Why isn't the the galois group of a polynomial with n distinct roots isomorphic to Sn?












2














If we consider the polynomial $f(x)=x^3-3x-1 in mathbb{Q}[x]$ which has 3 real roots ${x_1,x_2,x_3}$. I read that its galois group is isomorphic to $A_3$ and not to $S_3$. I don't really understand this.
If I consider the map (Here $E$ is the splitting field of $f$ over $mathbb{Q})$:
begin{align*}
sigma colon E to E\
x_1 mapsto x_2\
x_3 mapsto x_3\
mathbb{Q} mapsto mathbb{Q}
end{align*}



This would correspond to a transposition $(12)$ which is not in the alternating group.

But isn't this a $mathbb{Q}$-automorphism? I actually don't really grasp why the Galois group isn't in general isomorphic to Sn (for n distinct roots).






galois-theory







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  • 1




    What do you mean by $E$?
    – Servaes
    Nov 29 at 15:24










  • $E$ is the splitting field of $f$ over $mathbb{Q}$. Thanks for the comment, I edited.
    – roi_saumon
    Nov 29 at 15:41








  • 3




    There may be hidden relations among the roots. Your cubic is a case in point. If $r=x_1$ is one of the zeros, then the other zeros are $x_2=2-r^2$ and $x_3=r^2-r-2$. So the splitting field is $Bbb{Q}(r)$. If you know $sigma(x_1)=sigma(r)$ then $$sigma(x_2)=sigma(2-r^2)=2-sigma(r)^2$$ is uniquely determined. In other words, you cannot choose $sigma(x_1)$ and $sigma(x_2)$ independently from each other in the case of this polynomial. Looking at the discriminant as in Servaes' answer is a more common constraint, I simply wanted to shed more light to this.
    – Jyrki Lahtonen
    Nov 30 at 4:10












  • @JyrkiLahtonen Oh, I see, but how did you find this hidden relation?
    – roi_saumon
    Nov 30 at 11:09












  • The polynomial is kinda "famous". The zeros are $x_1=-2cos(2pi/9)$, $x_2=-2cos(4pi/9)$ and $x_3=-2cos(8pi/9)$. The implication $r$ is a root $implies$ $2-r^2$ is a root, is simple the double angle cosine formula $cos2alpha=2cos^2alpha-1$ in disguise. More generally, adjoining a single zero $r$ may allow us to factor the polynomial more than just splitting off a factor $x-r$ of $f(x)$, and such factorizations reduce the extension degree of the splitting field (and hence also the number of automorphisms).
    – Jyrki Lahtonen
    Nov 30 at 11:36


















2














If we consider the polynomial $f(x)=x^3-3x-1 in mathbb{Q}[x]$ which has 3 real roots ${x_1,x_2,x_3}$. I read that its galois group is isomorphic to $A_3$ and not to $S_3$. I don't really understand this.
If I consider the map (Here $E$ is the splitting field of $f$ over $mathbb{Q})$:
begin{align*}
sigma colon E to E\
x_1 mapsto x_2\
x_3 mapsto x_3\
mathbb{Q} mapsto mathbb{Q}
end{align*}



This would correspond to a transposition $(12)$ which is not in the alternating group.

But isn't this a $mathbb{Q}$-automorphism? I actually don't really grasp why the Galois group isn't in general isomorphic to Sn (for n distinct roots).






galois-theory







share|cite|improve this question




















  • 1




    What do you mean by $E$?
    – Servaes
    Nov 29 at 15:24










  • $E$ is the splitting field of $f$ over $mathbb{Q}$. Thanks for the comment, I edited.
    – roi_saumon
    Nov 29 at 15:41








  • 3




    There may be hidden relations among the roots. Your cubic is a case in point. If $r=x_1$ is one of the zeros, then the other zeros are $x_2=2-r^2$ and $x_3=r^2-r-2$. So the splitting field is $Bbb{Q}(r)$. If you know $sigma(x_1)=sigma(r)$ then $$sigma(x_2)=sigma(2-r^2)=2-sigma(r)^2$$ is uniquely determined. In other words, you cannot choose $sigma(x_1)$ and $sigma(x_2)$ independently from each other in the case of this polynomial. Looking at the discriminant as in Servaes' answer is a more common constraint, I simply wanted to shed more light to this.
    – Jyrki Lahtonen
    Nov 30 at 4:10












  • @JyrkiLahtonen Oh, I see, but how did you find this hidden relation?
    – roi_saumon
    Nov 30 at 11:09












  • The polynomial is kinda "famous". The zeros are $x_1=-2cos(2pi/9)$, $x_2=-2cos(4pi/9)$ and $x_3=-2cos(8pi/9)$. The implication $r$ is a root $implies$ $2-r^2$ is a root, is simple the double angle cosine formula $cos2alpha=2cos^2alpha-1$ in disguise. More generally, adjoining a single zero $r$ may allow us to factor the polynomial more than just splitting off a factor $x-r$ of $f(x)$, and such factorizations reduce the extension degree of the splitting field (and hence also the number of automorphisms).
    – Jyrki Lahtonen
    Nov 30 at 11:36
















2












2








2







If we consider the polynomial $f(x)=x^3-3x-1 in mathbb{Q}[x]$ which has 3 real roots ${x_1,x_2,x_3}$. I read that its galois group is isomorphic to $A_3$ and not to $S_3$. I don't really understand this.
If I consider the map (Here $E$ is the splitting field of $f$ over $mathbb{Q})$:
begin{align*}
sigma colon E to E\
x_1 mapsto x_2\
x_3 mapsto x_3\
mathbb{Q} mapsto mathbb{Q}
end{align*}



This would correspond to a transposition $(12)$ which is not in the alternating group.

But isn't this a $mathbb{Q}$-automorphism? I actually don't really grasp why the Galois group isn't in general isomorphic to Sn (for n distinct roots).






galois-theory







share|cite|improve this question















If we consider the polynomial $f(x)=x^3-3x-1 in mathbb{Q}[x]$ which has 3 real roots ${x_1,x_2,x_3}$. I read that its galois group is isomorphic to $A_3$ and not to $S_3$. I don't really understand this.
If I consider the map (Here $E$ is the splitting field of $f$ over $mathbb{Q})$:
begin{align*}
sigma colon E to E\
x_1 mapsto x_2\
x_3 mapsto x_3\
mathbb{Q} mapsto mathbb{Q}
end{align*}



This would correspond to a transposition $(12)$ which is not in the alternating group.

But isn't this a $mathbb{Q}$-automorphism? I actually don't really grasp why the Galois group isn't in general isomorphic to Sn (for n distinct roots).






galois-theory




galois-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 29 at 15:40

























asked Nov 29 at 15:21









roi_saumon

38517




38517








  • 1




    What do you mean by $E$?
    – Servaes
    Nov 29 at 15:24










  • $E$ is the splitting field of $f$ over $mathbb{Q}$. Thanks for the comment, I edited.
    – roi_saumon
    Nov 29 at 15:41








  • 3




    There may be hidden relations among the roots. Your cubic is a case in point. If $r=x_1$ is one of the zeros, then the other zeros are $x_2=2-r^2$ and $x_3=r^2-r-2$. So the splitting field is $Bbb{Q}(r)$. If you know $sigma(x_1)=sigma(r)$ then $$sigma(x_2)=sigma(2-r^2)=2-sigma(r)^2$$ is uniquely determined. In other words, you cannot choose $sigma(x_1)$ and $sigma(x_2)$ independently from each other in the case of this polynomial. Looking at the discriminant as in Servaes' answer is a more common constraint, I simply wanted to shed more light to this.
    – Jyrki Lahtonen
    Nov 30 at 4:10












  • @JyrkiLahtonen Oh, I see, but how did you find this hidden relation?
    – roi_saumon
    Nov 30 at 11:09












  • The polynomial is kinda "famous". The zeros are $x_1=-2cos(2pi/9)$, $x_2=-2cos(4pi/9)$ and $x_3=-2cos(8pi/9)$. The implication $r$ is a root $implies$ $2-r^2$ is a root, is simple the double angle cosine formula $cos2alpha=2cos^2alpha-1$ in disguise. More generally, adjoining a single zero $r$ may allow us to factor the polynomial more than just splitting off a factor $x-r$ of $f(x)$, and such factorizations reduce the extension degree of the splitting field (and hence also the number of automorphisms).
    – Jyrki Lahtonen
    Nov 30 at 11:36
















  • 1




    What do you mean by $E$?
    – Servaes
    Nov 29 at 15:24










  • $E$ is the splitting field of $f$ over $mathbb{Q}$. Thanks for the comment, I edited.
    – roi_saumon
    Nov 29 at 15:41








  • 3




    There may be hidden relations among the roots. Your cubic is a case in point. If $r=x_1$ is one of the zeros, then the other zeros are $x_2=2-r^2$ and $x_3=r^2-r-2$. So the splitting field is $Bbb{Q}(r)$. If you know $sigma(x_1)=sigma(r)$ then $$sigma(x_2)=sigma(2-r^2)=2-sigma(r)^2$$ is uniquely determined. In other words, you cannot choose $sigma(x_1)$ and $sigma(x_2)$ independently from each other in the case of this polynomial. Looking at the discriminant as in Servaes' answer is a more common constraint, I simply wanted to shed more light to this.
    – Jyrki Lahtonen
    Nov 30 at 4:10












  • @JyrkiLahtonen Oh, I see, but how did you find this hidden relation?
    – roi_saumon
    Nov 30 at 11:09












  • The polynomial is kinda "famous". The zeros are $x_1=-2cos(2pi/9)$, $x_2=-2cos(4pi/9)$ and $x_3=-2cos(8pi/9)$. The implication $r$ is a root $implies$ $2-r^2$ is a root, is simple the double angle cosine formula $cos2alpha=2cos^2alpha-1$ in disguise. More generally, adjoining a single zero $r$ may allow us to factor the polynomial more than just splitting off a factor $x-r$ of $f(x)$, and such factorizations reduce the extension degree of the splitting field (and hence also the number of automorphisms).
    – Jyrki Lahtonen
    Nov 30 at 11:36










1




1




What do you mean by $E$?
– Servaes
Nov 29 at 15:24




What do you mean by $E$?
– Servaes
Nov 29 at 15:24












$E$ is the splitting field of $f$ over $mathbb{Q}$. Thanks for the comment, I edited.
– roi_saumon
Nov 29 at 15:41






$E$ is the splitting field of $f$ over $mathbb{Q}$. Thanks for the comment, I edited.
– roi_saumon
Nov 29 at 15:41






3




3




There may be hidden relations among the roots. Your cubic is a case in point. If $r=x_1$ is one of the zeros, then the other zeros are $x_2=2-r^2$ and $x_3=r^2-r-2$. So the splitting field is $Bbb{Q}(r)$. If you know $sigma(x_1)=sigma(r)$ then $$sigma(x_2)=sigma(2-r^2)=2-sigma(r)^2$$ is uniquely determined. In other words, you cannot choose $sigma(x_1)$ and $sigma(x_2)$ independently from each other in the case of this polynomial. Looking at the discriminant as in Servaes' answer is a more common constraint, I simply wanted to shed more light to this.
– Jyrki Lahtonen
Nov 30 at 4:10






There may be hidden relations among the roots. Your cubic is a case in point. If $r=x_1$ is one of the zeros, then the other zeros are $x_2=2-r^2$ and $x_3=r^2-r-2$. So the splitting field is $Bbb{Q}(r)$. If you know $sigma(x_1)=sigma(r)$ then $$sigma(x_2)=sigma(2-r^2)=2-sigma(r)^2$$ is uniquely determined. In other words, you cannot choose $sigma(x_1)$ and $sigma(x_2)$ independently from each other in the case of this polynomial. Looking at the discriminant as in Servaes' answer is a more common constraint, I simply wanted to shed more light to this.
– Jyrki Lahtonen
Nov 30 at 4:10














@JyrkiLahtonen Oh, I see, but how did you find this hidden relation?
– roi_saumon
Nov 30 at 11:09






@JyrkiLahtonen Oh, I see, but how did you find this hidden relation?
– roi_saumon
Nov 30 at 11:09














The polynomial is kinda "famous". The zeros are $x_1=-2cos(2pi/9)$, $x_2=-2cos(4pi/9)$ and $x_3=-2cos(8pi/9)$. The implication $r$ is a root $implies$ $2-r^2$ is a root, is simple the double angle cosine formula $cos2alpha=2cos^2alpha-1$ in disguise. More generally, adjoining a single zero $r$ may allow us to factor the polynomial more than just splitting off a factor $x-r$ of $f(x)$, and such factorizations reduce the extension degree of the splitting field (and hence also the number of automorphisms).
– Jyrki Lahtonen
Nov 30 at 11:36






The polynomial is kinda "famous". The zeros are $x_1=-2cos(2pi/9)$, $x_2=-2cos(4pi/9)$ and $x_3=-2cos(8pi/9)$. The implication $r$ is a root $implies$ $2-r^2$ is a root, is simple the double angle cosine formula $cos2alpha=2cos^2alpha-1$ in disguise. More generally, adjoining a single zero $r$ may allow us to factor the polynomial more than just splitting off a factor $x-r$ of $f(x)$, and such factorizations reduce the extension degree of the splitting field (and hence also the number of automorphisms).
– Jyrki Lahtonen
Nov 30 at 11:36












1 Answer
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3














TL;DR The element $(x_1-x_2)(x_1-x_3)(x_2-x_3)$ is contained in $Bbb{Q}$ but not fixed by $sigma$.




The following theorem tells us when the Galois group of a polynomial over $Bbb{Q}$ is contained in $A_n$:




Let $finBbb{Q}[x]$ be irreducible of degree $n$, and let $E$ be a splitting field of $f$. Identify $operatorname{Gal}(E/Bbb{Q})$ with a subgroup of $S_n$ by enumerating the roots of $f$ in $E$. If $Delta(f)inBbb{Q}$ is a square in $Bbb{Q}$ then $operatorname{Gal}(E/Bbb{Q})$ is contained in $A_n$.




The proof shows in particular why your map is not a field automorphism of $E$:



Let $x_1,ldots,x_nin E$ be the roots of $f$ and let $S_n$ act on $E$ by its action on the indices of the roots, i.e. $sigma(x_i):=x_{sigma(i)}$ for all $i$. Consider the element
$$delta:=prod_{1leq i<jleq n}(x_i-x_j)in E.$$
Note that for all $sigmain S_n$ we have $sigma(delta)=operatorname{sgn}(sigma)delta$ and that $delta^2=Delta(f)$.



If $Delta(f)$ is a square in $Bbb{Q}$ then it $deltainBbb{Q}$. It follows that $sigma(delta)=delta$ for all $sigmainoperatorname{Gal}(EBbb{Q})$ and hence that $operatorname{sgn}(sigma)=1$ for all $sigmainoperatorname{Gal}(EBbb{Q})$. This means precisely that $operatorname{Gal}(EBbb{Q})$ is contained in $A_n$.




In this particular case we see that $Delta(x^3-3x-1)=81$, so the element
$$delta=(x_1-x_2)(x_1-x_3)(x_2-x_3)=pm9,$$
is not fixed by your map $sigma$, so $sigma$ cannot be a field automorphism of $E$.






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    1 Answer
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    1 Answer
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    active

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    3














    TL;DR The element $(x_1-x_2)(x_1-x_3)(x_2-x_3)$ is contained in $Bbb{Q}$ but not fixed by $sigma$.




    The following theorem tells us when the Galois group of a polynomial over $Bbb{Q}$ is contained in $A_n$:




    Let $finBbb{Q}[x]$ be irreducible of degree $n$, and let $E$ be a splitting field of $f$. Identify $operatorname{Gal}(E/Bbb{Q})$ with a subgroup of $S_n$ by enumerating the roots of $f$ in $E$. If $Delta(f)inBbb{Q}$ is a square in $Bbb{Q}$ then $operatorname{Gal}(E/Bbb{Q})$ is contained in $A_n$.




    The proof shows in particular why your map is not a field automorphism of $E$:



    Let $x_1,ldots,x_nin E$ be the roots of $f$ and let $S_n$ act on $E$ by its action on the indices of the roots, i.e. $sigma(x_i):=x_{sigma(i)}$ for all $i$. Consider the element
    $$delta:=prod_{1leq i<jleq n}(x_i-x_j)in E.$$
    Note that for all $sigmain S_n$ we have $sigma(delta)=operatorname{sgn}(sigma)delta$ and that $delta^2=Delta(f)$.



    If $Delta(f)$ is a square in $Bbb{Q}$ then it $deltainBbb{Q}$. It follows that $sigma(delta)=delta$ for all $sigmainoperatorname{Gal}(EBbb{Q})$ and hence that $operatorname{sgn}(sigma)=1$ for all $sigmainoperatorname{Gal}(EBbb{Q})$. This means precisely that $operatorname{Gal}(EBbb{Q})$ is contained in $A_n$.




    In this particular case we see that $Delta(x^3-3x-1)=81$, so the element
    $$delta=(x_1-x_2)(x_1-x_3)(x_2-x_3)=pm9,$$
    is not fixed by your map $sigma$, so $sigma$ cannot be a field automorphism of $E$.






    share|cite|improve this answer




























      3














      TL;DR The element $(x_1-x_2)(x_1-x_3)(x_2-x_3)$ is contained in $Bbb{Q}$ but not fixed by $sigma$.




      The following theorem tells us when the Galois group of a polynomial over $Bbb{Q}$ is contained in $A_n$:




      Let $finBbb{Q}[x]$ be irreducible of degree $n$, and let $E$ be a splitting field of $f$. Identify $operatorname{Gal}(E/Bbb{Q})$ with a subgroup of $S_n$ by enumerating the roots of $f$ in $E$. If $Delta(f)inBbb{Q}$ is a square in $Bbb{Q}$ then $operatorname{Gal}(E/Bbb{Q})$ is contained in $A_n$.




      The proof shows in particular why your map is not a field automorphism of $E$:



      Let $x_1,ldots,x_nin E$ be the roots of $f$ and let $S_n$ act on $E$ by its action on the indices of the roots, i.e. $sigma(x_i):=x_{sigma(i)}$ for all $i$. Consider the element
      $$delta:=prod_{1leq i<jleq n}(x_i-x_j)in E.$$
      Note that for all $sigmain S_n$ we have $sigma(delta)=operatorname{sgn}(sigma)delta$ and that $delta^2=Delta(f)$.



      If $Delta(f)$ is a square in $Bbb{Q}$ then it $deltainBbb{Q}$. It follows that $sigma(delta)=delta$ for all $sigmainoperatorname{Gal}(EBbb{Q})$ and hence that $operatorname{sgn}(sigma)=1$ for all $sigmainoperatorname{Gal}(EBbb{Q})$. This means precisely that $operatorname{Gal}(EBbb{Q})$ is contained in $A_n$.




      In this particular case we see that $Delta(x^3-3x-1)=81$, so the element
      $$delta=(x_1-x_2)(x_1-x_3)(x_2-x_3)=pm9,$$
      is not fixed by your map $sigma$, so $sigma$ cannot be a field automorphism of $E$.






      share|cite|improve this answer


























        3












        3








        3






        TL;DR The element $(x_1-x_2)(x_1-x_3)(x_2-x_3)$ is contained in $Bbb{Q}$ but not fixed by $sigma$.




        The following theorem tells us when the Galois group of a polynomial over $Bbb{Q}$ is contained in $A_n$:




        Let $finBbb{Q}[x]$ be irreducible of degree $n$, and let $E$ be a splitting field of $f$. Identify $operatorname{Gal}(E/Bbb{Q})$ with a subgroup of $S_n$ by enumerating the roots of $f$ in $E$. If $Delta(f)inBbb{Q}$ is a square in $Bbb{Q}$ then $operatorname{Gal}(E/Bbb{Q})$ is contained in $A_n$.




        The proof shows in particular why your map is not a field automorphism of $E$:



        Let $x_1,ldots,x_nin E$ be the roots of $f$ and let $S_n$ act on $E$ by its action on the indices of the roots, i.e. $sigma(x_i):=x_{sigma(i)}$ for all $i$. Consider the element
        $$delta:=prod_{1leq i<jleq n}(x_i-x_j)in E.$$
        Note that for all $sigmain S_n$ we have $sigma(delta)=operatorname{sgn}(sigma)delta$ and that $delta^2=Delta(f)$.



        If $Delta(f)$ is a square in $Bbb{Q}$ then it $deltainBbb{Q}$. It follows that $sigma(delta)=delta$ for all $sigmainoperatorname{Gal}(EBbb{Q})$ and hence that $operatorname{sgn}(sigma)=1$ for all $sigmainoperatorname{Gal}(EBbb{Q})$. This means precisely that $operatorname{Gal}(EBbb{Q})$ is contained in $A_n$.




        In this particular case we see that $Delta(x^3-3x-1)=81$, so the element
        $$delta=(x_1-x_2)(x_1-x_3)(x_2-x_3)=pm9,$$
        is not fixed by your map $sigma$, so $sigma$ cannot be a field automorphism of $E$.






        share|cite|improve this answer














        TL;DR The element $(x_1-x_2)(x_1-x_3)(x_2-x_3)$ is contained in $Bbb{Q}$ but not fixed by $sigma$.




        The following theorem tells us when the Galois group of a polynomial over $Bbb{Q}$ is contained in $A_n$:




        Let $finBbb{Q}[x]$ be irreducible of degree $n$, and let $E$ be a splitting field of $f$. Identify $operatorname{Gal}(E/Bbb{Q})$ with a subgroup of $S_n$ by enumerating the roots of $f$ in $E$. If $Delta(f)inBbb{Q}$ is a square in $Bbb{Q}$ then $operatorname{Gal}(E/Bbb{Q})$ is contained in $A_n$.




        The proof shows in particular why your map is not a field automorphism of $E$:



        Let $x_1,ldots,x_nin E$ be the roots of $f$ and let $S_n$ act on $E$ by its action on the indices of the roots, i.e. $sigma(x_i):=x_{sigma(i)}$ for all $i$. Consider the element
        $$delta:=prod_{1leq i<jleq n}(x_i-x_j)in E.$$
        Note that for all $sigmain S_n$ we have $sigma(delta)=operatorname{sgn}(sigma)delta$ and that $delta^2=Delta(f)$.



        If $Delta(f)$ is a square in $Bbb{Q}$ then it $deltainBbb{Q}$. It follows that $sigma(delta)=delta$ for all $sigmainoperatorname{Gal}(EBbb{Q})$ and hence that $operatorname{sgn}(sigma)=1$ for all $sigmainoperatorname{Gal}(EBbb{Q})$. This means precisely that $operatorname{Gal}(EBbb{Q})$ is contained in $A_n$.




        In this particular case we see that $Delta(x^3-3x-1)=81$, so the element
        $$delta=(x_1-x_2)(x_1-x_3)(x_2-x_3)=pm9,$$
        is not fixed by your map $sigma$, so $sigma$ cannot be a field automorphism of $E$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 29 at 16:17

























        answered Nov 29 at 15:56









        Servaes

        22.3k33793




        22.3k33793






























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            Wiesbaden

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            Wappen Deutschlandkarte 50.0825 8.24 117 Koordinaten: 50° 5′  N , 8° 14′  O Basisdaten Bundesland: Hessen Regierungsbezirk: Darmstadt Höhe: 117 m ü. NHN Fläche: 203,93 km 2 Einwohner: 278.654 (31. Dez. 2017) [1] Bevölkerungsdichte: 1366 Einwohner je km 2 Postleitzahlen: 65183–65207, 55246, 55252 Vorlage:Infobox Gemeinde in Deutschland/Wartung/PLZ enthält Text Vorwahlen: 0611, 06122, 06127, 06134 Kfz-Kennzeichen: WI Gemeindeschlüssel: 06 4 14 000 LOCODE: DE WIB NUTS: DE714 Stadtgliederung: 26 Ortsbezirke Adresse der Stadtverwaltung: Schlossplatz 6 65183 Wiesbaden Webpräsenz: www.wiesbaden.de Oberbürgermeister: Sven Gerich (SPD) Lage der Landeshauptstadt Wiesbaden in Hessen und im Regierungsbezirk Darmstadt Offizielles Logo der Landeshauptstadt Wiesbaden Flagge der Landeshauptstadt Wiesbaden Wiesbaden is
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            Marschland

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            Kilometerweite Marsch in den Vier- und Marschlanden am Elbdeich Die Wedeler Marschlandschaft in Schleswig-Holstein Der Hadelner Kanal zur Entwässerung der Marsch Die tief gelegene Marsch und ein Entwässerungskanal Das Schöpfwerk Otterndorf im Land Hadeln zur Entwässerung der Medem verfügte damals bereits über die größte Kreiselpumpe Europas. Das Schöpfwerk in Neuhaus für die Aue und des Neuhaus-Bülkauer Kanals Als Marsch(land) (v. niederdt., altsächs. mersc ) – auch Masch , Mersch oder Schwemmland genannt – bezeichnet man eine nacheiszeitlich entstandene geomorphologische Landform im Gebiet der nordwestdeutschen Küsten und Flüsse sowie vergleichbare Landformen weltweit. Inhaltsverzeichnis 1 Landschaft 2 Entstehung 3 Marschböden 4 Nutzung 5 Marschgebiete in Deutschland 6 Siehe auch 7 Literatur 8 Weblinks 9 Einzelnachweise Landschaft | Marschen sind generell flache Landstriche ohne natürliche Erhebungen. Si
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            Dieringhausen

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            Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
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