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

oldest

votes








2 Answers
2






active

oldest

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active

oldest

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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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Bundesstraße 106

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Dieser Artikel beschreibt die Bundesstraße 106 in Deutschland. Zur gleichnamigen Straße in Österreich siehe Mölltal Straße. Vorlage:Infobox hochrangige Straße/Wartung/DE-B Bundesstraße 106 in Deutschland Karte Basisdaten Betreiber: Deutschland  Bundesrepublik Deutschland Straßenbeginn: Wismar ( 53° 54′  N , 11° 24′  O 53.900536 11.396875 ) Straßenende: Schwerin ( 53° 36′  N , 11° 24′  O 53.592785 11.404892 ) Gesamtlänge: 40,7 km Bundesland: Mecklenburg-Vorpommern Ausbauzustand: zweistreifig [1] Bundesstraße 106 in Schwerin (Umgehungsstraße) Straßenverlauf Land Mecklenburg-Vorpommern Landkreis Nordwestmecklenburg Ortsumgehung Wismar Wallensteingraben über   Dorf Mecklenburg Wallensteingraben Groß Stieten Bad Kleinen OT Niendorf Bahnstrecke Lübeck–Bad Kleinen Zickhusen Lübstorf Klein Trebbow OT Kirch Stück
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Verónica Boquete

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Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl
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Ida-Boy-Ed-Garten

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Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
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