$bigcap_{ninmathbb{N}} I^n = (0)$ if and only if no zero divisor of $R$ is of the form $1-z$ with $zin I$.












4












$begingroup$



Full problem, suppose that $R$ is a commutative Noetherian ring and $I$ is an ideal of $R$. We wish to prove that
$$bigcap_{n=1}^{infty} I^n=(0)$$ if and only if no zerodivisor of $R$ is of the form $1-z$ with $zin I$.




First I'll suppose that the intersection is $(0)$. Let $zin I$ and let $0neq rin R$ such that $r(1-z)=0$. Then $r=rz$ and so $rin I$. Is this useful here? I'm not sure how to use the Noetherian condition of $R$ since the chain of $I^n$ is descending, not ascending.



Any help would be much appreciated! I'm studying for a qual and need all the help I can get.






abstract-algebra ring-theory commutative-algebra ideals noetherian







share|cite|improve this question











$endgroup$












  • $begingroup$
    Maybe take a look at Krulls intersection theorem?
    $endgroup$
    – red_trumpet
    Dec 7 '18 at 18:38










  • $begingroup$
    That assumes that R is local, though @red_trumpet
    $endgroup$
    – Gengar
    Dec 7 '18 at 18:47












  • $begingroup$
    Right, I missed that :(
    $endgroup$
    – red_trumpet
    Dec 7 '18 at 19:37
















4












$begingroup$



Full problem, suppose that $R$ is a commutative Noetherian ring and $I$ is an ideal of $R$. We wish to prove that
$$bigcap_{n=1}^{infty} I^n=(0)$$ if and only if no zerodivisor of $R$ is of the form $1-z$ with $zin I$.




First I'll suppose that the intersection is $(0)$. Let $zin I$ and let $0neq rin R$ such that $r(1-z)=0$. Then $r=rz$ and so $rin I$. Is this useful here? I'm not sure how to use the Noetherian condition of $R$ since the chain of $I^n$ is descending, not ascending.



Any help would be much appreciated! I'm studying for a qual and need all the help I can get.






abstract-algebra ring-theory commutative-algebra ideals noetherian







share|cite|improve this question











$endgroup$












  • $begingroup$
    Maybe take a look at Krulls intersection theorem?
    $endgroup$
    – red_trumpet
    Dec 7 '18 at 18:38










  • $begingroup$
    That assumes that R is local, though @red_trumpet
    $endgroup$
    – Gengar
    Dec 7 '18 at 18:47












  • $begingroup$
    Right, I missed that :(
    $endgroup$
    – red_trumpet
    Dec 7 '18 at 19:37














4












4








4





$begingroup$



Full problem, suppose that $R$ is a commutative Noetherian ring and $I$ is an ideal of $R$. We wish to prove that
$$bigcap_{n=1}^{infty} I^n=(0)$$ if and only if no zerodivisor of $R$ is of the form $1-z$ with $zin I$.




First I'll suppose that the intersection is $(0)$. Let $zin I$ and let $0neq rin R$ such that $r(1-z)=0$. Then $r=rz$ and so $rin I$. Is this useful here? I'm not sure how to use the Noetherian condition of $R$ since the chain of $I^n$ is descending, not ascending.



Any help would be much appreciated! I'm studying for a qual and need all the help I can get.






abstract-algebra ring-theory commutative-algebra ideals noetherian







share|cite|improve this question











$endgroup$





Full problem, suppose that $R$ is a commutative Noetherian ring and $I$ is an ideal of $R$. We wish to prove that
$$bigcap_{n=1}^{infty} I^n=(0)$$ if and only if no zerodivisor of $R$ is of the form $1-z$ with $zin I$.




First I'll suppose that the intersection is $(0)$. Let $zin I$ and let $0neq rin R$ such that $r(1-z)=0$. Then $r=rz$ and so $rin I$. Is this useful here? I'm not sure how to use the Noetherian condition of $R$ since the chain of $I^n$ is descending, not ascending.



Any help would be much appreciated! I'm studying for a qual and need all the help I can get.






abstract-algebra ring-theory commutative-algebra ideals noetherian




abstract-algebra ring-theory commutative-algebra ideals noetherian






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 7 '18 at 18:28









user10354138

7,3772925




7,3772925










asked Dec 7 '18 at 18:23









GengarGengar

1949




1949












  • $begingroup$
    Maybe take a look at Krulls intersection theorem?
    $endgroup$
    – red_trumpet
    Dec 7 '18 at 18:38










  • $begingroup$
    That assumes that R is local, though @red_trumpet
    $endgroup$
    – Gengar
    Dec 7 '18 at 18:47












  • $begingroup$
    Right, I missed that :(
    $endgroup$
    – red_trumpet
    Dec 7 '18 at 19:37


















  • $begingroup$
    Maybe take a look at Krulls intersection theorem?
    $endgroup$
    – red_trumpet
    Dec 7 '18 at 18:38










  • $begingroup$
    That assumes that R is local, though @red_trumpet
    $endgroup$
    – Gengar
    Dec 7 '18 at 18:47












  • $begingroup$
    Right, I missed that :(
    $endgroup$
    – red_trumpet
    Dec 7 '18 at 19:37
















$begingroup$
Maybe take a look at Krulls intersection theorem?
$endgroup$
– red_trumpet
Dec 7 '18 at 18:38




$begingroup$
Maybe take a look at Krulls intersection theorem?
$endgroup$
– red_trumpet
Dec 7 '18 at 18:38












$begingroup$
That assumes that R is local, though @red_trumpet
$endgroup$
– Gengar
Dec 7 '18 at 18:47






$begingroup$
That assumes that R is local, though @red_trumpet
$endgroup$
– Gengar
Dec 7 '18 at 18:47














$begingroup$
Right, I missed that :(
$endgroup$
– red_trumpet
Dec 7 '18 at 19:37




$begingroup$
Right, I missed that :(
$endgroup$
– red_trumpet
Dec 7 '18 at 19:37










2 Answers
2






active

oldest

votes


















3












$begingroup$

Yes, the observation that $r=rz$ $implies$ $rin I$ is very useful, since we can use the equation again to get that $rin I^2$, and thus $rin I^3$, and so on. Hence $rin bigcap_{n=1}^infty I^n$, so $r=0$, contradiction. No need to use Noetherianness here. (It may be necessary for the converse, but I haven't thought that far, since it's not clear from your question if you're also asking about that.)



Edit



Worked out my thoughts on the converse. I was being dumb. It's Nakayama's lemma (the general, not local ring version).



Let $$I^infty = bigcap_{n=1}^infty I^n.$$
Observe that clearly $I(I^infty) = I^infty$. Then, since $R$ is Noetherian, $I^infty$ is finitely generated, so Nakayama's lemma (Statement 1) applies.



Thus there exists $rin R$ with $r-1in I$ such that $rI^infty =0$. But then $r-1=i$ for some $iin I$, and $r=1+i$. Then $r$ is not a zero divisor by assumption, hence the fact that $rI^infty = 0$ implies that $I^infty=0$ as desired.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Do you have any ideas on the other direction? Thank you so much!
    $endgroup$
    – Gengar
    Dec 7 '18 at 18:41










  • $begingroup$
    @Gengar, for the converse red_trumpet's suggestion to use Krull's intersection theorem looks viable. I'll flesh out the details in my answer.
    $endgroup$
    – jgon
    Dec 7 '18 at 18:47










  • $begingroup$
    I appreciate it so much @jgon
    $endgroup$
    – Gengar
    Dec 7 '18 at 18:47










  • $begingroup$
    @Gengar actually I have to go, and I can't complete my thoughts rn, I'll post what I have.
    $endgroup$
    – jgon
    Dec 7 '18 at 19:09










  • $begingroup$
    @Gengar, got it, I was being silly.
    $endgroup$
    – jgon
    Dec 8 '18 at 0:30



















-1












$begingroup$

If there exists some y $in$ I such that (1-y) is a zero divisor, then there exists an x $not=$ 0 such that x(1-y) = 0. Then x = xy $in$ I. Thus x $in$ I $cap$ I$^2$. It follows that x $in$ $displaystylebigcap_n$ I$^n$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Yeah, jgon already said that
    $endgroup$
    – Gengar
    Dec 7 '18 at 19:53











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3030209%2fbigcap-n-in-mathbbn-in-0-if-and-only-if-no-zero-divisor-of-r-is-of%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

Yes, the observation that $r=rz$ $implies$ $rin I$ is very useful, since we can use the equation again to get that $rin I^2$, and thus $rin I^3$, and so on. Hence $rin bigcap_{n=1}^infty I^n$, so $r=0$, contradiction. No need to use Noetherianness here. (It may be necessary for the converse, but I haven't thought that far, since it's not clear from your question if you're also asking about that.)



Edit



Worked out my thoughts on the converse. I was being dumb. It's Nakayama's lemma (the general, not local ring version).



Let $$I^infty = bigcap_{n=1}^infty I^n.$$
Observe that clearly $I(I^infty) = I^infty$. Then, since $R$ is Noetherian, $I^infty$ is finitely generated, so Nakayama's lemma (Statement 1) applies.



Thus there exists $rin R$ with $r-1in I$ such that $rI^infty =0$. But then $r-1=i$ for some $iin I$, and $r=1+i$. Then $r$ is not a zero divisor by assumption, hence the fact that $rI^infty = 0$ implies that $I^infty=0$ as desired.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Do you have any ideas on the other direction? Thank you so much!
    $endgroup$
    – Gengar
    Dec 7 '18 at 18:41










  • $begingroup$
    @Gengar, for the converse red_trumpet's suggestion to use Krull's intersection theorem looks viable. I'll flesh out the details in my answer.
    $endgroup$
    – jgon
    Dec 7 '18 at 18:47










  • $begingroup$
    I appreciate it so much @jgon
    $endgroup$
    – Gengar
    Dec 7 '18 at 18:47










  • $begingroup$
    @Gengar actually I have to go, and I can't complete my thoughts rn, I'll post what I have.
    $endgroup$
    – jgon
    Dec 7 '18 at 19:09










  • $begingroup$
    @Gengar, got it, I was being silly.
    $endgroup$
    – jgon
    Dec 8 '18 at 0:30
















3












$begingroup$

Yes, the observation that $r=rz$ $implies$ $rin I$ is very useful, since we can use the equation again to get that $rin I^2$, and thus $rin I^3$, and so on. Hence $rin bigcap_{n=1}^infty I^n$, so $r=0$, contradiction. No need to use Noetherianness here. (It may be necessary for the converse, but I haven't thought that far, since it's not clear from your question if you're also asking about that.)



Edit



Worked out my thoughts on the converse. I was being dumb. It's Nakayama's lemma (the general, not local ring version).



Let $$I^infty = bigcap_{n=1}^infty I^n.$$
Observe that clearly $I(I^infty) = I^infty$. Then, since $R$ is Noetherian, $I^infty$ is finitely generated, so Nakayama's lemma (Statement 1) applies.



Thus there exists $rin R$ with $r-1in I$ such that $rI^infty =0$. But then $r-1=i$ for some $iin I$, and $r=1+i$. Then $r$ is not a zero divisor by assumption, hence the fact that $rI^infty = 0$ implies that $I^infty=0$ as desired.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Do you have any ideas on the other direction? Thank you so much!
    $endgroup$
    – Gengar
    Dec 7 '18 at 18:41










  • $begingroup$
    @Gengar, for the converse red_trumpet's suggestion to use Krull's intersection theorem looks viable. I'll flesh out the details in my answer.
    $endgroup$
    – jgon
    Dec 7 '18 at 18:47










  • $begingroup$
    I appreciate it so much @jgon
    $endgroup$
    – Gengar
    Dec 7 '18 at 18:47










  • $begingroup$
    @Gengar actually I have to go, and I can't complete my thoughts rn, I'll post what I have.
    $endgroup$
    – jgon
    Dec 7 '18 at 19:09










  • $begingroup$
    @Gengar, got it, I was being silly.
    $endgroup$
    – jgon
    Dec 8 '18 at 0:30














3












3








3





$begingroup$

Yes, the observation that $r=rz$ $implies$ $rin I$ is very useful, since we can use the equation again to get that $rin I^2$, and thus $rin I^3$, and so on. Hence $rin bigcap_{n=1}^infty I^n$, so $r=0$, contradiction. No need to use Noetherianness here. (It may be necessary for the converse, but I haven't thought that far, since it's not clear from your question if you're also asking about that.)



Edit



Worked out my thoughts on the converse. I was being dumb. It's Nakayama's lemma (the general, not local ring version).



Let $$I^infty = bigcap_{n=1}^infty I^n.$$
Observe that clearly $I(I^infty) = I^infty$. Then, since $R$ is Noetherian, $I^infty$ is finitely generated, so Nakayama's lemma (Statement 1) applies.



Thus there exists $rin R$ with $r-1in I$ such that $rI^infty =0$. But then $r-1=i$ for some $iin I$, and $r=1+i$. Then $r$ is not a zero divisor by assumption, hence the fact that $rI^infty = 0$ implies that $I^infty=0$ as desired.






share|cite|improve this answer











$endgroup$



Yes, the observation that $r=rz$ $implies$ $rin I$ is very useful, since we can use the equation again to get that $rin I^2$, and thus $rin I^3$, and so on. Hence $rin bigcap_{n=1}^infty I^n$, so $r=0$, contradiction. No need to use Noetherianness here. (It may be necessary for the converse, but I haven't thought that far, since it's not clear from your question if you're also asking about that.)



Edit



Worked out my thoughts on the converse. I was being dumb. It's Nakayama's lemma (the general, not local ring version).



Let $$I^infty = bigcap_{n=1}^infty I^n.$$
Observe that clearly $I(I^infty) = I^infty$. Then, since $R$ is Noetherian, $I^infty$ is finitely generated, so Nakayama's lemma (Statement 1) applies.



Thus there exists $rin R$ with $r-1in I$ such that $rI^infty =0$. But then $r-1=i$ for some $iin I$, and $r=1+i$. Then $r$ is not a zero divisor by assumption, hence the fact that $rI^infty = 0$ implies that $I^infty=0$ as desired.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 8 '18 at 0:30

























answered Dec 7 '18 at 18:35









jgonjgon

13.6k22041




13.6k22041












  • $begingroup$
    Do you have any ideas on the other direction? Thank you so much!
    $endgroup$
    – Gengar
    Dec 7 '18 at 18:41










  • $begingroup$
    @Gengar, for the converse red_trumpet's suggestion to use Krull's intersection theorem looks viable. I'll flesh out the details in my answer.
    $endgroup$
    – jgon
    Dec 7 '18 at 18:47










  • $begingroup$
    I appreciate it so much @jgon
    $endgroup$
    – Gengar
    Dec 7 '18 at 18:47










  • $begingroup$
    @Gengar actually I have to go, and I can't complete my thoughts rn, I'll post what I have.
    $endgroup$
    – jgon
    Dec 7 '18 at 19:09










  • $begingroup$
    @Gengar, got it, I was being silly.
    $endgroup$
    – jgon
    Dec 8 '18 at 0:30


















  • $begingroup$
    Do you have any ideas on the other direction? Thank you so much!
    $endgroup$
    – Gengar
    Dec 7 '18 at 18:41










  • $begingroup$
    @Gengar, for the converse red_trumpet's suggestion to use Krull's intersection theorem looks viable. I'll flesh out the details in my answer.
    $endgroup$
    – jgon
    Dec 7 '18 at 18:47










  • $begingroup$
    I appreciate it so much @jgon
    $endgroup$
    – Gengar
    Dec 7 '18 at 18:47










  • $begingroup$
    @Gengar actually I have to go, and I can't complete my thoughts rn, I'll post what I have.
    $endgroup$
    – jgon
    Dec 7 '18 at 19:09










  • $begingroup$
    @Gengar, got it, I was being silly.
    $endgroup$
    – jgon
    Dec 8 '18 at 0:30
















$begingroup$
Do you have any ideas on the other direction? Thank you so much!
$endgroup$
– Gengar
Dec 7 '18 at 18:41




$begingroup$
Do you have any ideas on the other direction? Thank you so much!
$endgroup$
– Gengar
Dec 7 '18 at 18:41












$begingroup$
@Gengar, for the converse red_trumpet's suggestion to use Krull's intersection theorem looks viable. I'll flesh out the details in my answer.
$endgroup$
– jgon
Dec 7 '18 at 18:47




$begingroup$
@Gengar, for the converse red_trumpet's suggestion to use Krull's intersection theorem looks viable. I'll flesh out the details in my answer.
$endgroup$
– jgon
Dec 7 '18 at 18:47












$begingroup$
I appreciate it so much @jgon
$endgroup$
– Gengar
Dec 7 '18 at 18:47




$begingroup$
I appreciate it so much @jgon
$endgroup$
– Gengar
Dec 7 '18 at 18:47












$begingroup$
@Gengar actually I have to go, and I can't complete my thoughts rn, I'll post what I have.
$endgroup$
– jgon
Dec 7 '18 at 19:09




$begingroup$
@Gengar actually I have to go, and I can't complete my thoughts rn, I'll post what I have.
$endgroup$
– jgon
Dec 7 '18 at 19:09












$begingroup$
@Gengar, got it, I was being silly.
$endgroup$
– jgon
Dec 8 '18 at 0:30




$begingroup$
@Gengar, got it, I was being silly.
$endgroup$
– jgon
Dec 8 '18 at 0:30











-1












$begingroup$

If there exists some y $in$ I such that (1-y) is a zero divisor, then there exists an x $not=$ 0 such that x(1-y) = 0. Then x = xy $in$ I. Thus x $in$ I $cap$ I$^2$. It follows that x $in$ $displaystylebigcap_n$ I$^n$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Yeah, jgon already said that
    $endgroup$
    – Gengar
    Dec 7 '18 at 19:53
















-1












$begingroup$

If there exists some y $in$ I such that (1-y) is a zero divisor, then there exists an x $not=$ 0 such that x(1-y) = 0. Then x = xy $in$ I. Thus x $in$ I $cap$ I$^2$. It follows that x $in$ $displaystylebigcap_n$ I$^n$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Yeah, jgon already said that
    $endgroup$
    – Gengar
    Dec 7 '18 at 19:53














-1












-1








-1





$begingroup$

If there exists some y $in$ I such that (1-y) is a zero divisor, then there exists an x $not=$ 0 such that x(1-y) = 0. Then x = xy $in$ I. Thus x $in$ I $cap$ I$^2$. It follows that x $in$ $displaystylebigcap_n$ I$^n$.






share|cite|improve this answer









$endgroup$



If there exists some y $in$ I such that (1-y) is a zero divisor, then there exists an x $not=$ 0 such that x(1-y) = 0. Then x = xy $in$ I. Thus x $in$ I $cap$ I$^2$. It follows that x $in$ $displaystylebigcap_n$ I$^n$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 7 '18 at 19:43









Joel PereiraJoel Pereira

74519




74519












  • $begingroup$
    Yeah, jgon already said that
    $endgroup$
    – Gengar
    Dec 7 '18 at 19:53


















  • $begingroup$
    Yeah, jgon already said that
    $endgroup$
    – Gengar
    Dec 7 '18 at 19:53
















$begingroup$
Yeah, jgon already said that
$endgroup$
– Gengar
Dec 7 '18 at 19:53




$begingroup$
Yeah, jgon already said that
$endgroup$
– Gengar
Dec 7 '18 at 19:53


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3030209%2fbigcap-n-in-mathbbn-in-0-if-and-only-if-no-zero-divisor-of-r-is-of%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Wiesbaden

Image
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
Read more

Marschland

Image
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
Read more

Dieringhausen

Image
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
Read more