$mathbb{R} ^ mathbb{R}$ is a commutative ring with identity that is neither noetherian nor artinian.












3












$begingroup$


let $R=mathbb{R}^ mathbb{R}$ (all the functions like $f:mathbb{R} rightarrow mathbb{R}$). For each $f, g in R$ and $a in R$:



$$(f+g)(a):=f(a)+g(a)$$
$$(fg)(a):=f(a)g(a)$$



I want to show that $R$ is a commutative ring with identity which is neither noetherian nor artinian.






abstract-algebra noetherian artinian







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    Just do it. Where are you stuck?
    $endgroup$
    – Berci
    Jan 8 at 8:11










  • $begingroup$
    I don't know how to show that it's neither noetherian nor atrinian. Don't have any idea about taking the ideals... @Berci
    $endgroup$
    – t.ysn
    Jan 8 at 8:14






  • 1




    $begingroup$
    Does an approach similar to this one math.stackexchange.com/a/4051/42969 work?
    $endgroup$
    – Martin R
    Jan 8 at 8:18










  • $begingroup$
    Hint: If $A$ is any subset of $mathbb{R}$, then $Z(A)={fin R:f(x)=0,forall xin A}$ is an ideal of $R$; if $Asubsetneq B$, then $Z(A)supsetneq Z(B)$.
    $endgroup$
    – egreg
    Jan 8 at 8:43










  • $begingroup$
    Same strategy as this works.
    $endgroup$
    – rschwieb
    Jan 8 at 14:58
















3












$begingroup$


let $R=mathbb{R}^ mathbb{R}$ (all the functions like $f:mathbb{R} rightarrow mathbb{R}$). For each $f, g in R$ and $a in R$:



$$(f+g)(a):=f(a)+g(a)$$
$$(fg)(a):=f(a)g(a)$$



I want to show that $R$ is a commutative ring with identity which is neither noetherian nor artinian.






abstract-algebra noetherian artinian







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    Just do it. Where are you stuck?
    $endgroup$
    – Berci
    Jan 8 at 8:11










  • $begingroup$
    I don't know how to show that it's neither noetherian nor atrinian. Don't have any idea about taking the ideals... @Berci
    $endgroup$
    – t.ysn
    Jan 8 at 8:14






  • 1




    $begingroup$
    Does an approach similar to this one math.stackexchange.com/a/4051/42969 work?
    $endgroup$
    – Martin R
    Jan 8 at 8:18










  • $begingroup$
    Hint: If $A$ is any subset of $mathbb{R}$, then $Z(A)={fin R:f(x)=0,forall xin A}$ is an ideal of $R$; if $Asubsetneq B$, then $Z(A)supsetneq Z(B)$.
    $endgroup$
    – egreg
    Jan 8 at 8:43










  • $begingroup$
    Same strategy as this works.
    $endgroup$
    – rschwieb
    Jan 8 at 14:58














3












3








3


1



$begingroup$


let $R=mathbb{R}^ mathbb{R}$ (all the functions like $f:mathbb{R} rightarrow mathbb{R}$). For each $f, g in R$ and $a in R$:



$$(f+g)(a):=f(a)+g(a)$$
$$(fg)(a):=f(a)g(a)$$



I want to show that $R$ is a commutative ring with identity which is neither noetherian nor artinian.






abstract-algebra noetherian artinian







share|cite|improve this question











$endgroup$




let $R=mathbb{R}^ mathbb{R}$ (all the functions like $f:mathbb{R} rightarrow mathbb{R}$). For each $f, g in R$ and $a in R$:



$$(f+g)(a):=f(a)+g(a)$$
$$(fg)(a):=f(a)g(a)$$



I want to show that $R$ is a commutative ring with identity which is neither noetherian nor artinian.






abstract-algebra noetherian artinian




abstract-algebra noetherian artinian






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 8 at 8:37







t.ysn

















asked Jan 8 at 8:07









t.ysnt.ysn

1397




1397








  • 4




    $begingroup$
    Just do it. Where are you stuck?
    $endgroup$
    – Berci
    Jan 8 at 8:11










  • $begingroup$
    I don't know how to show that it's neither noetherian nor atrinian. Don't have any idea about taking the ideals... @Berci
    $endgroup$
    – t.ysn
    Jan 8 at 8:14






  • 1




    $begingroup$
    Does an approach similar to this one math.stackexchange.com/a/4051/42969 work?
    $endgroup$
    – Martin R
    Jan 8 at 8:18










  • $begingroup$
    Hint: If $A$ is any subset of $mathbb{R}$, then $Z(A)={fin R:f(x)=0,forall xin A}$ is an ideal of $R$; if $Asubsetneq B$, then $Z(A)supsetneq Z(B)$.
    $endgroup$
    – egreg
    Jan 8 at 8:43










  • $begingroup$
    Same strategy as this works.
    $endgroup$
    – rschwieb
    Jan 8 at 14:58














  • 4




    $begingroup$
    Just do it. Where are you stuck?
    $endgroup$
    – Berci
    Jan 8 at 8:11










  • $begingroup$
    I don't know how to show that it's neither noetherian nor atrinian. Don't have any idea about taking the ideals... @Berci
    $endgroup$
    – t.ysn
    Jan 8 at 8:14






  • 1




    $begingroup$
    Does an approach similar to this one math.stackexchange.com/a/4051/42969 work?
    $endgroup$
    – Martin R
    Jan 8 at 8:18










  • $begingroup$
    Hint: If $A$ is any subset of $mathbb{R}$, then $Z(A)={fin R:f(x)=0,forall xin A}$ is an ideal of $R$; if $Asubsetneq B$, then $Z(A)supsetneq Z(B)$.
    $endgroup$
    – egreg
    Jan 8 at 8:43










  • $begingroup$
    Same strategy as this works.
    $endgroup$
    – rschwieb
    Jan 8 at 14:58








4




4




$begingroup$
Just do it. Where are you stuck?
$endgroup$
– Berci
Jan 8 at 8:11




$begingroup$
Just do it. Where are you stuck?
$endgroup$
– Berci
Jan 8 at 8:11












$begingroup$
I don't know how to show that it's neither noetherian nor atrinian. Don't have any idea about taking the ideals... @Berci
$endgroup$
– t.ysn
Jan 8 at 8:14




$begingroup$
I don't know how to show that it's neither noetherian nor atrinian. Don't have any idea about taking the ideals... @Berci
$endgroup$
– t.ysn
Jan 8 at 8:14




1




1




$begingroup$
Does an approach similar to this one math.stackexchange.com/a/4051/42969 work?
$endgroup$
– Martin R
Jan 8 at 8:18




$begingroup$
Does an approach similar to this one math.stackexchange.com/a/4051/42969 work?
$endgroup$
– Martin R
Jan 8 at 8:18












$begingroup$
Hint: If $A$ is any subset of $mathbb{R}$, then $Z(A)={fin R:f(x)=0,forall xin A}$ is an ideal of $R$; if $Asubsetneq B$, then $Z(A)supsetneq Z(B)$.
$endgroup$
– egreg
Jan 8 at 8:43




$begingroup$
Hint: If $A$ is any subset of $mathbb{R}$, then $Z(A)={fin R:f(x)=0,forall xin A}$ is an ideal of $R$; if $Asubsetneq B$, then $Z(A)supsetneq Z(B)$.
$endgroup$
– egreg
Jan 8 at 8:43












$begingroup$
Same strategy as this works.
$endgroup$
– rschwieb
Jan 8 at 14:58




$begingroup$
Same strategy as this works.
$endgroup$
– rschwieb
Jan 8 at 14:58










1 Answer
1






active

oldest

votes


















3












$begingroup$

Let's prove it's neither Noetherian nor Artinian.



To each $A subseteq mathbb{R}$, we can assign a set $varphi(A) subseteq mathbb{R}^mathbb{R}$ as follows:



$$varphi(A) = {f in mathbb{R}^mathbb{R} : forall x in A, f(x) = 0}.$$




Exercise 0. Show that $varphi(A)$ is an ideal for all sets $A subseteq mathbb{R}$.




Conclude that $$varphi : mathcal{P}(mathbb{R}) rightarrow mathcal{P}(mathbb{R}^mathbb{R})$$ can be viewed as a function $$mathcal{P}(mathbb{R}) rightarrow mathrm{Ideal}(mathbb{R}^mathbb{R}).$$




Exercise 1. Show that $varphi : mathcal{P}(mathbb{R}) rightarrow mathrm{Ideal}(mathbb{R}^mathbb{R})$ is injective and order-reversing.



Exercise 2. Find an order-reversing injection $psi : mathbb{Z} rightarrow mathcal{P}(mathbb{R})$.




Conclude that $varphi circ psi : mathbb{Z} rightarrow mathrm{Ideal}(mathbb{R}^mathbb{R})$ is an order-preserving injection.



Conclude that $mathbb{R}^mathbb{R}$ is neither Artinian nor Noetherian.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I get the distinct sense this website is slowly dying...
    $endgroup$
    – goblin
    Jan 8 at 9:44










  • $begingroup$
    Thank you so much. I'm still a little bit confused but I try to understand it...
    $endgroup$
    – t.ysn
    Jan 8 at 12:58












  • $begingroup$
    @t.ysn, which part is unclear? (I'll answer tomorrow since I'm going to bed now)
    $endgroup$
    – goblin
    Jan 8 at 14:15










  • $begingroup$
    This answer is interesting and OK... but it seems to be going out of its way to go over the posters head! I think a good compromise that keeps it a hint would be to add something like "consider what $phi$ does to the chain of sets of the form $[-1/n, 1/n]$ and $[-n,n]$ for varying $nin mathbb N^+$ ordered by inclusion."
    $endgroup$
    – rschwieb
    Jan 8 at 15:05












  • $begingroup$
    @rschwieb, I take your point, but for me, such hints were always very confusing, perhaps because my reading comprehension skills aren't very strong. I try to offer people a bit more formality than is usual in "hint" answers for this reason. Anyway, why not post your own answer referencing the $varphi$ function I've just defined and then reconceptualizing things in your own terms?
    $endgroup$
    – goblin
    Jan 9 at 8:49












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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

Let's prove it's neither Noetherian nor Artinian.



To each $A subseteq mathbb{R}$, we can assign a set $varphi(A) subseteq mathbb{R}^mathbb{R}$ as follows:



$$varphi(A) = {f in mathbb{R}^mathbb{R} : forall x in A, f(x) = 0}.$$




Exercise 0. Show that $varphi(A)$ is an ideal for all sets $A subseteq mathbb{R}$.




Conclude that $$varphi : mathcal{P}(mathbb{R}) rightarrow mathcal{P}(mathbb{R}^mathbb{R})$$ can be viewed as a function $$mathcal{P}(mathbb{R}) rightarrow mathrm{Ideal}(mathbb{R}^mathbb{R}).$$




Exercise 1. Show that $varphi : mathcal{P}(mathbb{R}) rightarrow mathrm{Ideal}(mathbb{R}^mathbb{R})$ is injective and order-reversing.



Exercise 2. Find an order-reversing injection $psi : mathbb{Z} rightarrow mathcal{P}(mathbb{R})$.




Conclude that $varphi circ psi : mathbb{Z} rightarrow mathrm{Ideal}(mathbb{R}^mathbb{R})$ is an order-preserving injection.



Conclude that $mathbb{R}^mathbb{R}$ is neither Artinian nor Noetherian.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I get the distinct sense this website is slowly dying...
    $endgroup$
    – goblin
    Jan 8 at 9:44










  • $begingroup$
    Thank you so much. I'm still a little bit confused but I try to understand it...
    $endgroup$
    – t.ysn
    Jan 8 at 12:58












  • $begingroup$
    @t.ysn, which part is unclear? (I'll answer tomorrow since I'm going to bed now)
    $endgroup$
    – goblin
    Jan 8 at 14:15










  • $begingroup$
    This answer is interesting and OK... but it seems to be going out of its way to go over the posters head! I think a good compromise that keeps it a hint would be to add something like "consider what $phi$ does to the chain of sets of the form $[-1/n, 1/n]$ and $[-n,n]$ for varying $nin mathbb N^+$ ordered by inclusion."
    $endgroup$
    – rschwieb
    Jan 8 at 15:05












  • $begingroup$
    @rschwieb, I take your point, but for me, such hints were always very confusing, perhaps because my reading comprehension skills aren't very strong. I try to offer people a bit more formality than is usual in "hint" answers for this reason. Anyway, why not post your own answer referencing the $varphi$ function I've just defined and then reconceptualizing things in your own terms?
    $endgroup$
    – goblin
    Jan 9 at 8:49
















3












$begingroup$

Let's prove it's neither Noetherian nor Artinian.



To each $A subseteq mathbb{R}$, we can assign a set $varphi(A) subseteq mathbb{R}^mathbb{R}$ as follows:



$$varphi(A) = {f in mathbb{R}^mathbb{R} : forall x in A, f(x) = 0}.$$




Exercise 0. Show that $varphi(A)$ is an ideal for all sets $A subseteq mathbb{R}$.




Conclude that $$varphi : mathcal{P}(mathbb{R}) rightarrow mathcal{P}(mathbb{R}^mathbb{R})$$ can be viewed as a function $$mathcal{P}(mathbb{R}) rightarrow mathrm{Ideal}(mathbb{R}^mathbb{R}).$$




Exercise 1. Show that $varphi : mathcal{P}(mathbb{R}) rightarrow mathrm{Ideal}(mathbb{R}^mathbb{R})$ is injective and order-reversing.



Exercise 2. Find an order-reversing injection $psi : mathbb{Z} rightarrow mathcal{P}(mathbb{R})$.




Conclude that $varphi circ psi : mathbb{Z} rightarrow mathrm{Ideal}(mathbb{R}^mathbb{R})$ is an order-preserving injection.



Conclude that $mathbb{R}^mathbb{R}$ is neither Artinian nor Noetherian.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I get the distinct sense this website is slowly dying...
    $endgroup$
    – goblin
    Jan 8 at 9:44










  • $begingroup$
    Thank you so much. I'm still a little bit confused but I try to understand it...
    $endgroup$
    – t.ysn
    Jan 8 at 12:58












  • $begingroup$
    @t.ysn, which part is unclear? (I'll answer tomorrow since I'm going to bed now)
    $endgroup$
    – goblin
    Jan 8 at 14:15










  • $begingroup$
    This answer is interesting and OK... but it seems to be going out of its way to go over the posters head! I think a good compromise that keeps it a hint would be to add something like "consider what $phi$ does to the chain of sets of the form $[-1/n, 1/n]$ and $[-n,n]$ for varying $nin mathbb N^+$ ordered by inclusion."
    $endgroup$
    – rschwieb
    Jan 8 at 15:05












  • $begingroup$
    @rschwieb, I take your point, but for me, such hints were always very confusing, perhaps because my reading comprehension skills aren't very strong. I try to offer people a bit more formality than is usual in "hint" answers for this reason. Anyway, why not post your own answer referencing the $varphi$ function I've just defined and then reconceptualizing things in your own terms?
    $endgroup$
    – goblin
    Jan 9 at 8:49














3












3








3





$begingroup$

Let's prove it's neither Noetherian nor Artinian.



To each $A subseteq mathbb{R}$, we can assign a set $varphi(A) subseteq mathbb{R}^mathbb{R}$ as follows:



$$varphi(A) = {f in mathbb{R}^mathbb{R} : forall x in A, f(x) = 0}.$$




Exercise 0. Show that $varphi(A)$ is an ideal for all sets $A subseteq mathbb{R}$.




Conclude that $$varphi : mathcal{P}(mathbb{R}) rightarrow mathcal{P}(mathbb{R}^mathbb{R})$$ can be viewed as a function $$mathcal{P}(mathbb{R}) rightarrow mathrm{Ideal}(mathbb{R}^mathbb{R}).$$




Exercise 1. Show that $varphi : mathcal{P}(mathbb{R}) rightarrow mathrm{Ideal}(mathbb{R}^mathbb{R})$ is injective and order-reversing.



Exercise 2. Find an order-reversing injection $psi : mathbb{Z} rightarrow mathcal{P}(mathbb{R})$.




Conclude that $varphi circ psi : mathbb{Z} rightarrow mathrm{Ideal}(mathbb{R}^mathbb{R})$ is an order-preserving injection.



Conclude that $mathbb{R}^mathbb{R}$ is neither Artinian nor Noetherian.






share|cite|improve this answer











$endgroup$



Let's prove it's neither Noetherian nor Artinian.



To each $A subseteq mathbb{R}$, we can assign a set $varphi(A) subseteq mathbb{R}^mathbb{R}$ as follows:



$$varphi(A) = {f in mathbb{R}^mathbb{R} : forall x in A, f(x) = 0}.$$




Exercise 0. Show that $varphi(A)$ is an ideal for all sets $A subseteq mathbb{R}$.




Conclude that $$varphi : mathcal{P}(mathbb{R}) rightarrow mathcal{P}(mathbb{R}^mathbb{R})$$ can be viewed as a function $$mathcal{P}(mathbb{R}) rightarrow mathrm{Ideal}(mathbb{R}^mathbb{R}).$$




Exercise 1. Show that $varphi : mathcal{P}(mathbb{R}) rightarrow mathrm{Ideal}(mathbb{R}^mathbb{R})$ is injective and order-reversing.



Exercise 2. Find an order-reversing injection $psi : mathbb{Z} rightarrow mathcal{P}(mathbb{R})$.




Conclude that $varphi circ psi : mathbb{Z} rightarrow mathrm{Ideal}(mathbb{R}^mathbb{R})$ is an order-preserving injection.



Conclude that $mathbb{R}^mathbb{R}$ is neither Artinian nor Noetherian.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 8 at 9:45

























answered Jan 8 at 9:05









goblingoblin

37.1k1159195




37.1k1159195












  • $begingroup$
    I get the distinct sense this website is slowly dying...
    $endgroup$
    – goblin
    Jan 8 at 9:44










  • $begingroup$
    Thank you so much. I'm still a little bit confused but I try to understand it...
    $endgroup$
    – t.ysn
    Jan 8 at 12:58












  • $begingroup$
    @t.ysn, which part is unclear? (I'll answer tomorrow since I'm going to bed now)
    $endgroup$
    – goblin
    Jan 8 at 14:15










  • $begingroup$
    This answer is interesting and OK... but it seems to be going out of its way to go over the posters head! I think a good compromise that keeps it a hint would be to add something like "consider what $phi$ does to the chain of sets of the form $[-1/n, 1/n]$ and $[-n,n]$ for varying $nin mathbb N^+$ ordered by inclusion."
    $endgroup$
    – rschwieb
    Jan 8 at 15:05












  • $begingroup$
    @rschwieb, I take your point, but for me, such hints were always very confusing, perhaps because my reading comprehension skills aren't very strong. I try to offer people a bit more formality than is usual in "hint" answers for this reason. Anyway, why not post your own answer referencing the $varphi$ function I've just defined and then reconceptualizing things in your own terms?
    $endgroup$
    – goblin
    Jan 9 at 8:49


















  • $begingroup$
    I get the distinct sense this website is slowly dying...
    $endgroup$
    – goblin
    Jan 8 at 9:44










  • $begingroup$
    Thank you so much. I'm still a little bit confused but I try to understand it...
    $endgroup$
    – t.ysn
    Jan 8 at 12:58












  • $begingroup$
    @t.ysn, which part is unclear? (I'll answer tomorrow since I'm going to bed now)
    $endgroup$
    – goblin
    Jan 8 at 14:15










  • $begingroup$
    This answer is interesting and OK... but it seems to be going out of its way to go over the posters head! I think a good compromise that keeps it a hint would be to add something like "consider what $phi$ does to the chain of sets of the form $[-1/n, 1/n]$ and $[-n,n]$ for varying $nin mathbb N^+$ ordered by inclusion."
    $endgroup$
    – rschwieb
    Jan 8 at 15:05












  • $begingroup$
    @rschwieb, I take your point, but for me, such hints were always very confusing, perhaps because my reading comprehension skills aren't very strong. I try to offer people a bit more formality than is usual in "hint" answers for this reason. Anyway, why not post your own answer referencing the $varphi$ function I've just defined and then reconceptualizing things in your own terms?
    $endgroup$
    – goblin
    Jan 9 at 8:49
















$begingroup$
I get the distinct sense this website is slowly dying...
$endgroup$
– goblin
Jan 8 at 9:44




$begingroup$
I get the distinct sense this website is slowly dying...
$endgroup$
– goblin
Jan 8 at 9:44












$begingroup$
Thank you so much. I'm still a little bit confused but I try to understand it...
$endgroup$
– t.ysn
Jan 8 at 12:58






$begingroup$
Thank you so much. I'm still a little bit confused but I try to understand it...
$endgroup$
– t.ysn
Jan 8 at 12:58














$begingroup$
@t.ysn, which part is unclear? (I'll answer tomorrow since I'm going to bed now)
$endgroup$
– goblin
Jan 8 at 14:15




$begingroup$
@t.ysn, which part is unclear? (I'll answer tomorrow since I'm going to bed now)
$endgroup$
– goblin
Jan 8 at 14:15












$begingroup$
This answer is interesting and OK... but it seems to be going out of its way to go over the posters head! I think a good compromise that keeps it a hint would be to add something like "consider what $phi$ does to the chain of sets of the form $[-1/n, 1/n]$ and $[-n,n]$ for varying $nin mathbb N^+$ ordered by inclusion."
$endgroup$
– rschwieb
Jan 8 at 15:05






$begingroup$
This answer is interesting and OK... but it seems to be going out of its way to go over the posters head! I think a good compromise that keeps it a hint would be to add something like "consider what $phi$ does to the chain of sets of the form $[-1/n, 1/n]$ and $[-n,n]$ for varying $nin mathbb N^+$ ordered by inclusion."
$endgroup$
– rschwieb
Jan 8 at 15:05














$begingroup$
@rschwieb, I take your point, but for me, such hints were always very confusing, perhaps because my reading comprehension skills aren't very strong. I try to offer people a bit more formality than is usual in "hint" answers for this reason. Anyway, why not post your own answer referencing the $varphi$ function I've just defined and then reconceptualizing things in your own terms?
$endgroup$
– goblin
Jan 9 at 8:49




$begingroup$
@rschwieb, I take your point, but for me, such hints were always very confusing, perhaps because my reading comprehension skills aren't very strong. I try to offer people a bit more formality than is usual in "hint" answers for this reason. Anyway, why not post your own answer referencing the $varphi$ function I've just defined and then reconceptualizing things in your own terms?
$endgroup$
– goblin
Jan 9 at 8:49


















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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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