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












1












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This question already has an answer here:




  • Quotient ring of Gaussian integers

    6 answers





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.










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Jan 10 at 3:49


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.


















  • $begingroup$
    Why are you unhappy with your solution? It looks perfectly reasonable to me.
    $endgroup$
    – Arthur
    Jan 3 at 15:31












  • $begingroup$
    I can't prove it directly, which makes me sad:(
    $endgroup$
    – yLccc
    Jan 3 at 16:06










  • $begingroup$
    What does "diectly" mean in this case?
    $endgroup$
    – Arthur
    Jan 3 at 16:17








  • 3




    $begingroup$
    $mathbb Z[i] / (1+2i)$ 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?
    $endgroup$
    – Lee Mosher
    Jan 3 at 20:31








  • 1




    $begingroup$
    @Mustafa, it is not generally true that $mathbb Z[i]/(a+bi)isom mathbb Z/N(a+bi)$. For example, by quadratic reciprocity, half the time when $N(a+bi)=p^2$ for a prime $p$, that quotient is a field with $p^2$ elements, which is not $mathbb Z/p^2$.
    $endgroup$
    – paul garrett
    Jan 4 at 1:11
















1












$begingroup$



This question already has an answer here:




  • Quotient ring of Gaussian integers

    6 answers





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.










share|cite|improve this question











$endgroup$



marked as duplicate by Watson, Lord Shark the Unknown abstract-algebra
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Jan 10 at 3:49


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.


















  • $begingroup$
    Why are you unhappy with your solution? It looks perfectly reasonable to me.
    $endgroup$
    – Arthur
    Jan 3 at 15:31












  • $begingroup$
    I can't prove it directly, which makes me sad:(
    $endgroup$
    – yLccc
    Jan 3 at 16:06










  • $begingroup$
    What does "diectly" mean in this case?
    $endgroup$
    – Arthur
    Jan 3 at 16:17








  • 3




    $begingroup$
    $mathbb Z[i] / (1+2i)$ 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?
    $endgroup$
    – Lee Mosher
    Jan 3 at 20:31








  • 1




    $begingroup$
    @Mustafa, it is not generally true that $mathbb Z[i]/(a+bi)isom mathbb Z/N(a+bi)$. For example, by quadratic reciprocity, half the time when $N(a+bi)=p^2$ for a prime $p$, that quotient is a field with $p^2$ elements, which is not $mathbb Z/p^2$.
    $endgroup$
    – paul garrett
    Jan 4 at 1:11














1












1








1





$begingroup$



This question already has an answer here:




  • Quotient ring of Gaussian integers

    6 answers





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.










share|cite|improve this question











$endgroup$





This question already has an answer here:




  • Quotient ring of Gaussian integers

    6 answers





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.





This question already has an answer here:




  • Quotient ring of Gaussian integers

    6 answers








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






share|cite|improve this question















share|cite|improve this question













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edited Jan 3 at 16:40







user593746

















asked Jan 3 at 15:21









yLcccyLccc

233




233




marked as duplicate by Watson, Lord Shark the Unknown abstract-algebra
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Jan 10 at 3:49


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Jan 10 at 3:49


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.














  • $begingroup$
    Why are you unhappy with your solution? It looks perfectly reasonable to me.
    $endgroup$
    – Arthur
    Jan 3 at 15:31












  • $begingroup$
    I can't prove it directly, which makes me sad:(
    $endgroup$
    – yLccc
    Jan 3 at 16:06










  • $begingroup$
    What does "diectly" mean in this case?
    $endgroup$
    – Arthur
    Jan 3 at 16:17








  • 3




    $begingroup$
    $mathbb Z[i] / (1+2i)$ 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?
    $endgroup$
    – Lee Mosher
    Jan 3 at 20:31








  • 1




    $begingroup$
    @Mustafa, it is not generally true that $mathbb Z[i]/(a+bi)isom mathbb Z/N(a+bi)$. For example, by quadratic reciprocity, half the time when $N(a+bi)=p^2$ for a prime $p$, that quotient is a field with $p^2$ elements, which is not $mathbb Z/p^2$.
    $endgroup$
    – paul garrett
    Jan 4 at 1:11


















  • $begingroup$
    Why are you unhappy with your solution? It looks perfectly reasonable to me.
    $endgroup$
    – Arthur
    Jan 3 at 15:31












  • $begingroup$
    I can't prove it directly, which makes me sad:(
    $endgroup$
    – yLccc
    Jan 3 at 16:06










  • $begingroup$
    What does "diectly" mean in this case?
    $endgroup$
    – Arthur
    Jan 3 at 16:17








  • 3




    $begingroup$
    $mathbb Z[i] / (1+2i)$ 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?
    $endgroup$
    – Lee Mosher
    Jan 3 at 20:31








  • 1




    $begingroup$
    @Mustafa, it is not generally true that $mathbb Z[i]/(a+bi)isom mathbb Z/N(a+bi)$. For example, by quadratic reciprocity, half the time when $N(a+bi)=p^2$ for a prime $p$, that quotient is a field with $p^2$ elements, which is not $mathbb Z/p^2$.
    $endgroup$
    – paul garrett
    Jan 4 at 1:11
















$begingroup$
Why are you unhappy with your solution? It looks perfectly reasonable to me.
$endgroup$
– Arthur
Jan 3 at 15:31






$begingroup$
Why are you unhappy with your solution? It looks perfectly reasonable to me.
$endgroup$
– Arthur
Jan 3 at 15:31














$begingroup$
I can't prove it directly, which makes me sad:(
$endgroup$
– yLccc
Jan 3 at 16:06




$begingroup$
I can't prove it directly, which makes me sad:(
$endgroup$
– yLccc
Jan 3 at 16:06












$begingroup$
What does "diectly" mean in this case?
$endgroup$
– Arthur
Jan 3 at 16:17






$begingroup$
What does "diectly" mean in this case?
$endgroup$
– Arthur
Jan 3 at 16:17






3




3




$begingroup$
$mathbb Z[i] / (1+2i)$ 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?
$endgroup$
– Lee Mosher
Jan 3 at 20:31






$begingroup$
$mathbb Z[i] / (1+2i)$ 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?
$endgroup$
– Lee Mosher
Jan 3 at 20:31






1




1




$begingroup$
@Mustafa, it is not generally true that $mathbb Z[i]/(a+bi)isom mathbb Z/N(a+bi)$. For example, by quadratic reciprocity, half the time when $N(a+bi)=p^2$ for a prime $p$, that quotient is a field with $p^2$ elements, which is not $mathbb Z/p^2$.
$endgroup$
– paul garrett
Jan 4 at 1:11




$begingroup$
@Mustafa, it is not generally true that $mathbb Z[i]/(a+bi)isom mathbb Z/N(a+bi)$. For example, by quadratic reciprocity, half the time when $N(a+bi)=p^2$ for a prime $p$, that quotient is a field with $p^2$ elements, which is not $mathbb Z/p^2$.
$endgroup$
– paul garrett
Jan 4 at 1:11










4 Answers
4






active

oldest

votes


















5












$begingroup$

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











$endgroup$













  • $begingroup$
    Could u explain why (5,x-2)/(x-2) is the ideal 5Z of Z for me? I don't understand it
    $endgroup$
    – yLccc
    Jan 3 at 16:55






  • 1




    $begingroup$
    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 $5Bbb Z$.
    $endgroup$
    – user593746
    Jan 3 at 20:00








  • 1




    $begingroup$
    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).
    $endgroup$
    – user593746
    Jan 3 at 20:00





















2












$begingroup$

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









$endgroup$





















    1












    $begingroup$

    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











    $endgroup$





















      0












      $begingroup$

      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









      $endgroup$













      • $begingroup$
        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?
        $endgroup$
        – yLccc
        Jan 3 at 16:05










      • $begingroup$
        @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.
        $endgroup$
        – Mark Bennet
        Jan 3 at 16:09


















      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      5












      $begingroup$

      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











      $endgroup$













      • $begingroup$
        Could u explain why (5,x-2)/(x-2) is the ideal 5Z of Z for me? I don't understand it
        $endgroup$
        – yLccc
        Jan 3 at 16:55






      • 1




        $begingroup$
        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 $5Bbb Z$.
        $endgroup$
        – user593746
        Jan 3 at 20:00








      • 1




        $begingroup$
        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).
        $endgroup$
        – user593746
        Jan 3 at 20:00


















      5












      $begingroup$

      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











      $endgroup$













      • $begingroup$
        Could u explain why (5,x-2)/(x-2) is the ideal 5Z of Z for me? I don't understand it
        $endgroup$
        – yLccc
        Jan 3 at 16:55






      • 1




        $begingroup$
        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 $5Bbb Z$.
        $endgroup$
        – user593746
        Jan 3 at 20:00








      • 1




        $begingroup$
        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).
        $endgroup$
        – user593746
        Jan 3 at 20:00
















      5












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      5





      $begingroup$

      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











      $endgroup$



      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 Jan 3 at 20:02

























      answered Jan 3 at 16:25







      user593746



















      • $begingroup$
        Could u explain why (5,x-2)/(x-2) is the ideal 5Z of Z for me? I don't understand it
        $endgroup$
        – yLccc
        Jan 3 at 16:55






      • 1




        $begingroup$
        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 $5Bbb Z$.
        $endgroup$
        – user593746
        Jan 3 at 20:00








      • 1




        $begingroup$
        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).
        $endgroup$
        – user593746
        Jan 3 at 20:00




















      • $begingroup$
        Could u explain why (5,x-2)/(x-2) is the ideal 5Z of Z for me? I don't understand it
        $endgroup$
        – yLccc
        Jan 3 at 16:55






      • 1




        $begingroup$
        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 $5Bbb Z$.
        $endgroup$
        – user593746
        Jan 3 at 20:00








      • 1




        $begingroup$
        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).
        $endgroup$
        – user593746
        Jan 3 at 20:00


















      $begingroup$
      Could u explain why (5,x-2)/(x-2) is the ideal 5Z of Z for me? I don't understand it
      $endgroup$
      – yLccc
      Jan 3 at 16:55




      $begingroup$
      Could u explain why (5,x-2)/(x-2) is the ideal 5Z of Z for me? I don't understand it
      $endgroup$
      – yLccc
      Jan 3 at 16:55




      1




      1




      $begingroup$
      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 $5Bbb Z$.
      $endgroup$
      – user593746
      Jan 3 at 20:00






      $begingroup$
      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 $5Bbb Z$.
      $endgroup$
      – user593746
      Jan 3 at 20:00






      1




      1




      $begingroup$
      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).
      $endgroup$
      – user593746
      Jan 3 at 20:00






      $begingroup$
      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).
      $endgroup$
      – user593746
      Jan 3 at 20:00













      2












      $begingroup$

      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









      $endgroup$


















        2












        $begingroup$

        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









        $endgroup$
















          2












          2








          2





          $begingroup$

          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









          $endgroup$



          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 Jan 3 at 22:02









          egregegreg

          185k1486206




          185k1486206























              1












              $begingroup$

              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











              $endgroup$


















                1












                $begingroup$

                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











                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  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











                  $endgroup$



                  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 Jan 4 at 1:07

























                  answered Jan 3 at 18:38









                  Chris CusterChris Custer

                  14.3k3827




                  14.3k3827























                      0












                      $begingroup$

                      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









                      $endgroup$













                      • $begingroup$
                        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?
                        $endgroup$
                        – yLccc
                        Jan 3 at 16:05










                      • $begingroup$
                        @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.
                        $endgroup$
                        – Mark Bennet
                        Jan 3 at 16:09
















                      0












                      $begingroup$

                      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









                      $endgroup$













                      • $begingroup$
                        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?
                        $endgroup$
                        – yLccc
                        Jan 3 at 16:05










                      • $begingroup$
                        @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.
                        $endgroup$
                        – Mark Bennet
                        Jan 3 at 16:09














                      0












                      0








                      0





                      $begingroup$

                      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









                      $endgroup$



                      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 Jan 3 at 15:41









                      Mark BennetMark Bennet

                      81.9k984183




                      81.9k984183












                      • $begingroup$
                        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?
                        $endgroup$
                        – yLccc
                        Jan 3 at 16:05










                      • $begingroup$
                        @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.
                        $endgroup$
                        – Mark Bennet
                        Jan 3 at 16:09


















                      • $begingroup$
                        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?
                        $endgroup$
                        – yLccc
                        Jan 3 at 16:05










                      • $begingroup$
                        @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.
                        $endgroup$
                        – Mark Bennet
                        Jan 3 at 16:09
















                      $begingroup$
                      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?
                      $endgroup$
                      – yLccc
                      Jan 3 at 16:05




                      $begingroup$
                      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?
                      $endgroup$
                      – yLccc
                      Jan 3 at 16:05












                      $begingroup$
                      @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.
                      $endgroup$
                      – Mark Bennet
                      Jan 3 at 16:09




                      $begingroup$
                      @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.
                      $endgroup$
                      – Mark Bennet
                      Jan 3 at 16:09



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