Describing Cosets in $R/A$
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
asked Nov 19 '18 at 4:09
CurioDidact
83
add a comment |
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
asked Nov 19 '18 at 4:09
CurioDidact
83
add a comment |
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
asked Nov 19 '18 at 4:09
CurioDidact
83
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
asked Nov 19 '18 at 4:09
CurioDidact
83
asked Nov 19 '18 at 4:09
CurioDidact
83
asked Nov 19 '18 at 4:09
CurioDidact
83
asked Nov 19 '18 at 4:09
CurioDidact
83
asked Nov 19 '18 at 4:09
CurioDidact
83
83
add a comment |
add a comment |
1 Answer
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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 $R/A$.)
answered Nov 19 '18 at 4:16
dyf
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
Nov 19 '18 at 4:52
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 $H$.
– dyf
Nov 19 '18 at 4:55
add a comment |
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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 $R/A$.)
answered Nov 19 '18 at 4:16
dyf
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
Nov 19 '18 at 4:52
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 $H$.
– dyf
Nov 19 '18 at 4:55
add a comment |
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 $R/A$.)
answered Nov 19 '18 at 4:16
dyf
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
Nov 19 '18 at 4:52
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 $H$.
– dyf
Nov 19 '18 at 4:55
add a comment |
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 $R/A$.)
answered Nov 19 '18 at 4:16
dyf
521110
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 $R/A$.)
answered Nov 19 '18 at 4:16
dyf
521110
answered Nov 19 '18 at 4:16
dyf
521110
answered Nov 19 '18 at 4:16
dyf
521110
answered Nov 19 '18 at 4:16
dyf
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
Nov 19 '18 at 4:52
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 $H$.
– dyf
Nov 19 '18 at 4:55
add a comment |
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
Nov 19 '18 at 4:52
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 $H$.
– dyf
Nov 19 '18 at 4:55
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
Nov 19 '18 at 4:52
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
Nov 19 '18 at 4:52
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 $H$.
– dyf
Nov 19 '18 at 4:55
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 $H$.
– dyf
Nov 19 '18 at 4:55
add a comment |
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