Describing Cosets in $R/A$











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In a worked example in my textbook, we are describing the cosets in $R/A$, where $R=mathbb{Z}[i]$, the Gaussian integers, and $A = (2+i)R$, the ideal of all multiples of $2 + i$.



It starts by stating that a typical coset $x$ in $R/A$ will have the form $x=(m + ni) + A$, with integers $m,n$. This makes sense to me, but I do not understand the following part:




"Since $2 + i in A$, we have $i + A = -2 + A$"




Can someone walk me through how we arrive at $i+A = -2+A$?






abstract-algebra ring-theory gaussian-integers







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    In a worked example in my textbook, we are describing the cosets in $R/A$, where $R=mathbb{Z}[i]$, the Gaussian integers, and $A = (2+i)R$, the ideal of all multiples of $2 + i$.



    It starts by stating that a typical coset $x$ in $R/A$ will have the form $x=(m + ni) + A$, with integers $m,n$. This makes sense to me, but I do not understand the following part:




    "Since $2 + i in A$, we have $i + A = -2 + A$"




    Can someone walk me through how we arrive at $i+A = -2+A$?






    abstract-algebra ring-theory gaussian-integers







    share|cite|improve this question







    New contributor




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






















      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      In a worked example in my textbook, we are describing the cosets in $R/A$, where $R=mathbb{Z}[i]$, the Gaussian integers, and $A = (2+i)R$, the ideal of all multiples of $2 + i$.



      It starts by stating that a typical coset $x$ in $R/A$ will have the form $x=(m + ni) + A$, with integers $m,n$. This makes sense to me, but I do not understand the following part:




      "Since $2 + i in A$, we have $i + A = -2 + A$"




      Can someone walk me through how we arrive at $i+A = -2+A$?






      abstract-algebra ring-theory gaussian-integers







      share|cite|improve this question







      New contributor




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











      In a worked example in my textbook, we are describing the cosets in $R/A$, where $R=mathbb{Z}[i]$, the Gaussian integers, and $A = (2+i)R$, the ideal of all multiples of $2 + i$.



      It starts by stating that a typical coset $x$ in $R/A$ will have the form $x=(m + ni) + A$, with integers $m,n$. This makes sense to me, but I do not understand the following part:




      "Since $2 + i in A$, we have $i + A = -2 + A$"




      Can someone walk me through how we arrive at $i+A = -2+A$?






      abstract-algebra ring-theory gaussian-integers




      abstract-algebra ring-theory gaussian-integers






      share|cite|improve this question







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      CurioDidact is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      share|cite|improve this question







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          Well, as you have said, an element in $R/A$ looks like $(m+ni) + A$. So $2+i in A$ means $2+i +A = A$ (this is an equality of elements in $R/A$). So $(2+i)+A - (2+A) = A - (2+A)$, giving $i+A = -2+A$. (If you are unsure how to get the last equality, think about what addition (and hence subtraction) means in $R/A$.)






          share|cite|improve this answer





















          • I'm starting to get it, can you elaborate more on why $2 + i + A = A$? Is it because every element of $A$ is a multiple of $2+i$, so adding $2+i$ to any element of A will just give you an element already in $A$?
            – CurioDidact
            21 hours ago










          • yes, that's essentially it. If you are familiar with quotient groups, here's an analogue: if $(G,+)$ is an abelian group, and $H$ is a subgroup and $h$ an element in it, then the coset $h+H$ is same as $H$.
            – dalbouvet
            21 hours ago











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          Well, as you have said, an element in $R/A$ looks like $(m+ni) + A$. So $2+i in A$ means $2+i +A = A$ (this is an equality of elements in $R/A$). So $(2+i)+A - (2+A) = A - (2+A)$, giving $i+A = -2+A$. (If you are unsure how to get the last equality, think about what addition (and hence subtraction) means in $R/A$.)






          share|cite|improve this answer





















          • I'm starting to get it, can you elaborate more on why $2 + i + A = A$? Is it because every element of $A$ is a multiple of $2+i$, so adding $2+i$ to any element of A will just give you an element already in $A$?
            – CurioDidact
            21 hours ago










          • yes, that's essentially it. If you are familiar with quotient groups, here's an analogue: if $(G,+)$ is an abelian group, and $H$ is a subgroup and $h$ an element in it, then the coset $h+H$ is same as $H$.
            – dalbouvet
            21 hours ago















          up vote
          0
          down vote



          accepted










          Well, as you have said, an element in $R/A$ looks like $(m+ni) + A$. So $2+i in A$ means $2+i +A = A$ (this is an equality of elements in $R/A$). So $(2+i)+A - (2+A) = A - (2+A)$, giving $i+A = -2+A$. (If you are unsure how to get the last equality, think about what addition (and hence subtraction) means in $R/A$.)






          share|cite|improve this answer





















          • I'm starting to get it, can you elaborate more on why $2 + i + A = A$? Is it because every element of $A$ is a multiple of $2+i$, so adding $2+i$ to any element of A will just give you an element already in $A$?
            – CurioDidact
            21 hours ago










          • yes, that's essentially it. If you are familiar with quotient groups, here's an analogue: if $(G,+)$ is an abelian group, and $H$ is a subgroup and $h$ an element in it, then the coset $h+H$ is same as $H$.
            – dalbouvet
            21 hours ago













          up vote
          0
          down vote



          accepted







          up vote
          0
          down vote



          accepted






          Well, as you have said, an element in $R/A$ looks like $(m+ni) + A$. So $2+i in A$ means $2+i +A = A$ (this is an equality of elements in $R/A$). So $(2+i)+A - (2+A) = A - (2+A)$, giving $i+A = -2+A$. (If you are unsure how to get the last equality, think about what addition (and hence subtraction) means in $R/A$.)






          share|cite|improve this answer












          Well, as you have said, an element in $R/A$ looks like $(m+ni) + A$. So $2+i in A$ means $2+i +A = A$ (this is an equality of elements in $R/A$). So $(2+i)+A - (2+A) = A - (2+A)$, giving $i+A = -2+A$. (If you are unsure how to get the last equality, think about what addition (and hence subtraction) means in $R/A$.)







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 22 hours ago









          dalbouvet

          521110




          521110












          • I'm starting to get it, can you elaborate more on why $2 + i + A = A$? Is it because every element of $A$ is a multiple of $2+i$, so adding $2+i$ to any element of A will just give you an element already in $A$?
            – CurioDidact
            21 hours ago










          • yes, that's essentially it. If you are familiar with quotient groups, here's an analogue: if $(G,+)$ is an abelian group, and $H$ is a subgroup and $h$ an element in it, then the coset $h+H$ is same as $H$.
            – dalbouvet
            21 hours ago


















          • I'm starting to get it, can you elaborate more on why $2 + i + A = A$? Is it because every element of $A$ is a multiple of $2+i$, so adding $2+i$ to any element of A will just give you an element already in $A$?
            – CurioDidact
            21 hours ago










          • yes, that's essentially it. If you are familiar with quotient groups, here's an analogue: if $(G,+)$ is an abelian group, and $H$ is a subgroup and $h$ an element in it, then the coset $h+H$ is same as $H$.
            – dalbouvet
            21 hours ago
















          I'm starting to get it, can you elaborate more on why $2 + i + A = A$? Is it because every element of $A$ is a multiple of $2+i$, so adding $2+i$ to any element of A will just give you an element already in $A$?
          – CurioDidact
          21 hours ago




          I'm starting to get it, can you elaborate more on why $2 + i + A = A$? Is it because every element of $A$ is a multiple of $2+i$, so adding $2+i$ to any element of A will just give you an element already in $A$?
          – CurioDidact
          21 hours ago












          yes, that's essentially it. If you are familiar with quotient groups, here's an analogue: if $(G,+)$ is an abelian group, and $H$ is a subgroup and $h$ an element in it, then the coset $h+H$ is same as $H$.
          – dalbouvet
          21 hours ago




          yes, that's essentially it. If you are familiar with quotient groups, here's an analogue: if $(G,+)$ is an abelian group, and $H$ is a subgroup and $h$ an element in it, then the coset $h+H$ is same as $H$.
          – dalbouvet
          21 hours ago










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