How to prove $Bbb Z[i]/(1+2i)cong Bbb Z_5$?












2















How to prove $Bbb Z[i]/(1+2i)cong Bbb Z_5$?




My method is



$$(1+2i)=big{a+bi丨a+2b≡0pmod 5big},$$



So any $a+bi$ in $Bbb Z(i)$,we got



$$a+bi=(b-2a)i+a(1+2i).$$



So $Bbb Z[i]/(1+2i)=big{0,[i],[2i],[3i],[4i]big}$.



I know how to prove this ring is isomorphic to $Bbb Z_5$, but how can I prove that $Bbb Z[i]/(1+2i)$ equals to $Bbb Z_5$ directly? Any suggestion ia appreciated.






abstract-algebra ring-theory ideals finite-rings gaussian-integers







share|cite|improve this question









New contributor




yLccc is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




















  • Why are you unhappy with your solution? It looks perfectly reasonable to me.
    – Arthur
    2 days ago












  • I can't prove it directly, which makes me sad:(
    – yLccc
    2 days ago










  • What does "diectly" mean in this case?
    – Arthur
    2 days ago












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






  • 3




    $mathbb Z[i] / (1+2i)$ is not equal to $mathbb Z_5$. By definition $mathbb Z_5$ is $mathbb Z / (5)$ and each element of $mathbb Z/(5)$ is a coset of the ideal $(5)$ in the ring $mathbb Z$. But each element of $mathbb Z [i] / (1+2i)$ is a coset of the ideal $(1+2i)$ in the ring $mathbb Z[i]$. No coset of the first type is equal to a coset of the second type. So the best you can hope for is an isomorphism, which you say you already have. Do you want some kind of "special" isomorphism? Can you state what special property you want that isomorphism to have?
    – Lee Mosher
    2 days ago


















2















How to prove $Bbb Z[i]/(1+2i)cong Bbb Z_5$?




My method is



$$(1+2i)=big{a+bi丨a+2b≡0pmod 5big},$$



So any $a+bi$ in $Bbb Z(i)$,we got



$$a+bi=(b-2a)i+a(1+2i).$$



So $Bbb Z[i]/(1+2i)=big{0,[i],[2i],[3i],[4i]big}$.



I know how to prove this ring is isomorphic to $Bbb Z_5$, but how can I prove that $Bbb Z[i]/(1+2i)$ equals to $Bbb Z_5$ directly? Any suggestion ia appreciated.






abstract-algebra ring-theory ideals finite-rings gaussian-integers







share|cite|improve this question









New contributor




yLccc is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




















  • Why are you unhappy with your solution? It looks perfectly reasonable to me.
    – Arthur
    2 days ago












  • I can't prove it directly, which makes me sad:(
    – yLccc
    2 days ago










  • What does "diectly" mean in this case?
    – Arthur
    2 days ago












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






  • 3




    $mathbb Z[i] / (1+2i)$ is not equal to $mathbb Z_5$. By definition $mathbb Z_5$ is $mathbb Z / (5)$ and each element of $mathbb Z/(5)$ is a coset of the ideal $(5)$ in the ring $mathbb Z$. But each element of $mathbb Z [i] / (1+2i)$ is a coset of the ideal $(1+2i)$ in the ring $mathbb Z[i]$. No coset of the first type is equal to a coset of the second type. So the best you can hope for is an isomorphism, which you say you already have. Do you want some kind of "special" isomorphism? Can you state what special property you want that isomorphism to have?
    – Lee Mosher
    2 days ago
















2












2








2








How to prove $Bbb Z[i]/(1+2i)cong Bbb Z_5$?




My method is



$$(1+2i)=big{a+bi丨a+2b≡0pmod 5big},$$



So any $a+bi$ in $Bbb Z(i)$,we got



$$a+bi=(b-2a)i+a(1+2i).$$



So $Bbb Z[i]/(1+2i)=big{0,[i],[2i],[3i],[4i]big}$.



I know how to prove this ring is isomorphic to $Bbb Z_5$, but how can I prove that $Bbb Z[i]/(1+2i)$ equals to $Bbb Z_5$ directly? Any suggestion ia appreciated.






abstract-algebra ring-theory ideals finite-rings gaussian-integers







share|cite|improve this question









New contributor




yLccc is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












How to prove $Bbb Z[i]/(1+2i)cong Bbb Z_5$?




My method is



$$(1+2i)=big{a+bi丨a+2b≡0pmod 5big},$$



So any $a+bi$ in $Bbb Z(i)$,we got



$$a+bi=(b-2a)i+a(1+2i).$$



So $Bbb Z[i]/(1+2i)=big{0,[i],[2i],[3i],[4i]big}$.



I know how to prove this ring is isomorphic to $Bbb Z_5$, but how can I prove that $Bbb Z[i]/(1+2i)$ equals to $Bbb Z_5$ directly? Any suggestion ia appreciated.






abstract-algebra ring-theory ideals finite-rings gaussian-integers




abstract-algebra ring-theory ideals finite-rings gaussian-integers






share|cite|improve this question









New contributor




yLccc is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




yLccc is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 2 days ago









Zvi

5,050430




5,050430






New contributor




yLccc is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 2 days ago









yLccc

283




283




New contributor




yLccc is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





yLccc is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






yLccc is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • Why are you unhappy with your solution? It looks perfectly reasonable to me.
    – Arthur
    2 days ago












  • I can't prove it directly, which makes me sad:(
    – yLccc
    2 days ago










  • What does "diectly" mean in this case?
    – Arthur
    2 days ago












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






  • 3




    $mathbb Z[i] / (1+2i)$ is not equal to $mathbb Z_5$. By definition $mathbb Z_5$ is $mathbb Z / (5)$ and each element of $mathbb Z/(5)$ is a coset of the ideal $(5)$ in the ring $mathbb Z$. But each element of $mathbb Z [i] / (1+2i)$ is a coset of the ideal $(1+2i)$ in the ring $mathbb Z[i]$. No coset of the first type is equal to a coset of the second type. So the best you can hope for is an isomorphism, which you say you already have. Do you want some kind of "special" isomorphism? Can you state what special property you want that isomorphism to have?
    – Lee Mosher
    2 days ago




















  • Why are you unhappy with your solution? It looks perfectly reasonable to me.
    – Arthur
    2 days ago












  • I can't prove it directly, which makes me sad:(
    – yLccc
    2 days ago










  • What does "diectly" mean in this case?
    – Arthur
    2 days ago












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






  • 3




    $mathbb Z[i] / (1+2i)$ is not equal to $mathbb Z_5$. By definition $mathbb Z_5$ is $mathbb Z / (5)$ and each element of $mathbb Z/(5)$ is a coset of the ideal $(5)$ in the ring $mathbb Z$. But each element of $mathbb Z [i] / (1+2i)$ is a coset of the ideal $(1+2i)$ in the ring $mathbb Z[i]$. No coset of the first type is equal to a coset of the second type. So the best you can hope for is an isomorphism, which you say you already have. Do you want some kind of "special" isomorphism? Can you state what special property you want that isomorphism to have?
    – Lee Mosher
    2 days ago


















Why are you unhappy with your solution? It looks perfectly reasonable to me.
– Arthur
2 days ago






Why are you unhappy with your solution? It looks perfectly reasonable to me.
– Arthur
2 days ago














I can't prove it directly, which makes me sad:(
– yLccc
2 days ago




I can't prove it directly, which makes me sad:(
– yLccc
2 days ago












What does "diectly" mean in this case?
– Arthur
2 days ago






What does "diectly" mean in this case?
– Arthur
2 days ago














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




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




3




3




$mathbb Z[i] / (1+2i)$ is not equal to $mathbb Z_5$. By definition $mathbb Z_5$ is $mathbb Z / (5)$ and each element of $mathbb Z/(5)$ is a coset of the ideal $(5)$ in the ring $mathbb Z$. But each element of $mathbb Z [i] / (1+2i)$ is a coset of the ideal $(1+2i)$ in the ring $mathbb Z[i]$. No coset of the first type is equal to a coset of the second type. So the best you can hope for is an isomorphism, which you say you already have. Do you want some kind of "special" isomorphism? Can you state what special property you want that isomorphism to have?
– Lee Mosher
2 days ago






$mathbb Z[i] / (1+2i)$ is not equal to $mathbb Z_5$. By definition $mathbb Z_5$ is $mathbb Z / (5)$ and each element of $mathbb Z/(5)$ is a coset of the ideal $(5)$ in the ring $mathbb Z$. But each element of $mathbb Z [i] / (1+2i)$ is a coset of the ideal $(1+2i)$ in the ring $mathbb Z[i]$. No coset of the first type is equal to a coset of the second type. So the best you can hope for is an isomorphism, which you say you already have. Do you want some kind of "special" isomorphism? Can you state what special property you want that isomorphism to have?
– Lee Mosher
2 days ago












4 Answers
4






active

oldest

votes


















6














I think it is easier to do it this way. Let $varphi:Bbb Z[x]to Bbb Z[i]$ sending $xmapsto i$. Then the preimage of the ideal $I=(1+2i)=(i-2)$ of $Bbb Z[i]$ under $varphi$ is the ideal $J=(x^2+1,x-2)$ of $Bbb Z[x]$. Thus $varphi$ induces the isomorphism $$Bbb{Z}[i]/Icong Bbb{Z}[x]/J.$$

Now observe that
$$J=(x^2+1,x-2)=(x^2+1,x-2,x^2-4)=(5,x-2).$$
This shows that$$frac{Bbb Z[i]}{(1+2i)}=frac{Bbb{Z}[i]}{(i-2)}cong frac{Bbb Z[x]}{(x^2+1,x-2)}=frac{Bbb{Z}[x]}{(5,x-2)}congfrac{Bbb{Z}[x]/(x-2)}{(5,x-2)/(x-2)}.$$
Because under the isomorphism $Bbb{Z}[x]/(x-2)cong Bbb Z$ (which sends $p(x)+{color{red}{(x-2)}}$ to $p(2)$), $(5,x-2)/(x-2)$ is the ideal $5Bbb{Z}$ of $Bbb Z$, we get
$$frac{Bbb Z[i]}{(1+2i)}congfrac{Bbb Z}{5Bbb Z},$$
as required.






share|cite|improve this answer























  • Could u explain why (5,x-2)/(x-2) is the ideal 5Z of Z for me? I don't understand it
    – yLccc
    2 days ago










  • Anything in $(5,x-2)$ is of the form $5p(x)+(x-2)q(x)$. So $(5,x-2)/(x-2)$ contains elements of the form $5p(x)+(x-2)q(x)+{color{red}{(x-2)}}=5p(x)+{color{red}{(x-2)}}$. Therefore, under the given isomorphism, $5p(x)+{color{red}{(x-2)}}$ is mapped to $5p(2)$ which is an integer multiple of $5$. Clearly $5$ is in the image (by taking $p$ to be the constant polynomial $p=1$). Therefore, $(5,x-2)/(x-2)$ is precisely $5Bbb Z$.
    – Zvi
    2 days ago












  • In my answer and in my previous comment, anything that appears in $color{red}{mathrm{red}}$ is an ideal (i.e., not a polynomial in parentheses).
    – Zvi
    2 days ago





















2














A ring homomorphism $varphicolonmathbb{Z}[i]tomathbb{Z}_5$ is uniquely determined by assigning $q=varphi(i)$, which should be an element satisfying $q^2=-1$; thus you have two choices: $q=2$ or $q=-2$.



With the second choice, you have $varphi(a+bi)=[a-2b]$. Can you determine $kervarphi$?






share|cite|improve this answer





























    1














    Hint: $(1+2i)(1-2i)=5$. Hence $(1+2i)supset (5)$. Hence $mathbb Z[i]/(1+2i)subset mathbb Z_5[i]/(1+2i)$. But $Z_5[i]/(1+2i)cong mathbb Z_5$, since $a+bi=a+2b+(1+2i)$.



    To show there are $5$ distinct elements (cosets), just consider (the equivalence classes of) ${0,1,2,3,4}$. It's easy to see none of them are the same.






    share|cite|improve this answer































      0














      If you know you have five cosets here, then try the representatives $0,1,2,3,4$ which you can prove are not equal. Because $(1+2i)(1-2i)=5$ is in your ideal then the arithmetic with these representatives is just that in $mathbb Z_5$ - you can reduce any integer representative of a coset modulo $5$.




      Note: If you want to do the whole thing, it is easy to show that every coset contains an integer because $1+2i$ and $i(1+2i)=i-2$ are both in your ideal. Then $5$ is in your ideal, so every coset contains one of $0,1,2,3,4$.






      share|cite|improve this answer





















      • So next step I prove 5 is in my ideal, so 1 is in my ideal, and then I get 2,3,4 are the same, and then prove these integers are in different cosets. Is that right?
        – yLccc
        2 days ago










      • @yLccc How do you get $1$ in the ideal generated by $1+2i$? If $1$ is in the ideal is the whole ring. The basic thing is you have five cosets, so choose five representatives which look as much like $mathbb Z_5$ as possible. Everything divisible by $5$ gets absorbed in the ideal, just as it does when you are working with the ideal generated by $5$ over the integers.
        – Mark Bennet
        2 days ago











      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
      });


      }
      });






      yLccc is a new contributor. Be nice, and check out our Code of Conduct.










      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3060664%2fhow-to-prove-bbb-zi-12i-cong-bbb-z-5%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      6














      I think it is easier to do it this way. Let $varphi:Bbb Z[x]to Bbb Z[i]$ sending $xmapsto i$. Then the preimage of the ideal $I=(1+2i)=(i-2)$ of $Bbb Z[i]$ under $varphi$ is the ideal $J=(x^2+1,x-2)$ of $Bbb Z[x]$. Thus $varphi$ induces the isomorphism $$Bbb{Z}[i]/Icong Bbb{Z}[x]/J.$$

      Now observe that
      $$J=(x^2+1,x-2)=(x^2+1,x-2,x^2-4)=(5,x-2).$$
      This shows that$$frac{Bbb Z[i]}{(1+2i)}=frac{Bbb{Z}[i]}{(i-2)}cong frac{Bbb Z[x]}{(x^2+1,x-2)}=frac{Bbb{Z}[x]}{(5,x-2)}congfrac{Bbb{Z}[x]/(x-2)}{(5,x-2)/(x-2)}.$$
      Because under the isomorphism $Bbb{Z}[x]/(x-2)cong Bbb Z$ (which sends $p(x)+{color{red}{(x-2)}}$ to $p(2)$), $(5,x-2)/(x-2)$ is the ideal $5Bbb{Z}$ of $Bbb Z$, we get
      $$frac{Bbb Z[i]}{(1+2i)}congfrac{Bbb Z}{5Bbb Z},$$
      as required.






      share|cite|improve this answer























      • Could u explain why (5,x-2)/(x-2) is the ideal 5Z of Z for me? I don't understand it
        – yLccc
        2 days ago










      • Anything in $(5,x-2)$ is of the form $5p(x)+(x-2)q(x)$. So $(5,x-2)/(x-2)$ contains elements of the form $5p(x)+(x-2)q(x)+{color{red}{(x-2)}}=5p(x)+{color{red}{(x-2)}}$. Therefore, under the given isomorphism, $5p(x)+{color{red}{(x-2)}}$ is mapped to $5p(2)$ which is an integer multiple of $5$. Clearly $5$ is in the image (by taking $p$ to be the constant polynomial $p=1$). Therefore, $(5,x-2)/(x-2)$ is precisely $5Bbb Z$.
        – Zvi
        2 days ago












      • In my answer and in my previous comment, anything that appears in $color{red}{mathrm{red}}$ is an ideal (i.e., not a polynomial in parentheses).
        – Zvi
        2 days ago


















      6














      I think it is easier to do it this way. Let $varphi:Bbb Z[x]to Bbb Z[i]$ sending $xmapsto i$. Then the preimage of the ideal $I=(1+2i)=(i-2)$ of $Bbb Z[i]$ under $varphi$ is the ideal $J=(x^2+1,x-2)$ of $Bbb Z[x]$. Thus $varphi$ induces the isomorphism $$Bbb{Z}[i]/Icong Bbb{Z}[x]/J.$$

      Now observe that
      $$J=(x^2+1,x-2)=(x^2+1,x-2,x^2-4)=(5,x-2).$$
      This shows that$$frac{Bbb Z[i]}{(1+2i)}=frac{Bbb{Z}[i]}{(i-2)}cong frac{Bbb Z[x]}{(x^2+1,x-2)}=frac{Bbb{Z}[x]}{(5,x-2)}congfrac{Bbb{Z}[x]/(x-2)}{(5,x-2)/(x-2)}.$$
      Because under the isomorphism $Bbb{Z}[x]/(x-2)cong Bbb Z$ (which sends $p(x)+{color{red}{(x-2)}}$ to $p(2)$), $(5,x-2)/(x-2)$ is the ideal $5Bbb{Z}$ of $Bbb Z$, we get
      $$frac{Bbb Z[i]}{(1+2i)}congfrac{Bbb Z}{5Bbb Z},$$
      as required.






      share|cite|improve this answer























      • Could u explain why (5,x-2)/(x-2) is the ideal 5Z of Z for me? I don't understand it
        – yLccc
        2 days ago










      • Anything in $(5,x-2)$ is of the form $5p(x)+(x-2)q(x)$. So $(5,x-2)/(x-2)$ contains elements of the form $5p(x)+(x-2)q(x)+{color{red}{(x-2)}}=5p(x)+{color{red}{(x-2)}}$. Therefore, under the given isomorphism, $5p(x)+{color{red}{(x-2)}}$ is mapped to $5p(2)$ which is an integer multiple of $5$. Clearly $5$ is in the image (by taking $p$ to be the constant polynomial $p=1$). Therefore, $(5,x-2)/(x-2)$ is precisely $5Bbb Z$.
        – Zvi
        2 days ago












      • In my answer and in my previous comment, anything that appears in $color{red}{mathrm{red}}$ is an ideal (i.e., not a polynomial in parentheses).
        – Zvi
        2 days ago
















      6












      6








      6






      I think it is easier to do it this way. Let $varphi:Bbb Z[x]to Bbb Z[i]$ sending $xmapsto i$. Then the preimage of the ideal $I=(1+2i)=(i-2)$ of $Bbb Z[i]$ under $varphi$ is the ideal $J=(x^2+1,x-2)$ of $Bbb Z[x]$. Thus $varphi$ induces the isomorphism $$Bbb{Z}[i]/Icong Bbb{Z}[x]/J.$$

      Now observe that
      $$J=(x^2+1,x-2)=(x^2+1,x-2,x^2-4)=(5,x-2).$$
      This shows that$$frac{Bbb Z[i]}{(1+2i)}=frac{Bbb{Z}[i]}{(i-2)}cong frac{Bbb Z[x]}{(x^2+1,x-2)}=frac{Bbb{Z}[x]}{(5,x-2)}congfrac{Bbb{Z}[x]/(x-2)}{(5,x-2)/(x-2)}.$$
      Because under the isomorphism $Bbb{Z}[x]/(x-2)cong Bbb Z$ (which sends $p(x)+{color{red}{(x-2)}}$ to $p(2)$), $(5,x-2)/(x-2)$ is the ideal $5Bbb{Z}$ of $Bbb Z$, we get
      $$frac{Bbb Z[i]}{(1+2i)}congfrac{Bbb Z}{5Bbb Z},$$
      as required.






      share|cite|improve this answer














      I think it is easier to do it this way. Let $varphi:Bbb Z[x]to Bbb Z[i]$ sending $xmapsto i$. Then the preimage of the ideal $I=(1+2i)=(i-2)$ of $Bbb Z[i]$ under $varphi$ is the ideal $J=(x^2+1,x-2)$ of $Bbb Z[x]$. Thus $varphi$ induces the isomorphism $$Bbb{Z}[i]/Icong Bbb{Z}[x]/J.$$

      Now observe that
      $$J=(x^2+1,x-2)=(x^2+1,x-2,x^2-4)=(5,x-2).$$
      This shows that$$frac{Bbb Z[i]}{(1+2i)}=frac{Bbb{Z}[i]}{(i-2)}cong frac{Bbb Z[x]}{(x^2+1,x-2)}=frac{Bbb{Z}[x]}{(5,x-2)}congfrac{Bbb{Z}[x]/(x-2)}{(5,x-2)/(x-2)}.$$
      Because under the isomorphism $Bbb{Z}[x]/(x-2)cong Bbb Z$ (which sends $p(x)+{color{red}{(x-2)}}$ to $p(2)$), $(5,x-2)/(x-2)$ is the ideal $5Bbb{Z}$ of $Bbb Z$, we get
      $$frac{Bbb Z[i]}{(1+2i)}congfrac{Bbb Z}{5Bbb Z},$$
      as required.







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited 2 days ago

























      answered 2 days ago









      Zvi

      5,050430




      5,050430












      • Could u explain why (5,x-2)/(x-2) is the ideal 5Z of Z for me? I don't understand it
        – yLccc
        2 days ago










      • Anything in $(5,x-2)$ is of the form $5p(x)+(x-2)q(x)$. So $(5,x-2)/(x-2)$ contains elements of the form $5p(x)+(x-2)q(x)+{color{red}{(x-2)}}=5p(x)+{color{red}{(x-2)}}$. Therefore, under the given isomorphism, $5p(x)+{color{red}{(x-2)}}$ is mapped to $5p(2)$ which is an integer multiple of $5$. Clearly $5$ is in the image (by taking $p$ to be the constant polynomial $p=1$). Therefore, $(5,x-2)/(x-2)$ is precisely $5Bbb Z$.
        – Zvi
        2 days ago












      • In my answer and in my previous comment, anything that appears in $color{red}{mathrm{red}}$ is an ideal (i.e., not a polynomial in parentheses).
        – Zvi
        2 days ago




















      • Could u explain why (5,x-2)/(x-2) is the ideal 5Z of Z for me? I don't understand it
        – yLccc
        2 days ago










      • Anything in $(5,x-2)$ is of the form $5p(x)+(x-2)q(x)$. So $(5,x-2)/(x-2)$ contains elements of the form $5p(x)+(x-2)q(x)+{color{red}{(x-2)}}=5p(x)+{color{red}{(x-2)}}$. Therefore, under the given isomorphism, $5p(x)+{color{red}{(x-2)}}$ is mapped to $5p(2)$ which is an integer multiple of $5$. Clearly $5$ is in the image (by taking $p$ to be the constant polynomial $p=1$). Therefore, $(5,x-2)/(x-2)$ is precisely $5Bbb Z$.
        – Zvi
        2 days ago












      • In my answer and in my previous comment, anything that appears in $color{red}{mathrm{red}}$ is an ideal (i.e., not a polynomial in parentheses).
        – Zvi
        2 days ago


















      Could u explain why (5,x-2)/(x-2) is the ideal 5Z of Z for me? I don't understand it
      – yLccc
      2 days ago




      Could u explain why (5,x-2)/(x-2) is the ideal 5Z of Z for me? I don't understand it
      – yLccc
      2 days ago












      Anything in $(5,x-2)$ is of the form $5p(x)+(x-2)q(x)$. So $(5,x-2)/(x-2)$ contains elements of the form $5p(x)+(x-2)q(x)+{color{red}{(x-2)}}=5p(x)+{color{red}{(x-2)}}$. Therefore, under the given isomorphism, $5p(x)+{color{red}{(x-2)}}$ is mapped to $5p(2)$ which is an integer multiple of $5$. Clearly $5$ is in the image (by taking $p$ to be the constant polynomial $p=1$). Therefore, $(5,x-2)/(x-2)$ is precisely $5Bbb Z$.
      – Zvi
      2 days ago






      Anything in $(5,x-2)$ is of the form $5p(x)+(x-2)q(x)$. So $(5,x-2)/(x-2)$ contains elements of the form $5p(x)+(x-2)q(x)+{color{red}{(x-2)}}=5p(x)+{color{red}{(x-2)}}$. Therefore, under the given isomorphism, $5p(x)+{color{red}{(x-2)}}$ is mapped to $5p(2)$ which is an integer multiple of $5$. Clearly $5$ is in the image (by taking $p$ to be the constant polynomial $p=1$). Therefore, $(5,x-2)/(x-2)$ is precisely $5Bbb Z$.
      – Zvi
      2 days ago














      In my answer and in my previous comment, anything that appears in $color{red}{mathrm{red}}$ is an ideal (i.e., not a polynomial in parentheses).
      – Zvi
      2 days ago






      In my answer and in my previous comment, anything that appears in $color{red}{mathrm{red}}$ is an ideal (i.e., not a polynomial in parentheses).
      – Zvi
      2 days ago













      2














      A ring homomorphism $varphicolonmathbb{Z}[i]tomathbb{Z}_5$ is uniquely determined by assigning $q=varphi(i)$, which should be an element satisfying $q^2=-1$; thus you have two choices: $q=2$ or $q=-2$.



      With the second choice, you have $varphi(a+bi)=[a-2b]$. Can you determine $kervarphi$?






      share|cite|improve this answer


























        2














        A ring homomorphism $varphicolonmathbb{Z}[i]tomathbb{Z}_5$ is uniquely determined by assigning $q=varphi(i)$, which should be an element satisfying $q^2=-1$; thus you have two choices: $q=2$ or $q=-2$.



        With the second choice, you have $varphi(a+bi)=[a-2b]$. Can you determine $kervarphi$?






        share|cite|improve this answer
























          2












          2








          2






          A ring homomorphism $varphicolonmathbb{Z}[i]tomathbb{Z}_5$ is uniquely determined by assigning $q=varphi(i)$, which should be an element satisfying $q^2=-1$; thus you have two choices: $q=2$ or $q=-2$.



          With the second choice, you have $varphi(a+bi)=[a-2b]$. Can you determine $kervarphi$?






          share|cite|improve this answer












          A ring homomorphism $varphicolonmathbb{Z}[i]tomathbb{Z}_5$ is uniquely determined by assigning $q=varphi(i)$, which should be an element satisfying $q^2=-1$; thus you have two choices: $q=2$ or $q=-2$.



          With the second choice, you have $varphi(a+bi)=[a-2b]$. Can you determine $kervarphi$?







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 2 days ago









          egreg

          178k1484201




          178k1484201























              1














              Hint: $(1+2i)(1-2i)=5$. Hence $(1+2i)supset (5)$. Hence $mathbb Z[i]/(1+2i)subset mathbb Z_5[i]/(1+2i)$. But $Z_5[i]/(1+2i)cong mathbb Z_5$, since $a+bi=a+2b+(1+2i)$.



              To show there are $5$ distinct elements (cosets), just consider (the equivalence classes of) ${0,1,2,3,4}$. It's easy to see none of them are the same.






              share|cite|improve this answer




























                1














                Hint: $(1+2i)(1-2i)=5$. Hence $(1+2i)supset (5)$. Hence $mathbb Z[i]/(1+2i)subset mathbb Z_5[i]/(1+2i)$. But $Z_5[i]/(1+2i)cong mathbb Z_5$, since $a+bi=a+2b+(1+2i)$.



                To show there are $5$ distinct elements (cosets), just consider (the equivalence classes of) ${0,1,2,3,4}$. It's easy to see none of them are the same.






                share|cite|improve this answer


























                  1












                  1








                  1






                  Hint: $(1+2i)(1-2i)=5$. Hence $(1+2i)supset (5)$. Hence $mathbb Z[i]/(1+2i)subset mathbb Z_5[i]/(1+2i)$. But $Z_5[i]/(1+2i)cong mathbb Z_5$, since $a+bi=a+2b+(1+2i)$.



                  To show there are $5$ distinct elements (cosets), just consider (the equivalence classes of) ${0,1,2,3,4}$. It's easy to see none of them are the same.






                  share|cite|improve this answer














                  Hint: $(1+2i)(1-2i)=5$. Hence $(1+2i)supset (5)$. Hence $mathbb Z[i]/(1+2i)subset mathbb Z_5[i]/(1+2i)$. But $Z_5[i]/(1+2i)cong mathbb Z_5$, since $a+bi=a+2b+(1+2i)$.



                  To show there are $5$ distinct elements (cosets), just consider (the equivalence classes of) ${0,1,2,3,4}$. It's easy to see none of them are the same.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited 2 days ago

























                  answered 2 days ago









                  Chris Custer

                  10.9k3824




                  10.9k3824























                      0














                      If you know you have five cosets here, then try the representatives $0,1,2,3,4$ which you can prove are not equal. Because $(1+2i)(1-2i)=5$ is in your ideal then the arithmetic with these representatives is just that in $mathbb Z_5$ - you can reduce any integer representative of a coset modulo $5$.




                      Note: If you want to do the whole thing, it is easy to show that every coset contains an integer because $1+2i$ and $i(1+2i)=i-2$ are both in your ideal. Then $5$ is in your ideal, so every coset contains one of $0,1,2,3,4$.






                      share|cite|improve this answer





















                      • So next step I prove 5 is in my ideal, so 1 is in my ideal, and then I get 2,3,4 are the same, and then prove these integers are in different cosets. Is that right?
                        – yLccc
                        2 days ago










                      • @yLccc How do you get $1$ in the ideal generated by $1+2i$? If $1$ is in the ideal is the whole ring. The basic thing is you have five cosets, so choose five representatives which look as much like $mathbb Z_5$ as possible. Everything divisible by $5$ gets absorbed in the ideal, just as it does when you are working with the ideal generated by $5$ over the integers.
                        – Mark Bennet
                        2 days ago
















                      0














                      If you know you have five cosets here, then try the representatives $0,1,2,3,4$ which you can prove are not equal. Because $(1+2i)(1-2i)=5$ is in your ideal then the arithmetic with these representatives is just that in $mathbb Z_5$ - you can reduce any integer representative of a coset modulo $5$.




                      Note: If you want to do the whole thing, it is easy to show that every coset contains an integer because $1+2i$ and $i(1+2i)=i-2$ are both in your ideal. Then $5$ is in your ideal, so every coset contains one of $0,1,2,3,4$.






                      share|cite|improve this answer





















                      • So next step I prove 5 is in my ideal, so 1 is in my ideal, and then I get 2,3,4 are the same, and then prove these integers are in different cosets. Is that right?
                        – yLccc
                        2 days ago










                      • @yLccc How do you get $1$ in the ideal generated by $1+2i$? If $1$ is in the ideal is the whole ring. The basic thing is you have five cosets, so choose five representatives which look as much like $mathbb Z_5$ as possible. Everything divisible by $5$ gets absorbed in the ideal, just as it does when you are working with the ideal generated by $5$ over the integers.
                        – Mark Bennet
                        2 days ago














                      0












                      0








                      0






                      If you know you have five cosets here, then try the representatives $0,1,2,3,4$ which you can prove are not equal. Because $(1+2i)(1-2i)=5$ is in your ideal then the arithmetic with these representatives is just that in $mathbb Z_5$ - you can reduce any integer representative of a coset modulo $5$.




                      Note: If you want to do the whole thing, it is easy to show that every coset contains an integer because $1+2i$ and $i(1+2i)=i-2$ are both in your ideal. Then $5$ is in your ideal, so every coset contains one of $0,1,2,3,4$.






                      share|cite|improve this answer












                      If you know you have five cosets here, then try the representatives $0,1,2,3,4$ which you can prove are not equal. Because $(1+2i)(1-2i)=5$ is in your ideal then the arithmetic with these representatives is just that in $mathbb Z_5$ - you can reduce any integer representative of a coset modulo $5$.




                      Note: If you want to do the whole thing, it is easy to show that every coset contains an integer because $1+2i$ and $i(1+2i)=i-2$ are both in your ideal. Then $5$ is in your ideal, so every coset contains one of $0,1,2,3,4$.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered 2 days ago









                      Mark Bennet

                      80.6k981179




                      80.6k981179












                      • So next step I prove 5 is in my ideal, so 1 is in my ideal, and then I get 2,3,4 are the same, and then prove these integers are in different cosets. Is that right?
                        – yLccc
                        2 days ago










                      • @yLccc How do you get $1$ in the ideal generated by $1+2i$? If $1$ is in the ideal is the whole ring. The basic thing is you have five cosets, so choose five representatives which look as much like $mathbb Z_5$ as possible. Everything divisible by $5$ gets absorbed in the ideal, just as it does when you are working with the ideal generated by $5$ over the integers.
                        – Mark Bennet
                        2 days ago


















                      • So next step I prove 5 is in my ideal, so 1 is in my ideal, and then I get 2,3,4 are the same, and then prove these integers are in different cosets. Is that right?
                        – yLccc
                        2 days ago










                      • @yLccc How do you get $1$ in the ideal generated by $1+2i$? If $1$ is in the ideal is the whole ring. The basic thing is you have five cosets, so choose five representatives which look as much like $mathbb Z_5$ as possible. Everything divisible by $5$ gets absorbed in the ideal, just as it does when you are working with the ideal generated by $5$ over the integers.
                        – Mark Bennet
                        2 days ago
















                      So next step I prove 5 is in my ideal, so 1 is in my ideal, and then I get 2,3,4 are the same, and then prove these integers are in different cosets. Is that right?
                      – yLccc
                      2 days ago




                      So next step I prove 5 is in my ideal, so 1 is in my ideal, and then I get 2,3,4 are the same, and then prove these integers are in different cosets. Is that right?
                      – yLccc
                      2 days ago












                      @yLccc How do you get $1$ in the ideal generated by $1+2i$? If $1$ is in the ideal is the whole ring. The basic thing is you have five cosets, so choose five representatives which look as much like $mathbb Z_5$ as possible. Everything divisible by $5$ gets absorbed in the ideal, just as it does when you are working with the ideal generated by $5$ over the integers.
                      – Mark Bennet
                      2 days ago




                      @yLccc How do you get $1$ in the ideal generated by $1+2i$? If $1$ is in the ideal is the whole ring. The basic thing is you have five cosets, so choose five representatives which look as much like $mathbb Z_5$ as possible. Everything divisible by $5$ gets absorbed in the ideal, just as it does when you are working with the ideal generated by $5$ over the integers.
                      – Mark Bennet
                      2 days ago










                      yLccc is a new contributor. Be nice, and check out our Code of Conduct.










                      draft saved

                      draft discarded


















                      yLccc is a new contributor. Be nice, and check out our Code of Conduct.













                      yLccc is a new contributor. Be nice, and check out our Code of Conduct.












                      yLccc is a new contributor. Be nice, and check out our Code of Conduct.
















                      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%2f3060664%2fhow-to-prove-bbb-zi-12i-cong-bbb-z-5%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

                      An IMO inspired problem

                      Image
                      14 4 This problem from IMO 1988 is said to be one of the most elegant ones in functional equations. Problem : The function $f$ is defined on the set of all positive integers as follows: begin{align} f(1) = 1, f(3) &= 3, f(2n) = f(n), \ f(4n+1) &= 2f(2n+1) - f(n), \ f(4n+3) &= 3 f(2n+1) - 2f(n) end{align} Find the number of $n$ with $f(n) = n, 1 leq n leq 1988$. The main idea towards the solution is realizing that these conditions stand for the fact that $f(n)$ just reverses the digits in the binary representation of the number $n$. So, essentially the solution is finding the number of binary pallindromes $leq 1988_{10}$. My question is the following : How to reformulate the problem so that $f(n)$ reverses the digits of $n$ in its ternary representation. Or even better, can we reformulate it for a
                      Read more

                      Management

                      Image
                      Management From Wikipedia, the free encyclopedia Jump to navigation Jump to search "Manager" redirects here. For other uses, see Management (disambiguation) and Manager (disambiguation). An organization chart for the United States Coast Guard shows the hierarchy of managerial roles in that organization. Business administration Management of a business Accounting Management accounting Financial accounting Financial audit Business entities Cooperative Corporation Limited liability company Partnership Sole proprietorship State-owned enterprise Corporate governance Annual general meeting Board of directors Supervisory board Advisory board Audit committee Corporate law Commercial law Constitutional documents Contract Corporate crime Corporate liability Insolvency law International trade law Mergers and acquisitions Economics Commodity Pu
                      Read more

                      Has there ever been an instance of an active nuclear power plant within or near a war zone?

                      Image
                      36 3 The question Has there ever been a military combat going on around an active (running or in outage but not decommissioned) commercial nuclear power plant? Or dangerously near it? If yes, how did the plant and its personnel fare in such situation? Why I ask Let's assume for the scope of this question modern commercial nuclear power plants (CANDU, PWR, BWR of generation II plus all of gen III and newer) are usually reasonably safe to operate given that the engineering has been done right and there is a whole cohort of very well trained personnel on-site at all times, well rested and with considerable resources on their hands. Things hard to get with an armed conflict like the recent Syrian war raging around. I can't seem to find any reference or a comprehensive article about this, hence this question.
                      Read more