Why $R/(a+bi)$ is a finite field with characteristic $p$ where $a^2+b^2=p$ [duplicate]












0















This question already has an answer here:




  • Quotient ring of Gaussian integers

    6 answers




Suppose $R=mathbb{Z}[i]$ is Gaussian domain , and $a^2+b^2=p$ is prime. Denote $alpha=a+bi$ and $I=(alpha)$. Prove $R/I$ is a finite field with characteristic p.



My attempts: First prove $alpha $ is irreducible, hence $I$ is a maximal ideal and it follows $R/I$ is a field. And $R=mathbb{Z}[i]$ is an Euclidean domain. Hence for $xin mathbb{Z}[i],x=qalpha+r$ where $N(r)lt N(alpha)$ , so there are finite elements in $R/I$. Thus $R/I$ is a finite filed. But I can't figure out why the characteristic is $p$.



Thanks for your hints.






abstract-algebra finite-fields







share|cite|improve this question













marked as duplicate by André 3000, amWhy abstract-algebra
Users with the  abstract-algebra badge can single-handedly close abstract-algebra questions as duplicates and reopen them as needed.

StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;

$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');

$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
 2 days ago


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.











  • 1




    $R/I$ has $p$ elements.
    – Lord Shark the Unknown
    2 days ago










  • always $mathbb Z[i]/(a+bi)cong mathbb Z/N(a+bi)mathbb Z$ where $(a,b)=1$
    – Mustafa
    2 days ago










  • How to determine the number of elements in $R/I$?
    – Jaqen Chou
    2 days ago
















0















This question already has an answer here:




  • Quotient ring of Gaussian integers

    6 answers




Suppose $R=mathbb{Z}[i]$ is Gaussian domain , and $a^2+b^2=p$ is prime. Denote $alpha=a+bi$ and $I=(alpha)$. Prove $R/I$ is a finite field with characteristic p.



My attempts: First prove $alpha $ is irreducible, hence $I$ is a maximal ideal and it follows $R/I$ is a field. And $R=mathbb{Z}[i]$ is an Euclidean domain. Hence for $xin mathbb{Z}[i],x=qalpha+r$ where $N(r)lt N(alpha)$ , so there are finite elements in $R/I$. Thus $R/I$ is a finite filed. But I can't figure out why the characteristic is $p$.



Thanks for your hints.






abstract-algebra finite-fields







share|cite|improve this question













marked as duplicate by André 3000, amWhy abstract-algebra
Users with the  abstract-algebra badge can single-handedly close abstract-algebra questions as duplicates and reopen them as needed.

StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;

$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');

$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
 2 days ago


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.











  • 1




    $R/I$ has $p$ elements.
    – Lord Shark the Unknown
    2 days ago










  • always $mathbb Z[i]/(a+bi)cong mathbb Z/N(a+bi)mathbb Z$ where $(a,b)=1$
    – Mustafa
    2 days ago










  • How to determine the number of elements in $R/I$?
    – Jaqen Chou
    2 days ago














0












0








0








This question already has an answer here:




  • Quotient ring of Gaussian integers

    6 answers




Suppose $R=mathbb{Z}[i]$ is Gaussian domain , and $a^2+b^2=p$ is prime. Denote $alpha=a+bi$ and $I=(alpha)$. Prove $R/I$ is a finite field with characteristic p.



My attempts: First prove $alpha $ is irreducible, hence $I$ is a maximal ideal and it follows $R/I$ is a field. And $R=mathbb{Z}[i]$ is an Euclidean domain. Hence for $xin mathbb{Z}[i],x=qalpha+r$ where $N(r)lt N(alpha)$ , so there are finite elements in $R/I$. Thus $R/I$ is a finite filed. But I can't figure out why the characteristic is $p$.



Thanks for your hints.






abstract-algebra finite-fields







share|cite|improve this question














This question already has an answer here:




  • Quotient ring of Gaussian integers

    6 answers




Suppose $R=mathbb{Z}[i]$ is Gaussian domain , and $a^2+b^2=p$ is prime. Denote $alpha=a+bi$ and $I=(alpha)$. Prove $R/I$ is a finite field with characteristic p.



My attempts: First prove $alpha $ is irreducible, hence $I$ is a maximal ideal and it follows $R/I$ is a field. And $R=mathbb{Z}[i]$ is an Euclidean domain. Hence for $xin mathbb{Z}[i],x=qalpha+r$ where $N(r)lt N(alpha)$ , so there are finite elements in $R/I$. Thus $R/I$ is a finite filed. But I can't figure out why the characteristic is $p$.



Thanks for your hints.





This question already has an answer here:




  • Quotient ring of Gaussian integers

    6 answers







abstract-algebra finite-fields




abstract-algebra finite-fields






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 2 days ago









Jaqen Chou

409110




409110




marked as duplicate by André 3000, amWhy abstract-algebra
Users with the  abstract-algebra badge can single-handedly close abstract-algebra questions as duplicates and reopen them as needed.

StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;

$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');

$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
 2 days ago


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.






marked as duplicate by André 3000, amWhy abstract-algebra
Users with the  abstract-algebra badge can single-handedly close abstract-algebra questions as duplicates and reopen them as needed.

StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;

$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');

$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
 2 days ago


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.










  • 1




    $R/I$ has $p$ elements.
    – Lord Shark the Unknown
    2 days ago










  • always $mathbb Z[i]/(a+bi)cong mathbb Z/N(a+bi)mathbb Z$ where $(a,b)=1$
    – Mustafa
    2 days ago










  • How to determine the number of elements in $R/I$?
    – Jaqen Chou
    2 days ago














  • 1




    $R/I$ has $p$ elements.
    – Lord Shark the Unknown
    2 days ago










  • always $mathbb Z[i]/(a+bi)cong mathbb Z/N(a+bi)mathbb Z$ where $(a,b)=1$
    – Mustafa
    2 days ago










  • How to determine the number of elements in $R/I$?
    – Jaqen Chou
    2 days ago








1




1




$R/I$ has $p$ elements.
– Lord Shark the Unknown
2 days ago




$R/I$ has $p$ elements.
– Lord Shark the Unknown
2 days ago












always $mathbb Z[i]/(a+bi)cong mathbb Z/N(a+bi)mathbb Z$ where $(a,b)=1$
– Mustafa
2 days ago




always $mathbb Z[i]/(a+bi)cong mathbb Z/N(a+bi)mathbb Z$ where $(a,b)=1$
– Mustafa
2 days ago












How to determine the number of elements in $R/I$?
– Jaqen Chou
2 days ago




How to determine the number of elements in $R/I$?
– Jaqen Chou
2 days ago










1 Answer
1






active

oldest

votes


















2














$p = a^2+b^2 = bar alpha alpha in I$ implies $p=0$ in $R/I$.



If you know that $R/I$ is a field, then it must have characteristic $p$.






share|cite|improve this answer




























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2














    $p = a^2+b^2 = bar alpha alpha in I$ implies $p=0$ in $R/I$.



    If you know that $R/I$ is a field, then it must have characteristic $p$.






    share|cite|improve this answer


























      2














      $p = a^2+b^2 = bar alpha alpha in I$ implies $p=0$ in $R/I$.



      If you know that $R/I$ is a field, then it must have characteristic $p$.






      share|cite|improve this answer
























        2












        2








        2






        $p = a^2+b^2 = bar alpha alpha in I$ implies $p=0$ in $R/I$.



        If you know that $R/I$ is a field, then it must have characteristic $p$.






        share|cite|improve this answer












        $p = a^2+b^2 = bar alpha alpha in I$ implies $p=0$ in $R/I$.



        If you know that $R/I$ is a field, then it must have characteristic $p$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 2 days ago









        lhf

        163k10167387




        163k10167387















            -

            Popular posts from this blog

            1300-talet

            Image
            Den här artikeln handlar om hundraårsperioden 1300–1399. För decenniet 1300–1309, se 1300-talet (decennium). .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output ...
            Read more

            1300-talet

            Image
            Den här artikeln handlar om hundraårsperioden 1300–1399. För decenniet 1300–1309, se 1300-talet (decennium). .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output ...
            Read more

            Display a custom attribute below product name in the front-end Magento 1.9.3.8

            Image
            0 I need to display a custom attribute bellow the product name and I don't know how can I call this attribute. I am editing the file /app/design/frontend/templatemela/MAG100219_1/template/catalog/product/view.phtml But how can I call this attribute in php? I tried calling a string echo with the name I put in the attribute, but it didn't work. Code is as below: <div class="page-title product-title"> <h1><?php echo $_helper->productAttribute($_product, $_product->getName(), 'name') ?></h1> </div> <div class="codigoinmetro"> <h2><?php echo $_helper->productAttribute($_product, $_product->getName(), 'registroinmetro') ?></h2> </div> php attrib...
            Read more