Let $R=Bbb Z(i)$ be the ring of gaussian integers. Describe the cosets of the factor ring $R$ $A$ where...
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Problem: Let $R=Bbb Z(i)$ be the ring of gaussian integers
Describe the cosets of the factor ring $R$ $A$ where $A=Ri$.
Thoughts:The elements of $R$ $A$ will be of the form: $(a+bi)+Ri$. I'm not really sure how to derive some sort of equivalence classes/cosets out of this. Any insights appreciated.
ring-theory
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add a comment |
$begingroup$
Problem: Let $R=Bbb Z(i)$ be the ring of gaussian integers
Describe the cosets of the factor ring $R$ $A$ where $A=Ri$.
Thoughts:The elements of $R$ $A$ will be of the form: $(a+bi)+Ri$. I'm not really sure how to derive some sort of equivalence classes/cosets out of this. Any insights appreciated.
ring-theory
$endgroup$
1
$begingroup$
Hint: $i$ is a unit of $R$.
$endgroup$
– Jyrki Lahtonen
Feb 2 at 20:54
add a comment |
$begingroup$
Problem: Let $R=Bbb Z(i)$ be the ring of gaussian integers
Describe the cosets of the factor ring $R$ $A$ where $A=Ri$.
Thoughts:The elements of $R$ $A$ will be of the form: $(a+bi)+Ri$. I'm not really sure how to derive some sort of equivalence classes/cosets out of this. Any insights appreciated.
ring-theory
$endgroup$
Problem: Let $R=Bbb Z(i)$ be the ring of gaussian integers
Describe the cosets of the factor ring $R$ $A$ where $A=Ri$.
Thoughts:The elements of $R$ $A$ will be of the form: $(a+bi)+Ri$. I'm not really sure how to derive some sort of equivalence classes/cosets out of this. Any insights appreciated.
ring-theory
ring-theory
edited Feb 3 at 2:50
IntegrateThis
asked Feb 2 at 20:47
IntegrateThisIntegrateThis
1,9661818
1,9661818
1
$begingroup$
Hint: $i$ is a unit of $R$.
$endgroup$
– Jyrki Lahtonen
Feb 2 at 20:54
add a comment |
1
$begingroup$
Hint: $i$ is a unit of $R$.
$endgroup$
– Jyrki Lahtonen
Feb 2 at 20:54
1
1
$begingroup$
Hint: $i$ is a unit of $R$.
$endgroup$
– Jyrki Lahtonen
Feb 2 at 20:54
$begingroup$
Hint: $i$ is a unit of $R$.
$endgroup$
– Jyrki Lahtonen
Feb 2 at 20:54
add a comment |
1 Answer
1
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$begingroup$
Observe that $A=iR=R$ and so $R/A={A}$.
Let $a+bi+Ain R/A$, then
$$a+bi+A=a+bi+i(-b+ai)+A=a+bi-bi-a+A=0+A=A$$
$endgroup$
add a comment |
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$begingroup$
Observe that $A=iR=R$ and so $R/A={A}$.
Let $a+bi+Ain R/A$, then
$$a+bi+A=a+bi+i(-b+ai)+A=a+bi-bi-a+A=0+A=A$$
$endgroup$
add a comment |
$begingroup$
Observe that $A=iR=R$ and so $R/A={A}$.
Let $a+bi+Ain R/A$, then
$$a+bi+A=a+bi+i(-b+ai)+A=a+bi-bi-a+A=0+A=A$$
$endgroup$
add a comment |
$begingroup$
Observe that $A=iR=R$ and so $R/A={A}$.
Let $a+bi+Ain R/A$, then
$$a+bi+A=a+bi+i(-b+ai)+A=a+bi-bi-a+A=0+A=A$$
$endgroup$
Observe that $A=iR=R$ and so $R/A={A}$.
Let $a+bi+Ain R/A$, then
$$a+bi+A=a+bi+i(-b+ai)+A=a+bi-bi-a+A=0+A=A$$
answered Feb 2 at 21:01


Don FanucciDon Fanucci
1,325521
1,325521
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$begingroup$
Hint: $i$ is a unit of $R$.
$endgroup$
– Jyrki Lahtonen
Feb 2 at 20:54