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






active

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2 Answers
2






active

oldest

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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

25.1k33994




25.1k33994












  • $begingroup$
    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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Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

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up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx
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Can I use Tabulator js library in my java Spring + Thymeleaf project?

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0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"
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