Number of roots of a quadratic polynomial with coefficients in ring $mathbb{Z_{18}}$












3












$begingroup$


I am trying to solve the following problem:




How many roots can a polynomial $P(x) = ax^2 + bx + c $, where $a$, $b$, $c in mathbb{Z_{18}}$, have?

($mathbb{Z_{18}} = {0, 1, 2, ldots, 17}$)




Obviously, they can not have more than 18 roots.



If it was in real numbers the answer would be either 2 or 0 as real polynomials have as many roots (can be pairs of complex numbers) as is their exponent.



I tried this...



Let $x_1$ be a root of $p(x)$.
Another root $x_2$ may exist;
$ 0 = a(x_1)^2 + bx_1 + c = a(x_2)^2 + bx_2 + c$
$ a(x_1)^2 + bx_1 = a(x_2)^2 + bx_2$
$ a((x_1)^2 - (x_2)^2) = b(x_2 - x_1)$



I can't just say $(x_1)^2 - (x_2)^2 = 0$ as there are zero divisors in $mathbb{Z_{18}}$.



Is any of this in the right direction? I could use some help.




And is there a more general way I could describe possible roots of polynomials from that ring?







abstract-algebra group-theory polynomials ring-theory cyclic-groups







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    In the second and third line you don't get that these expressions are equal to zero.
    $endgroup$
    – Lukas Geyer
    Dec 17 '18 at 21:51










  • $begingroup$
    Oh, of course, thanks. I will correct it.
    $endgroup$
    – Coupeau
    Dec 17 '18 at 22:07










  • $begingroup$
    You get $(a(x_1+x_2)+b)(x_1-x_2)=0$.
    $endgroup$
    – Berci
    Dec 17 '18 at 22:37












  • $begingroup$
    It's unclear whether the question is asking for the maximum possible number of roots, or for the set of all possible numbers of roots.
    $endgroup$
    – Daniel Schepler
    Dec 17 '18 at 23:17
















3












$begingroup$


I am trying to solve the following problem:




How many roots can a polynomial $P(x) = ax^2 + bx + c $, where $a$, $b$, $c in mathbb{Z_{18}}$, have?

($mathbb{Z_{18}} = {0, 1, 2, ldots, 17}$)




Obviously, they can not have more than 18 roots.



If it was in real numbers the answer would be either 2 or 0 as real polynomials have as many roots (can be pairs of complex numbers) as is their exponent.



I tried this...



Let $x_1$ be a root of $p(x)$.
Another root $x_2$ may exist;
$ 0 = a(x_1)^2 + bx_1 + c = a(x_2)^2 + bx_2 + c$
$ a(x_1)^2 + bx_1 = a(x_2)^2 + bx_2$
$ a((x_1)^2 - (x_2)^2) = b(x_2 - x_1)$



I can't just say $(x_1)^2 - (x_2)^2 = 0$ as there are zero divisors in $mathbb{Z_{18}}$.



Is any of this in the right direction? I could use some help.




And is there a more general way I could describe possible roots of polynomials from that ring?







abstract-algebra group-theory polynomials ring-theory cyclic-groups







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    In the second and third line you don't get that these expressions are equal to zero.
    $endgroup$
    – Lukas Geyer
    Dec 17 '18 at 21:51










  • $begingroup$
    Oh, of course, thanks. I will correct it.
    $endgroup$
    – Coupeau
    Dec 17 '18 at 22:07










  • $begingroup$
    You get $(a(x_1+x_2)+b)(x_1-x_2)=0$.
    $endgroup$
    – Berci
    Dec 17 '18 at 22:37












  • $begingroup$
    It's unclear whether the question is asking for the maximum possible number of roots, or for the set of all possible numbers of roots.
    $endgroup$
    – Daniel Schepler
    Dec 17 '18 at 23:17














3












3








3





$begingroup$


I am trying to solve the following problem:




How many roots can a polynomial $P(x) = ax^2 + bx + c $, where $a$, $b$, $c in mathbb{Z_{18}}$, have?

($mathbb{Z_{18}} = {0, 1, 2, ldots, 17}$)




Obviously, they can not have more than 18 roots.



If it was in real numbers the answer would be either 2 or 0 as real polynomials have as many roots (can be pairs of complex numbers) as is their exponent.



I tried this...



Let $x_1$ be a root of $p(x)$.
Another root $x_2$ may exist;
$ 0 = a(x_1)^2 + bx_1 + c = a(x_2)^2 + bx_2 + c$
$ a(x_1)^2 + bx_1 = a(x_2)^2 + bx_2$
$ a((x_1)^2 - (x_2)^2) = b(x_2 - x_1)$



I can't just say $(x_1)^2 - (x_2)^2 = 0$ as there are zero divisors in $mathbb{Z_{18}}$.



Is any of this in the right direction? I could use some help.




And is there a more general way I could describe possible roots of polynomials from that ring?







abstract-algebra group-theory polynomials ring-theory cyclic-groups







share|cite|improve this question











$endgroup$




I am trying to solve the following problem:




How many roots can a polynomial $P(x) = ax^2 + bx + c $, where $a$, $b$, $c in mathbb{Z_{18}}$, have?

($mathbb{Z_{18}} = {0, 1, 2, ldots, 17}$)




Obviously, they can not have more than 18 roots.



If it was in real numbers the answer would be either 2 or 0 as real polynomials have as many roots (can be pairs of complex numbers) as is their exponent.



I tried this...



Let $x_1$ be a root of $p(x)$.
Another root $x_2$ may exist;
$ 0 = a(x_1)^2 + bx_1 + c = a(x_2)^2 + bx_2 + c$
$ a(x_1)^2 + bx_1 = a(x_2)^2 + bx_2$
$ a((x_1)^2 - (x_2)^2) = b(x_2 - x_1)$



I can't just say $(x_1)^2 - (x_2)^2 = 0$ as there are zero divisors in $mathbb{Z_{18}}$.



Is any of this in the right direction? I could use some help.




And is there a more general way I could describe possible roots of polynomials from that ring?







abstract-algebra group-theory polynomials ring-theory cyclic-groups




abstract-algebra group-theory polynomials ring-theory cyclic-groups






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 17 '18 at 22:16









Blue

48.5k870154




48.5k870154










asked Dec 17 '18 at 21:18









CoupeauCoupeau

1296




1296








  • 1




    $begingroup$
    In the second and third line you don't get that these expressions are equal to zero.
    $endgroup$
    – Lukas Geyer
    Dec 17 '18 at 21:51










  • $begingroup$
    Oh, of course, thanks. I will correct it.
    $endgroup$
    – Coupeau
    Dec 17 '18 at 22:07










  • $begingroup$
    You get $(a(x_1+x_2)+b)(x_1-x_2)=0$.
    $endgroup$
    – Berci
    Dec 17 '18 at 22:37












  • $begingroup$
    It's unclear whether the question is asking for the maximum possible number of roots, or for the set of all possible numbers of roots.
    $endgroup$
    – Daniel Schepler
    Dec 17 '18 at 23:17














  • 1




    $begingroup$
    In the second and third line you don't get that these expressions are equal to zero.
    $endgroup$
    – Lukas Geyer
    Dec 17 '18 at 21:51










  • $begingroup$
    Oh, of course, thanks. I will correct it.
    $endgroup$
    – Coupeau
    Dec 17 '18 at 22:07










  • $begingroup$
    You get $(a(x_1+x_2)+b)(x_1-x_2)=0$.
    $endgroup$
    – Berci
    Dec 17 '18 at 22:37












  • $begingroup$
    It's unclear whether the question is asking for the maximum possible number of roots, or for the set of all possible numbers of roots.
    $endgroup$
    – Daniel Schepler
    Dec 17 '18 at 23:17








1




1




$begingroup$
In the second and third line you don't get that these expressions are equal to zero.
$endgroup$
– Lukas Geyer
Dec 17 '18 at 21:51




$begingroup$
In the second and third line you don't get that these expressions are equal to zero.
$endgroup$
– Lukas Geyer
Dec 17 '18 at 21:51












$begingroup$
Oh, of course, thanks. I will correct it.
$endgroup$
– Coupeau
Dec 17 '18 at 22:07




$begingroup$
Oh, of course, thanks. I will correct it.
$endgroup$
– Coupeau
Dec 17 '18 at 22:07












$begingroup$
You get $(a(x_1+x_2)+b)(x_1-x_2)=0$.
$endgroup$
– Berci
Dec 17 '18 at 22:37






$begingroup$
You get $(a(x_1+x_2)+b)(x_1-x_2)=0$.
$endgroup$
– Berci
Dec 17 '18 at 22:37














$begingroup$
It's unclear whether the question is asking for the maximum possible number of roots, or for the set of all possible numbers of roots.
$endgroup$
– Daniel Schepler
Dec 17 '18 at 23:17




$begingroup$
It's unclear whether the question is asking for the maximum possible number of roots, or for the set of all possible numbers of roots.
$endgroup$
– Daniel Schepler
Dec 17 '18 at 23:17










1 Answer
1






active

oldest

votes


















3












$begingroup$

[I assume that $a$ is required to be nonzero so that we really have a quadratic; taking the problem statement literally you could just let $a=b=c=0$ to trivially get $18$ roots.]



The way I would approach this is using the Chinese remainder theorem. Since $mathbb{Z}_{18}cong mathbb{Z}_2timesmathbb{Z}_9$, we can think about the problem separately in $mathbb{Z}_2$ and $mathbb{Z}_9$.



Now $mathbb{Z}_2$ is easy, since it only has two elements: $x(x-1)=x^2-x$ has two roots, and that is the maximum possible.



Notice now that any quadratic $ax^2+bx+c$ over $mathbb{Z}_{18}$ which reduces to $x^2-x$ mod $2$ will have $aneq 0$. This means that we are free to have $a=b=c=0$ when we reduce mod $9$, so that over $mathbb{Z}_9$ we will have $9$ roots. That will get us a quadratic over $mathbb{Z}_{18}$ with all $18$ elements as roots!



Explicitly, we want $ax^2+bx+c$ which reduces to $x^2-x$ mod $2$ and reduces to $0$ mod $9$. We can do this with $a=b=9$ and $c=0$, so the quadratic $9x^2+9x$ has $18$ roots over $mathbb{Z}_{18}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    $9x^2+9x$ is the only quadratic that has $18$ roots over $mathbb{Z}_{18}$.
    $endgroup$
    – lhf
    Dec 17 '18 at 23:10










  • $begingroup$
    Sorry to bother you, but I was wondering if you could help with this question. It might be trivial, but I'm not sure.
    $endgroup$
    – MathematicsStudent1122
    Dec 18 '18 at 3:25













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%2f3044451%2fnumber-of-roots-of-a-quadratic-polynomial-with-coefficients-in-ring-mathbbz%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

[I assume that $a$ is required to be nonzero so that we really have a quadratic; taking the problem statement literally you could just let $a=b=c=0$ to trivially get $18$ roots.]



The way I would approach this is using the Chinese remainder theorem. Since $mathbb{Z}_{18}cong mathbb{Z}_2timesmathbb{Z}_9$, we can think about the problem separately in $mathbb{Z}_2$ and $mathbb{Z}_9$.



Now $mathbb{Z}_2$ is easy, since it only has two elements: $x(x-1)=x^2-x$ has two roots, and that is the maximum possible.



Notice now that any quadratic $ax^2+bx+c$ over $mathbb{Z}_{18}$ which reduces to $x^2-x$ mod $2$ will have $aneq 0$. This means that we are free to have $a=b=c=0$ when we reduce mod $9$, so that over $mathbb{Z}_9$ we will have $9$ roots. That will get us a quadratic over $mathbb{Z}_{18}$ with all $18$ elements as roots!



Explicitly, we want $ax^2+bx+c$ which reduces to $x^2-x$ mod $2$ and reduces to $0$ mod $9$. We can do this with $a=b=9$ and $c=0$, so the quadratic $9x^2+9x$ has $18$ roots over $mathbb{Z}_{18}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    $9x^2+9x$ is the only quadratic that has $18$ roots over $mathbb{Z}_{18}$.
    $endgroup$
    – lhf
    Dec 17 '18 at 23:10










  • $begingroup$
    Sorry to bother you, but I was wondering if you could help with this question. It might be trivial, but I'm not sure.
    $endgroup$
    – MathematicsStudent1122
    Dec 18 '18 at 3:25


















3












$begingroup$

[I assume that $a$ is required to be nonzero so that we really have a quadratic; taking the problem statement literally you could just let $a=b=c=0$ to trivially get $18$ roots.]



The way I would approach this is using the Chinese remainder theorem. Since $mathbb{Z}_{18}cong mathbb{Z}_2timesmathbb{Z}_9$, we can think about the problem separately in $mathbb{Z}_2$ and $mathbb{Z}_9$.



Now $mathbb{Z}_2$ is easy, since it only has two elements: $x(x-1)=x^2-x$ has two roots, and that is the maximum possible.



Notice now that any quadratic $ax^2+bx+c$ over $mathbb{Z}_{18}$ which reduces to $x^2-x$ mod $2$ will have $aneq 0$. This means that we are free to have $a=b=c=0$ when we reduce mod $9$, so that over $mathbb{Z}_9$ we will have $9$ roots. That will get us a quadratic over $mathbb{Z}_{18}$ with all $18$ elements as roots!



Explicitly, we want $ax^2+bx+c$ which reduces to $x^2-x$ mod $2$ and reduces to $0$ mod $9$. We can do this with $a=b=9$ and $c=0$, so the quadratic $9x^2+9x$ has $18$ roots over $mathbb{Z}_{18}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    $9x^2+9x$ is the only quadratic that has $18$ roots over $mathbb{Z}_{18}$.
    $endgroup$
    – lhf
    Dec 17 '18 at 23:10










  • $begingroup$
    Sorry to bother you, but I was wondering if you could help with this question. It might be trivial, but I'm not sure.
    $endgroup$
    – MathematicsStudent1122
    Dec 18 '18 at 3:25
















3












3








3





$begingroup$

[I assume that $a$ is required to be nonzero so that we really have a quadratic; taking the problem statement literally you could just let $a=b=c=0$ to trivially get $18$ roots.]



The way I would approach this is using the Chinese remainder theorem. Since $mathbb{Z}_{18}cong mathbb{Z}_2timesmathbb{Z}_9$, we can think about the problem separately in $mathbb{Z}_2$ and $mathbb{Z}_9$.



Now $mathbb{Z}_2$ is easy, since it only has two elements: $x(x-1)=x^2-x$ has two roots, and that is the maximum possible.



Notice now that any quadratic $ax^2+bx+c$ over $mathbb{Z}_{18}$ which reduces to $x^2-x$ mod $2$ will have $aneq 0$. This means that we are free to have $a=b=c=0$ when we reduce mod $9$, so that over $mathbb{Z}_9$ we will have $9$ roots. That will get us a quadratic over $mathbb{Z}_{18}$ with all $18$ elements as roots!



Explicitly, we want $ax^2+bx+c$ which reduces to $x^2-x$ mod $2$ and reduces to $0$ mod $9$. We can do this with $a=b=9$ and $c=0$, so the quadratic $9x^2+9x$ has $18$ roots over $mathbb{Z}_{18}$.






share|cite|improve this answer











$endgroup$



[I assume that $a$ is required to be nonzero so that we really have a quadratic; taking the problem statement literally you could just let $a=b=c=0$ to trivially get $18$ roots.]



The way I would approach this is using the Chinese remainder theorem. Since $mathbb{Z}_{18}cong mathbb{Z}_2timesmathbb{Z}_9$, we can think about the problem separately in $mathbb{Z}_2$ and $mathbb{Z}_9$.



Now $mathbb{Z}_2$ is easy, since it only has two elements: $x(x-1)=x^2-x$ has two roots, and that is the maximum possible.



Notice now that any quadratic $ax^2+bx+c$ over $mathbb{Z}_{18}$ which reduces to $x^2-x$ mod $2$ will have $aneq 0$. This means that we are free to have $a=b=c=0$ when we reduce mod $9$, so that over $mathbb{Z}_9$ we will have $9$ roots. That will get us a quadratic over $mathbb{Z}_{18}$ with all $18$ elements as roots!



Explicitly, we want $ax^2+bx+c$ which reduces to $x^2-x$ mod $2$ and reduces to $0$ mod $9$. We can do this with $a=b=9$ and $c=0$, so the quadratic $9x^2+9x$ has $18$ roots over $mathbb{Z}_{18}$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 17 '18 at 23:07

























answered Dec 17 '18 at 23:02









Eric WofseyEric Wofsey

187k14215344




187k14215344












  • $begingroup$
    $9x^2+9x$ is the only quadratic that has $18$ roots over $mathbb{Z}_{18}$.
    $endgroup$
    – lhf
    Dec 17 '18 at 23:10










  • $begingroup$
    Sorry to bother you, but I was wondering if you could help with this question. It might be trivial, but I'm not sure.
    $endgroup$
    – MathematicsStudent1122
    Dec 18 '18 at 3:25




















  • $begingroup$
    $9x^2+9x$ is the only quadratic that has $18$ roots over $mathbb{Z}_{18}$.
    $endgroup$
    – lhf
    Dec 17 '18 at 23:10










  • $begingroup$
    Sorry to bother you, but I was wondering if you could help with this question. It might be trivial, but I'm not sure.
    $endgroup$
    – MathematicsStudent1122
    Dec 18 '18 at 3:25


















$begingroup$
$9x^2+9x$ is the only quadratic that has $18$ roots over $mathbb{Z}_{18}$.
$endgroup$
– lhf
Dec 17 '18 at 23:10




$begingroup$
$9x^2+9x$ is the only quadratic that has $18$ roots over $mathbb{Z}_{18}$.
$endgroup$
– lhf
Dec 17 '18 at 23:10












$begingroup$
Sorry to bother you, but I was wondering if you could help with this question. It might be trivial, but I'm not sure.
$endgroup$
– MathematicsStudent1122
Dec 18 '18 at 3:25






$begingroup$
Sorry to bother you, but I was wondering if you could help with this question. It might be trivial, but I'm not sure.
$endgroup$
– MathematicsStudent1122
Dec 18 '18 at 3:25




















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%2f3044451%2fnumber-of-roots-of-a-quadratic-polynomial-with-coefficients-in-ring-mathbbz%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