Inverses of elements for (Part One) an ideal generated by a reducible polynomial and (Part Two) an...











up vote
0
down vote

favorite












Let $langle x^2-4 rangle$ denote the ideal generated by the polynomial $x^2-4$. (Artin uses $(x^2-4)$.)



In connection with this Visualizing quotient polynomial rings are fields for maximal ideals which are generated by irreducible monic



Part One: Inverses in the quotient ring $mathbb R[x]/langle x^2-4 rangle$.




  1. Consider the quotient ring $mathbb R[x]/langle x^2-4 rangle$ and one of its elements $overline {x^2-2x} = [x^2+2x+langle x^2-4 rangle]$. I think this element has no multiplicative inverse because $x^2-2x=0$ for $x=2$. Is that correct?


  2. In the quotient ring, $mathbb R[x]/langle x-2 rangle$, the relation is "x=2". This relation is used to compute inverses. What exactly is the relation in the quotient ring $mathbb R[x]/langle x^2-4 rangle$? I think it is "$x=2$ or $x=-2$".


  3. In the quotient ring $mathbb R[x]/langle x^2-4 rangle$, does the element $overline {x+1} = [x+1+langle x^2-4 rangle]$ have an inverse, and why? If so, then what is it? (I think I can answer these if I know the answer for 2.)



Part Two: Inverses in the quotient ring $mathbb R[x]/langle x^2+1 rangle$.



For the quotient ring $mathbb R[x]/langle x-2 rangle$, the relation is $x=2$ so inverse of $overline p = [p(x)+langle x-2 rangle]$ is $overline {frac{1}{p(2)}}$.



For the quotient ring $mathbb R[x]/langle x^2+1 rangle$, I believe $x^2+1$ is irreducible in $mathbb R[x]$, so $mathbb R[x]/langle x^2+1 rangle$ is a field.



To compute the inverse of $overline p = [p(x)+langle x^2+1 rangle]$, I believe I need the relation in the quotient ring $mathbb R[x]/langle x^2+1 rangle$.




  1. What's the relation in the quotient ring $mathbb R[x]/langle x^2+1 rangle$? I think it is "$x= i$ or $x=-i$".


  2. What is the inverse of $overline p = [p(x)+langle x^2+1 rangle]$?



( I think I will know the answer if I know the answer to 2 in Part One.)






abstract-algebra ring-theory modular-arithmetic ideals maximal-and-prime-ideals







share|cite|improve this question




















  • 1




    Generally a post should have one clearly outlined question. It's not even clear what the questions are here.
    – Matt Samuel
    Nov 20 at 11:46






  • 1




    $(x+1)(ax+b)=ad+(a+b)x+a(x^2+b/a-d)=1+c(x^2-4)$. $a=-b$, $d=3$, $a=1/3$. Inverse is $(x-1)/3$. Check, we get $x^2/3-1/3=(x^2-4)/3+1$.
    – Matt Samuel
    Nov 20 at 11:58






  • 1




    $x^2-2x=x(x-2)$ is a zero divisor. No inverse.
    – Matt Samuel
    Nov 20 at 13:54






  • 1




    You can't get this ring by substituting a real number, so it's neither $2$ nor $-2$. We have $(x-2)(x+2)=0$, and there's no real number satisfying this.
    – Matt Samuel
    Nov 20 at 14:21






  • 1




    For the quotient by $(x^2+1)$, note that $x^2=-1$. Again there's no real number satisfying this, but there is a complex number satisfying it.
    – Matt Samuel
    Nov 20 at 14:25















up vote
0
down vote

favorite












Let $langle x^2-4 rangle$ denote the ideal generated by the polynomial $x^2-4$. (Artin uses $(x^2-4)$.)



In connection with this Visualizing quotient polynomial rings are fields for maximal ideals which are generated by irreducible monic



Part One: Inverses in the quotient ring $mathbb R[x]/langle x^2-4 rangle$.




  1. Consider the quotient ring $mathbb R[x]/langle x^2-4 rangle$ and one of its elements $overline {x^2-2x} = [x^2+2x+langle x^2-4 rangle]$. I think this element has no multiplicative inverse because $x^2-2x=0$ for $x=2$. Is that correct?


  2. In the quotient ring, $mathbb R[x]/langle x-2 rangle$, the relation is "x=2". This relation is used to compute inverses. What exactly is the relation in the quotient ring $mathbb R[x]/langle x^2-4 rangle$? I think it is "$x=2$ or $x=-2$".


  3. In the quotient ring $mathbb R[x]/langle x^2-4 rangle$, does the element $overline {x+1} = [x+1+langle x^2-4 rangle]$ have an inverse, and why? If so, then what is it? (I think I can answer these if I know the answer for 2.)



Part Two: Inverses in the quotient ring $mathbb R[x]/langle x^2+1 rangle$.



For the quotient ring $mathbb R[x]/langle x-2 rangle$, the relation is $x=2$ so inverse of $overline p = [p(x)+langle x-2 rangle]$ is $overline {frac{1}{p(2)}}$.



For the quotient ring $mathbb R[x]/langle x^2+1 rangle$, I believe $x^2+1$ is irreducible in $mathbb R[x]$, so $mathbb R[x]/langle x^2+1 rangle$ is a field.



To compute the inverse of $overline p = [p(x)+langle x^2+1 rangle]$, I believe I need the relation in the quotient ring $mathbb R[x]/langle x^2+1 rangle$.




  1. What's the relation in the quotient ring $mathbb R[x]/langle x^2+1 rangle$? I think it is "$x= i$ or $x=-i$".


  2. What is the inverse of $overline p = [p(x)+langle x^2+1 rangle]$?



( I think I will know the answer if I know the answer to 2 in Part One.)






abstract-algebra ring-theory modular-arithmetic ideals maximal-and-prime-ideals







share|cite|improve this question




















  • 1




    Generally a post should have one clearly outlined question. It's not even clear what the questions are here.
    – Matt Samuel
    Nov 20 at 11:46






  • 1




    $(x+1)(ax+b)=ad+(a+b)x+a(x^2+b/a-d)=1+c(x^2-4)$. $a=-b$, $d=3$, $a=1/3$. Inverse is $(x-1)/3$. Check, we get $x^2/3-1/3=(x^2-4)/3+1$.
    – Matt Samuel
    Nov 20 at 11:58






  • 1




    $x^2-2x=x(x-2)$ is a zero divisor. No inverse.
    – Matt Samuel
    Nov 20 at 13:54






  • 1




    You can't get this ring by substituting a real number, so it's neither $2$ nor $-2$. We have $(x-2)(x+2)=0$, and there's no real number satisfying this.
    – Matt Samuel
    Nov 20 at 14:21






  • 1




    For the quotient by $(x^2+1)$, note that $x^2=-1$. Again there's no real number satisfying this, but there is a complex number satisfying it.
    – Matt Samuel
    Nov 20 at 14:25













up vote
0
down vote

favorite









up vote
0
down vote

favorite











Let $langle x^2-4 rangle$ denote the ideal generated by the polynomial $x^2-4$. (Artin uses $(x^2-4)$.)



In connection with this Visualizing quotient polynomial rings are fields for maximal ideals which are generated by irreducible monic



Part One: Inverses in the quotient ring $mathbb R[x]/langle x^2-4 rangle$.




  1. Consider the quotient ring $mathbb R[x]/langle x^2-4 rangle$ and one of its elements $overline {x^2-2x} = [x^2+2x+langle x^2-4 rangle]$. I think this element has no multiplicative inverse because $x^2-2x=0$ for $x=2$. Is that correct?


  2. In the quotient ring, $mathbb R[x]/langle x-2 rangle$, the relation is "x=2". This relation is used to compute inverses. What exactly is the relation in the quotient ring $mathbb R[x]/langle x^2-4 rangle$? I think it is "$x=2$ or $x=-2$".


  3. In the quotient ring $mathbb R[x]/langle x^2-4 rangle$, does the element $overline {x+1} = [x+1+langle x^2-4 rangle]$ have an inverse, and why? If so, then what is it? (I think I can answer these if I know the answer for 2.)



Part Two: Inverses in the quotient ring $mathbb R[x]/langle x^2+1 rangle$.



For the quotient ring $mathbb R[x]/langle x-2 rangle$, the relation is $x=2$ so inverse of $overline p = [p(x)+langle x-2 rangle]$ is $overline {frac{1}{p(2)}}$.



For the quotient ring $mathbb R[x]/langle x^2+1 rangle$, I believe $x^2+1$ is irreducible in $mathbb R[x]$, so $mathbb R[x]/langle x^2+1 rangle$ is a field.



To compute the inverse of $overline p = [p(x)+langle x^2+1 rangle]$, I believe I need the relation in the quotient ring $mathbb R[x]/langle x^2+1 rangle$.




  1. What's the relation in the quotient ring $mathbb R[x]/langle x^2+1 rangle$? I think it is "$x= i$ or $x=-i$".


  2. What is the inverse of $overline p = [p(x)+langle x^2+1 rangle]$?



( I think I will know the answer if I know the answer to 2 in Part One.)






abstract-algebra ring-theory modular-arithmetic ideals maximal-and-prime-ideals







share|cite|improve this question















Let $langle x^2-4 rangle$ denote the ideal generated by the polynomial $x^2-4$. (Artin uses $(x^2-4)$.)



In connection with this Visualizing quotient polynomial rings are fields for maximal ideals which are generated by irreducible monic



Part One: Inverses in the quotient ring $mathbb R[x]/langle x^2-4 rangle$.




  1. Consider the quotient ring $mathbb R[x]/langle x^2-4 rangle$ and one of its elements $overline {x^2-2x} = [x^2+2x+langle x^2-4 rangle]$. I think this element has no multiplicative inverse because $x^2-2x=0$ for $x=2$. Is that correct?


  2. In the quotient ring, $mathbb R[x]/langle x-2 rangle$, the relation is "x=2". This relation is used to compute inverses. What exactly is the relation in the quotient ring $mathbb R[x]/langle x^2-4 rangle$? I think it is "$x=2$ or $x=-2$".


  3. In the quotient ring $mathbb R[x]/langle x^2-4 rangle$, does the element $overline {x+1} = [x+1+langle x^2-4 rangle]$ have an inverse, and why? If so, then what is it? (I think I can answer these if I know the answer for 2.)



Part Two: Inverses in the quotient ring $mathbb R[x]/langle x^2+1 rangle$.



For the quotient ring $mathbb R[x]/langle x-2 rangle$, the relation is $x=2$ so inverse of $overline p = [p(x)+langle x-2 rangle]$ is $overline {frac{1}{p(2)}}$.



For the quotient ring $mathbb R[x]/langle x^2+1 rangle$, I believe $x^2+1$ is irreducible in $mathbb R[x]$, so $mathbb R[x]/langle x^2+1 rangle$ is a field.



To compute the inverse of $overline p = [p(x)+langle x^2+1 rangle]$, I believe I need the relation in the quotient ring $mathbb R[x]/langle x^2+1 rangle$.




  1. What's the relation in the quotient ring $mathbb R[x]/langle x^2+1 rangle$? I think it is "$x= i$ or $x=-i$".


  2. What is the inverse of $overline p = [p(x)+langle x^2+1 rangle]$?



( I think I will know the answer if I know the answer to 2 in Part One.)






abstract-algebra ring-theory modular-arithmetic ideals maximal-and-prime-ideals




abstract-algebra ring-theory modular-arithmetic ideals maximal-and-prime-ideals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 20 at 13:47

























asked Nov 20 at 11:42









Jack Bauer

1,2531631




1,2531631








  • 1




    Generally a post should have one clearly outlined question. It's not even clear what the questions are here.
    – Matt Samuel
    Nov 20 at 11:46






  • 1




    $(x+1)(ax+b)=ad+(a+b)x+a(x^2+b/a-d)=1+c(x^2-4)$. $a=-b$, $d=3$, $a=1/3$. Inverse is $(x-1)/3$. Check, we get $x^2/3-1/3=(x^2-4)/3+1$.
    – Matt Samuel
    Nov 20 at 11:58






  • 1




    $x^2-2x=x(x-2)$ is a zero divisor. No inverse.
    – Matt Samuel
    Nov 20 at 13:54






  • 1




    You can't get this ring by substituting a real number, so it's neither $2$ nor $-2$. We have $(x-2)(x+2)=0$, and there's no real number satisfying this.
    – Matt Samuel
    Nov 20 at 14:21






  • 1




    For the quotient by $(x^2+1)$, note that $x^2=-1$. Again there's no real number satisfying this, but there is a complex number satisfying it.
    – Matt Samuel
    Nov 20 at 14:25














  • 1




    Generally a post should have one clearly outlined question. It's not even clear what the questions are here.
    – Matt Samuel
    Nov 20 at 11:46






  • 1




    $(x+1)(ax+b)=ad+(a+b)x+a(x^2+b/a-d)=1+c(x^2-4)$. $a=-b$, $d=3$, $a=1/3$. Inverse is $(x-1)/3$. Check, we get $x^2/3-1/3=(x^2-4)/3+1$.
    – Matt Samuel
    Nov 20 at 11:58






  • 1




    $x^2-2x=x(x-2)$ is a zero divisor. No inverse.
    – Matt Samuel
    Nov 20 at 13:54






  • 1




    You can't get this ring by substituting a real number, so it's neither $2$ nor $-2$. We have $(x-2)(x+2)=0$, and there's no real number satisfying this.
    – Matt Samuel
    Nov 20 at 14:21






  • 1




    For the quotient by $(x^2+1)$, note that $x^2=-1$. Again there's no real number satisfying this, but there is a complex number satisfying it.
    – Matt Samuel
    Nov 20 at 14:25








1




1




Generally a post should have one clearly outlined question. It's not even clear what the questions are here.
– Matt Samuel
Nov 20 at 11:46




Generally a post should have one clearly outlined question. It's not even clear what the questions are here.
– Matt Samuel
Nov 20 at 11:46




1




1




$(x+1)(ax+b)=ad+(a+b)x+a(x^2+b/a-d)=1+c(x^2-4)$. $a=-b$, $d=3$, $a=1/3$. Inverse is $(x-1)/3$. Check, we get $x^2/3-1/3=(x^2-4)/3+1$.
– Matt Samuel
Nov 20 at 11:58




$(x+1)(ax+b)=ad+(a+b)x+a(x^2+b/a-d)=1+c(x^2-4)$. $a=-b$, $d=3$, $a=1/3$. Inverse is $(x-1)/3$. Check, we get $x^2/3-1/3=(x^2-4)/3+1$.
– Matt Samuel
Nov 20 at 11:58




1




1




$x^2-2x=x(x-2)$ is a zero divisor. No inverse.
– Matt Samuel
Nov 20 at 13:54




$x^2-2x=x(x-2)$ is a zero divisor. No inverse.
– Matt Samuel
Nov 20 at 13:54




1




1




You can't get this ring by substituting a real number, so it's neither $2$ nor $-2$. We have $(x-2)(x+2)=0$, and there's no real number satisfying this.
– Matt Samuel
Nov 20 at 14:21




You can't get this ring by substituting a real number, so it's neither $2$ nor $-2$. We have $(x-2)(x+2)=0$, and there's no real number satisfying this.
– Matt Samuel
Nov 20 at 14:21




1




1




For the quotient by $(x^2+1)$, note that $x^2=-1$. Again there's no real number satisfying this, but there is a complex number satisfying it.
– Matt Samuel
Nov 20 at 14:25




For the quotient by $(x^2+1)$, note that $x^2=-1$. Again there's no real number satisfying this, but there is a complex number satisfying it.
– Matt Samuel
Nov 20 at 14:25















active

oldest

votes











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',
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%2f3006217%2finverses-of-elements-for-part-one-an-ideal-generated-by-a-reducible-polynomial%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown






























active

oldest

votes













active

oldest

votes









active

oldest

votes






active

oldest

votes
















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.





Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


Please pay close attention to the following guidance:


  • 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.


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%2f3006217%2finverses-of-elements-for-part-one-an-ideal-generated-by-a-reducible-polynomial%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

Ellipse (mathématiques)

Image
Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
Read more

Quarter-circle Tiles

Image
up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
Read more

Mont Emei

Image
Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
Read more