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







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  • 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
    yesterday






  • 2




    $mathbb Z[i] / (1+2i)$ 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
    yesterday


















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
    yesterday






  • 2




    $mathbb Z[i] / (1+2i)$ 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
    yesterday
















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

4,960430




4,960430






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asked 2 days ago









yLccc

283




283




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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.






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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
    yesterday






  • 2




    $mathbb Z[i] / (1+2i)$ 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
    yesterday




















  • 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
    yesterday






  • 2




    $mathbb Z[i] / (1+2i)$ 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
    yesterday


















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
yesterday




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




2




2




$mathbb Z[i] / (1+2i)$ 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
yesterday






$mathbb Z[i] / (1+2i)$ 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
yesterday












4 Answers
4






active

oldest

votes


















5














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
    yesterday












  • 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
    yesterday





















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 $kervarphi$?






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    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











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      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      5














      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
        yesterday












      • 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
        yesterday


















      5














      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
        yesterday












      • 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
        yesterday
















      5












      5








      5






      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 yesterday

























      answered 2 days ago









      Zvi

      4,960430




      4,960430












      • 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
        yesterday












      • 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
        yesterday




















      • 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
        yesterday












      • 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
        yesterday


















      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
      yesterday






      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
      yesterday














      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
      yesterday






      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
      yesterday













      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 yesterday









          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 yesterday

























                  answered yesterday









                  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










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