Mixing
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
asked 22 hours ago
CurioDidact
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up vote
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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
asked 22 hours ago
CurioDidact
83
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.
add a comment |
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
asked 22 hours ago
CurioDidact
83
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
asked 22 hours ago
CurioDidact
83
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.
asked 22 hours ago
CurioDidact
83
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.
asked 22 hours ago
CurioDidact
83
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.
asked 22 hours ago
CurioDidact
83
asked 22 hours ago
CurioDidact
83
83
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CurioDidact is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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.
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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 22 hours ago
dalbouvet
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
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 $R/A$.)
answered 22 hours ago
dalbouvet
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
add a comment |
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 $R/A$.)
answered 22 hours ago
dalbouvet
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
add a comment |
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 $R/A$.)
answered 22 hours ago
dalbouvet
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 22 hours ago
dalbouvet
521110
answered 22 hours ago
dalbouvet
521110
answered 22 hours ago
dalbouvet
521110
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
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
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
add a comment |
CurioDidact is a new contributor. Be nice, and check out our Code of Conduct.
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