Fdtxjr

Sword in the Stone story where the sword was held in place by electromagnets

Happy pi day, everyone!

What is this large pipe coming out of my roof?

Is a lawful good "antagonist" effective?

Identifying the interval from A♭ to D♯

How could a scammer know the apps on my phone / iTunes account?

An Accountant Seeks the Help of a Mathematician

Employee lack of ownership

Informing my boss about remarks from a nasty colleague

Possible Leak In Concrete

Why does Deadpool say "You're welcome, Canada," after shooting Ryan Reynolds in the end credits?

Latest web browser compatible with Windows 98

What are the possible solutions of the given equation?

Is it possible to upcast ritual spells?

Ban on all campaign finance?

Do I need life insurance if I can cover my own funeral costs?

Why did it take so long to abandon sail after steamships were demonstrated?

How to deal with taxi scam when on vacation?

Can anyone tell me why this program fails?

Will a pinhole camera work with instant film?

Why are there 40 737 Max planes in flight when they have been grounded as not airworthy?

How could a female member of a species produce eggs unto death?

Replacing Windows 7 security updates with anti-virus?

Where is the 1/8 CR apprentice in Volo's Guide to Monsters?

Describing the elements of quotient ring of $mathbbZ[sqrtD]$.

Why is $mathbbZ[sqrt-n], nge 3$ not a UFD?Does $mathbbZ[sqrt[3]2]=mathbbZ[sqrt[3]2]^mathrmint$?In which extensions of type $mathbbQ(sqrtd)$ is the number 3 reducible?Why is quadratic integer ring defined in that way?Maximal ideal in a polynomial ring over $mathbb Z$Quotient ring $R/I$, $R= mathbbZ[sqrt-10]$, $I =(sqrt-10)$About the ring $mathbbZ[sqrt4]$Tensor algebra on non free modulesMethod to adjoin elements to a ring - is the symbol $x$ overloaded?Prime ideals in certain quadratic ring

1

$begingroup$

For Gaussian integer ring $mathbbZ[i]$, there is a method describing distinct elements of certain quotient ring of $mathbbZ[i]$ using 'the visualization'.

The images(in K. Conrad's note) below are examples of the method using visualization.

So, i wonder about that:

Is there a similar method for quotient ring of $mathbbZ[sqrtD]$, where $D$ is square-free?

I guess it is possible in imaginary quadratic field $mathbbQ[sqrt-d]$, where $d>0$.

Give some advice! Thank you!

abstract-algebra gaussian-integers

share|cite|improve this question

edited Mar 11 at 11:06

Primavera

asked Mar 11 at 11:01

PrimaveraPrimavera

30719

$endgroup$

add a comment | 

1

$begingroup$

For Gaussian integer ring $mathbbZ[i]$, there is a method describing distinct elements of certain quotient ring of $mathbbZ[i]$ using 'the visualization'.

The images(in K. Conrad's note) below are examples of the method using visualization.

So, i wonder about that:

Is there a similar method for quotient ring of $mathbbZ[sqrtD]$, where $D$ is square-free?

I guess it is possible in imaginary quadratic field $mathbbQ[sqrt-d]$, where $d>0$.

Give some advice! Thank you!

abstract-algebra gaussian-integers

share|cite|improve this question

edited Mar 11 at 11:06

Primavera

asked Mar 11 at 11:01

PrimaveraPrimavera

30719

$endgroup$

add a comment | 

$begingroup$

For Gaussian integer ring $mathbbZ[i]$, there is a method describing distinct elements of certain quotient ring of $mathbbZ[i]$ using 'the visualization'.

The images(in K. Conrad's note) below are examples of the method using visualization.

So, i wonder about that:

Is there a similar method for quotient ring of $mathbbZ[sqrtD]$, where $D$ is square-free?

I guess it is possible in imaginary quadratic field $mathbbQ[sqrt-d]$, where $d>0$.

Give some advice! Thank you!

abstract-algebra gaussian-integers

share|cite|improve this question

edited Mar 11 at 11:06

Primavera

asked Mar 11 at 11:01

PrimaveraPrimavera

30719

$endgroup$

share|cite|improve this question

edited Mar 11 at 11:06

Primavera

asked Mar 11 at 11:01

PrimaveraPrimavera

30719

share|cite|improve this question

edited Mar 11 at 11:06

Primavera

asked Mar 11 at 11:01

PrimaveraPrimavera

30719

asked Mar 11 at 11:01

PrimaveraPrimavera

30719

asked Mar 11 at 11:01

PrimaveraPrimavera

30719

1 Answer
1

active

oldest

votes

0

$begingroup$

I'll make a try...

Let's say we have $mathbbZ[isqrtk]=a+bisqrtk:a,binmathbbZ$ where $kinmathbbN $ is square-free.

So for the multiples of $(1+2i)$ we see that
$$(1+2i)(a+bisqrtk)=a(1+2i)+b(-2sqrtk+sqrtki)$$

and $(1+2i),(-2sqrtk+sqrtki)$ correspond to perpendicular vectors, so you can proceed as before (now you will not have squares but rectangles).

share|cite|improve this answer

answered Mar 11 at 12:05

giannispapavgiannispapav

1,794324

$endgroup$

add a comment | 

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

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3143557%2fdescribing-the-elements-of-quotient-ring-of-mathbbz-sqrtd%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

0

$begingroup$

I'll make a try...

Let's say we have $mathbbZ[isqrtk]=a+bisqrtk:a,binmathbbZ$ where $kinmathbbN $ is square-free.

So for the multiples of $(1+2i)$ we see that
$$(1+2i)(a+bisqrtk)=a(1+2i)+b(-2sqrtk+sqrtki)$$

and $(1+2i),(-2sqrtk+sqrtki)$ correspond to perpendicular vectors, so you can proceed as before (now you will not have squares but rectangles).

share|cite|improve this answer

answered Mar 11 at 12:05

giannispapavgiannispapav

1,794324

$endgroup$

add a comment | 

0

$begingroup$

I'll make a try...

Let's say we have $mathbbZ[isqrtk]=a+bisqrtk:a,binmathbbZ$ where $kinmathbbN $ is square-free.

So for the multiples of $(1+2i)$ we see that
$$(1+2i)(a+bisqrtk)=a(1+2i)+b(-2sqrtk+sqrtki)$$

and $(1+2i),(-2sqrtk+sqrtki)$ correspond to perpendicular vectors, so you can proceed as before (now you will not have squares but rectangles).

share|cite|improve this answer

answered Mar 11 at 12:05

giannispapavgiannispapav

1,794324

$endgroup$

add a comment | 

$begingroup$

I'll make a try...

Let's say we have $mathbbZ[isqrtk]=a+bisqrtk:a,binmathbbZ$ where $kinmathbbN $ is square-free.

So for the multiples of $(1+2i)$ we see that
$$(1+2i)(a+bisqrtk)=a(1+2i)+b(-2sqrtk+sqrtki)$$

and $(1+2i),(-2sqrtk+sqrtki)$ correspond to perpendicular vectors, so you can proceed as before (now you will not have squares but rectangles).

share|cite|improve this answer

answered Mar 11 at 12:05

giannispapavgiannispapav

1,794324

$endgroup$

share|cite|improve this answer

answered Mar 11 at 12:05

giannispapavgiannispapav

1,794324

answered Mar 11 at 12:05

giannispapavgiannispapav

1,794324

answered Mar 11 at 12:05

giannispapavgiannispapav

1,794324