the answer is the cyclic group of order 5.



any idea will be helpful!






abstract-algebra galois-theory finite-fields







share|cite|improve this question









$endgroup$












  • $begingroup$
    Since $f$ doesn't have root in $Bbb Z_3$, it is reducible iff an irreducible monic quadratic polynomial divides it. Without any further idea, I'd list all these quadratics and see if they divide $f$..
    $endgroup$
    – Berci
    Dec 1 '18 at 8:57








  • 1




    $begingroup$
    yes, we can prove that $f$ is irreducible
    $endgroup$
    – yufeng lu
    Dec 2 '18 at 13:09
















2












$begingroup$


consider $f(x)=x^5+2x+1inmathbb{Z}_3[x]$,what is the splitting field of $f$ and its Galois group?



I know it is a Galois extension. But i do not know the degree of the extension for i have no idea of the root of $f$,
the answer is the cyclic group of order 5.



any idea will be helpful!






abstract-algebra galois-theory finite-fields







share|cite|improve this question









$endgroup$












  • $begingroup$
    Since $f$ doesn't have root in $Bbb Z_3$, it is reducible iff an irreducible monic quadratic polynomial divides it. Without any further idea, I'd list all these quadratics and see if they divide $f$..
    $endgroup$
    – Berci
    Dec 1 '18 at 8:57








  • 1




    $begingroup$
    yes, we can prove that $f$ is irreducible
    $endgroup$
    – yufeng lu
    Dec 2 '18 at 13:09














2












2








2


0



$begingroup$


consider $f(x)=x^5+2x+1inmathbb{Z}_3[x]$,what is the splitting field of $f$ and its Galois group?



I know it is a Galois extension. But i do not know the degree of the extension for i have no idea of the root of $f$,
the answer is the cyclic group of order 5.



any idea will be helpful!






abstract-algebra galois-theory finite-fields







share|cite|improve this question









$endgroup$




consider $f(x)=x^5+2x+1inmathbb{Z}_3[x]$,what is the splitting field of $f$ and its Galois group?



I know it is a Galois extension. But i do not know the degree of the extension for i have no idea of the root of $f$,
the answer is the cyclic group of order 5.



any idea will be helpful!






abstract-algebra galois-theory finite-fields




abstract-algebra galois-theory finite-fields






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 1 '18 at 8:39









yufeng luyufeng lu

353




353












  • $begingroup$
    Since $f$ doesn't have root in $Bbb Z_3$, it is reducible iff an irreducible monic quadratic polynomial divides it. Without any further idea, I'd list all these quadratics and see if they divide $f$..
    $endgroup$
    – Berci
    Dec 1 '18 at 8:57








  • 1




    $begingroup$
    yes, we can prove that $f$ is irreducible
    $endgroup$
    – yufeng lu
    Dec 2 '18 at 13:09


















  • $begingroup$
    Since $f$ doesn't have root in $Bbb Z_3$, it is reducible iff an irreducible monic quadratic polynomial divides it. Without any further idea, I'd list all these quadratics and see if they divide $f$..
    $endgroup$
    – Berci
    Dec 1 '18 at 8:57








  • 1




    $begingroup$
    yes, we can prove that $f$ is irreducible
    $endgroup$
    – yufeng lu
    Dec 2 '18 at 13:09
















$begingroup$
Since $f$ doesn't have root in $Bbb Z_3$, it is reducible iff an irreducible monic quadratic polynomial divides it. Without any further idea, I'd list all these quadratics and see if they divide $f$..
$endgroup$
– Berci
Dec 1 '18 at 8:57






$begingroup$
Since $f$ doesn't have root in $Bbb Z_3$, it is reducible iff an irreducible monic quadratic polynomial divides it. Without any further idea, I'd list all these quadratics and see if they divide $f$..
$endgroup$
– Berci
Dec 1 '18 at 8:57






1




1




$begingroup$
yes, we can prove that $f$ is irreducible
$endgroup$
– yufeng lu
Dec 2 '18 at 13:09




$begingroup$
yes, we can prove that $f$ is irreducible
$endgroup$
– yufeng lu
Dec 2 '18 at 13:09










2 Answers
2






active

oldest

votes


















1












$begingroup$

If $f$ is irreducible then $Bbb{F}_3[x]/(f)$ is a field of $3^5$ elements. Since all fields of $3^5$ elements are isomorphic, adjoining any root of $f$ yields the same field (up to isomorphism) and hence this field is a splitting field of $f$, and so
$$|operatorname{Gal}(f)|=[Bbb{F}_3[x]/(f):Bbb{F}_3]=5.$$
Every group of order $5$ is cyclic.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    if $alpha beta$are roots of $f$, then $mathbb{F}_3[alpha]$ is isomorphism to $mathbb{F}_3[beta]$.but it does not mean that $alpha in mathbb{F}_3[beta]$.
    $endgroup$
    – yufeng lu
    Dec 3 '18 at 14:54












  • $begingroup$
    @yufenglu Actually, it does; if $alphanotinBbb{F}_3[beta]$ then $Bbb{F}_3[alpha,beta]$ is a field containing two distinct subfields of $3^5$ elements. But any element of a field of $3^5$ elements satisfies $x^{3^5}=x$, so $$Bbb{F}_3[alpha]={xinBbb{F}_3[alpha,beta]: x^{3^5}=x}=Bbb{F}_3[beta],$$ because any field contains at most $3^5$ elements satisfying $x^{3^5}=x$.
    $endgroup$
    – Servaes
    Dec 3 '18 at 15:58












  • $begingroup$
    i got the point. thanks
    $endgroup$
    – yufeng lu
    Dec 4 '18 at 14:56



















0












$begingroup$

Hint: if $f$ is irriducible then you can consider the field exstension $F'=Bbb Z_3[x]/I$, where $I:=(f(x))$, the ideal generated by $f$. Now for the theorem of Kronecker exist a root $alpha$ of $f$ ( $alpha:= x+I$). The elements of $F'$ are $$F'={a+balpha+calpha ^2+dalpha ^3+ealpha ^4 : a,b,c,d,e in Bbb Z_3[x]}.$$ Using the euclidean division we obtain: $$x^5+2x+1=(x^4+alpha x^3+alpha ^2x^2+alpha ^3x+(2+alpha ^4))(x-alpha ).$$ Now yuo can prove that the other roots belong to $F'$. This show that $F'$ is the splitting field of $f$ and that $G(F'/F)=Z_5$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    what is the root of $(x^5+2x+1)/(x-alpha)$?
    $endgroup$
    – yufeng lu
    Dec 3 '18 at 14:47










  • $begingroup$
    @yufenglu use the fact that $alpha ^5+2alpha+1=0$
    $endgroup$
    – Domenico Vuono
    Dec 3 '18 at 20:16










  • $begingroup$
    what are the other roots besides the $alpha$?
    $endgroup$
    – yufeng lu
    Dec 4 '18 at 14:55











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2 Answers
2






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2 Answers
2






active

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active

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active

oldest

votes









1












$begingroup$

If $f$ is irreducible then $Bbb{F}_3[x]/(f)$ is a field of $3^5$ elements. Since all fields of $3^5$ elements are isomorphic, adjoining any root of $f$ yields the same field (up to isomorphism) and hence this field is a splitting field of $f$, and so
$$|operatorname{Gal}(f)|=[Bbb{F}_3[x]/(f):Bbb{F}_3]=5.$$
Every group of order $5$ is cyclic.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    if $alpha beta$are roots of $f$, then $mathbb{F}_3[alpha]$ is isomorphism to $mathbb{F}_3[beta]$.but it does not mean that $alpha in mathbb{F}_3[beta]$.
    $endgroup$
    – yufeng lu
    Dec 3 '18 at 14:54












  • $begingroup$
    @yufenglu Actually, it does; if $alphanotinBbb{F}_3[beta]$ then $Bbb{F}_3[alpha,beta]$ is a field containing two distinct subfields of $3^5$ elements. But any element of a field of $3^5$ elements satisfies $x^{3^5}=x$, so $$Bbb{F}_3[alpha]={xinBbb{F}_3[alpha,beta]: x^{3^5}=x}=Bbb{F}_3[beta],$$ because any field contains at most $3^5$ elements satisfying $x^{3^5}=x$.
    $endgroup$
    – Servaes
    Dec 3 '18 at 15:58












  • $begingroup$
    i got the point. thanks
    $endgroup$
    – yufeng lu
    Dec 4 '18 at 14:56
















1












$begingroup$

If $f$ is irreducible then $Bbb{F}_3[x]/(f)$ is a field of $3^5$ elements. Since all fields of $3^5$ elements are isomorphic, adjoining any root of $f$ yields the same field (up to isomorphism) and hence this field is a splitting field of $f$, and so
$$|operatorname{Gal}(f)|=[Bbb{F}_3[x]/(f):Bbb{F}_3]=5.$$
Every group of order $5$ is cyclic.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    if $alpha beta$are roots of $f$, then $mathbb{F}_3[alpha]$ is isomorphism to $mathbb{F}_3[beta]$.but it does not mean that $alpha in mathbb{F}_3[beta]$.
    $endgroup$
    – yufeng lu
    Dec 3 '18 at 14:54












  • $begingroup$
    @yufenglu Actually, it does; if $alphanotinBbb{F}_3[beta]$ then $Bbb{F}_3[alpha,beta]$ is a field containing two distinct subfields of $3^5$ elements. But any element of a field of $3^5$ elements satisfies $x^{3^5}=x$, so $$Bbb{F}_3[alpha]={xinBbb{F}_3[alpha,beta]: x^{3^5}=x}=Bbb{F}_3[beta],$$ because any field contains at most $3^5$ elements satisfying $x^{3^5}=x$.
    $endgroup$
    – Servaes
    Dec 3 '18 at 15:58












  • $begingroup$
    i got the point. thanks
    $endgroup$
    – yufeng lu
    Dec 4 '18 at 14:56














1












1








1





$begingroup$

If $f$ is irreducible then $Bbb{F}_3[x]/(f)$ is a field of $3^5$ elements. Since all fields of $3^5$ elements are isomorphic, adjoining any root of $f$ yields the same field (up to isomorphism) and hence this field is a splitting field of $f$, and so
$$|operatorname{Gal}(f)|=[Bbb{F}_3[x]/(f):Bbb{F}_3]=5.$$
Every group of order $5$ is cyclic.






share|cite|improve this answer











$endgroup$



If $f$ is irreducible then $Bbb{F}_3[x]/(f)$ is a field of $3^5$ elements. Since all fields of $3^5$ elements are isomorphic, adjoining any root of $f$ yields the same field (up to isomorphism) and hence this field is a splitting field of $f$, and so
$$|operatorname{Gal}(f)|=[Bbb{F}_3[x]/(f):Bbb{F}_3]=5.$$
Every group of order $5$ is cyclic.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 2 '18 at 14:42

























answered Dec 2 '18 at 14:37









ServaesServaes

22.5k33793




22.5k33793












  • $begingroup$
    if $alpha beta$are roots of $f$, then $mathbb{F}_3[alpha]$ is isomorphism to $mathbb{F}_3[beta]$.but it does not mean that $alpha in mathbb{F}_3[beta]$.
    $endgroup$
    – yufeng lu
    Dec 3 '18 at 14:54












  • $begingroup$
    @yufenglu Actually, it does; if $alphanotinBbb{F}_3[beta]$ then $Bbb{F}_3[alpha,beta]$ is a field containing two distinct subfields of $3^5$ elements. But any element of a field of $3^5$ elements satisfies $x^{3^5}=x$, so $$Bbb{F}_3[alpha]={xinBbb{F}_3[alpha,beta]: x^{3^5}=x}=Bbb{F}_3[beta],$$ because any field contains at most $3^5$ elements satisfying $x^{3^5}=x$.
    $endgroup$
    – Servaes
    Dec 3 '18 at 15:58












  • $begingroup$
    i got the point. thanks
    $endgroup$
    – yufeng lu
    Dec 4 '18 at 14:56


















  • $begingroup$
    if $alpha beta$are roots of $f$, then $mathbb{F}_3[alpha]$ is isomorphism to $mathbb{F}_3[beta]$.but it does not mean that $alpha in mathbb{F}_3[beta]$.
    $endgroup$
    – yufeng lu
    Dec 3 '18 at 14:54












  • $begingroup$
    @yufenglu Actually, it does; if $alphanotinBbb{F}_3[beta]$ then $Bbb{F}_3[alpha,beta]$ is a field containing two distinct subfields of $3^5$ elements. But any element of a field of $3^5$ elements satisfies $x^{3^5}=x$, so $$Bbb{F}_3[alpha]={xinBbb{F}_3[alpha,beta]: x^{3^5}=x}=Bbb{F}_3[beta],$$ because any field contains at most $3^5$ elements satisfying $x^{3^5}=x$.
    $endgroup$
    – Servaes
    Dec 3 '18 at 15:58












  • $begingroup$
    i got the point. thanks
    $endgroup$
    – yufeng lu
    Dec 4 '18 at 14:56
















$begingroup$
if $alpha beta$are roots of $f$, then $mathbb{F}_3[alpha]$ is isomorphism to $mathbb{F}_3[beta]$.but it does not mean that $alpha in mathbb{F}_3[beta]$.
$endgroup$
– yufeng lu
Dec 3 '18 at 14:54






$begingroup$
if $alpha beta$are roots of $f$, then $mathbb{F}_3[alpha]$ is isomorphism to $mathbb{F}_3[beta]$.but it does not mean that $alpha in mathbb{F}_3[beta]$.
$endgroup$
– yufeng lu
Dec 3 '18 at 14:54














$begingroup$
@yufenglu Actually, it does; if $alphanotinBbb{F}_3[beta]$ then $Bbb{F}_3[alpha,beta]$ is a field containing two distinct subfields of $3^5$ elements. But any element of a field of $3^5$ elements satisfies $x^{3^5}=x$, so $$Bbb{F}_3[alpha]={xinBbb{F}_3[alpha,beta]: x^{3^5}=x}=Bbb{F}_3[beta],$$ because any field contains at most $3^5$ elements satisfying $x^{3^5}=x$.
$endgroup$
– Servaes
Dec 3 '18 at 15:58






$begingroup$
@yufenglu Actually, it does; if $alphanotinBbb{F}_3[beta]$ then $Bbb{F}_3[alpha,beta]$ is a field containing two distinct subfields of $3^5$ elements. But any element of a field of $3^5$ elements satisfies $x^{3^5}=x$, so $$Bbb{F}_3[alpha]={xinBbb{F}_3[alpha,beta]: x^{3^5}=x}=Bbb{F}_3[beta],$$ because any field contains at most $3^5$ elements satisfying $x^{3^5}=x$.
$endgroup$
– Servaes
Dec 3 '18 at 15:58














$begingroup$
i got the point. thanks
$endgroup$
– yufeng lu
Dec 4 '18 at 14:56




$begingroup$
i got the point. thanks
$endgroup$
– yufeng lu
Dec 4 '18 at 14:56











0












$begingroup$

Hint: if $f$ is irriducible then you can consider the field exstension $F'=Bbb Z_3[x]/I$, where $I:=(f(x))$, the ideal generated by $f$. Now for the theorem of Kronecker exist a root $alpha$ of $f$ ( $alpha:= x+I$). The elements of $F'$ are $$F'={a+balpha+calpha ^2+dalpha ^3+ealpha ^4 : a,b,c,d,e in Bbb Z_3[x]}.$$ Using the euclidean division we obtain: $$x^5+2x+1=(x^4+alpha x^3+alpha ^2x^2+alpha ^3x+(2+alpha ^4))(x-alpha ).$$ Now yuo can prove that the other roots belong to $F'$. This show that $F'$ is the splitting field of $f$ and that $G(F'/F)=Z_5$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    what is the root of $(x^5+2x+1)/(x-alpha)$?
    $endgroup$
    – yufeng lu
    Dec 3 '18 at 14:47










  • $begingroup$
    @yufenglu use the fact that $alpha ^5+2alpha+1=0$
    $endgroup$
    – Domenico Vuono
    Dec 3 '18 at 20:16










  • $begingroup$
    what are the other roots besides the $alpha$?
    $endgroup$
    – yufeng lu
    Dec 4 '18 at 14:55
















0












$begingroup$

Hint: if $f$ is irriducible then you can consider the field exstension $F'=Bbb Z_3[x]/I$, where $I:=(f(x))$, the ideal generated by $f$. Now for the theorem of Kronecker exist a root $alpha$ of $f$ ( $alpha:= x+I$). The elements of $F'$ are $$F'={a+balpha+calpha ^2+dalpha ^3+ealpha ^4 : a,b,c,d,e in Bbb Z_3[x]}.$$ Using the euclidean division we obtain: $$x^5+2x+1=(x^4+alpha x^3+alpha ^2x^2+alpha ^3x+(2+alpha ^4))(x-alpha ).$$ Now yuo can prove that the other roots belong to $F'$. This show that $F'$ is the splitting field of $f$ and that $G(F'/F)=Z_5$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    what is the root of $(x^5+2x+1)/(x-alpha)$?
    $endgroup$
    – yufeng lu
    Dec 3 '18 at 14:47










  • $begingroup$
    @yufenglu use the fact that $alpha ^5+2alpha+1=0$
    $endgroup$
    – Domenico Vuono
    Dec 3 '18 at 20:16










  • $begingroup$
    what are the other roots besides the $alpha$?
    $endgroup$
    – yufeng lu
    Dec 4 '18 at 14:55














0












0








0





$begingroup$

Hint: if $f$ is irriducible then you can consider the field exstension $F'=Bbb Z_3[x]/I$, where $I:=(f(x))$, the ideal generated by $f$. Now for the theorem of Kronecker exist a root $alpha$ of $f$ ( $alpha:= x+I$). The elements of $F'$ are $$F'={a+balpha+calpha ^2+dalpha ^3+ealpha ^4 : a,b,c,d,e in Bbb Z_3[x]}.$$ Using the euclidean division we obtain: $$x^5+2x+1=(x^4+alpha x^3+alpha ^2x^2+alpha ^3x+(2+alpha ^4))(x-alpha ).$$ Now yuo can prove that the other roots belong to $F'$. This show that $F'$ is the splitting field of $f$ and that $G(F'/F)=Z_5$.






share|cite|improve this answer









$endgroup$



Hint: if $f$ is irriducible then you can consider the field exstension $F'=Bbb Z_3[x]/I$, where $I:=(f(x))$, the ideal generated by $f$. Now for the theorem of Kronecker exist a root $alpha$ of $f$ ( $alpha:= x+I$). The elements of $F'$ are $$F'={a+balpha+calpha ^2+dalpha ^3+ealpha ^4 : a,b,c,d,e in Bbb Z_3[x]}.$$ Using the euclidean division we obtain: $$x^5+2x+1=(x^4+alpha x^3+alpha ^2x^2+alpha ^3x+(2+alpha ^4))(x-alpha ).$$ Now yuo can prove that the other roots belong to $F'$. This show that $F'$ is the splitting field of $f$ and that $G(F'/F)=Z_5$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 2 '18 at 13:30









Domenico VuonoDomenico Vuono

2,3161523




2,3161523












  • $begingroup$
    what is the root of $(x^5+2x+1)/(x-alpha)$?
    $endgroup$
    – yufeng lu
    Dec 3 '18 at 14:47










  • $begingroup$
    @yufenglu use the fact that $alpha ^5+2alpha+1=0$
    $endgroup$
    – Domenico Vuono
    Dec 3 '18 at 20:16










  • $begingroup$
    what are the other roots besides the $alpha$?
    $endgroup$
    – yufeng lu
    Dec 4 '18 at 14:55


















  • $begingroup$
    what is the root of $(x^5+2x+1)/(x-alpha)$?
    $endgroup$
    – yufeng lu
    Dec 3 '18 at 14:47










  • $begingroup$
    @yufenglu use the fact that $alpha ^5+2alpha+1=0$
    $endgroup$
    – Domenico Vuono
    Dec 3 '18 at 20:16










  • $begingroup$
    what are the other roots besides the $alpha$?
    $endgroup$
    – yufeng lu
    Dec 4 '18 at 14:55
















$begingroup$
what is the root of $(x^5+2x+1)/(x-alpha)$?
$endgroup$
– yufeng lu
Dec 3 '18 at 14:47




$begingroup$
what is the root of $(x^5+2x+1)/(x-alpha)$?
$endgroup$
– yufeng lu
Dec 3 '18 at 14:47












$begingroup$
@yufenglu use the fact that $alpha ^5+2alpha+1=0$
$endgroup$
– Domenico Vuono
Dec 3 '18 at 20:16




$begingroup$
@yufenglu use the fact that $alpha ^5+2alpha+1=0$
$endgroup$
– Domenico Vuono
Dec 3 '18 at 20:16












$begingroup$
what are the other roots besides the $alpha$?
$endgroup$
– yufeng lu
Dec 4 '18 at 14:55




$begingroup$
what are the other roots besides the $alpha$?
$endgroup$
– yufeng lu
Dec 4 '18 at 14:55


















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build a pushdown automaton that recognizes the reverse language of a given pushdown automaton?

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0 I think it's similiar to NFAs. I replace the finite states of the given automaton for start-states for my new automaton. I do it with $epsilon$ -transition from the start state to the actual finite states of the given automaton. The start state of the given automaton will be my accepting(finite) state of my new one. All transitions will be reversed of course to recognize the reversed word. But how do I have to change the stack of the pushdown automaton? That's tricky. formal-languages automata context-free-grammar share | cite | improve this question edited Nov 29 '18 at 14:20 joee
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Mont Emei

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Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog
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