Mixing
Quotient Ring And Cosets
$begingroup$
I know that for a Ring $R$, the Quotient Ring $R/I$ is defined as the set of all cosets of $I$ in $R$.
And the definition of a coset being ${r+I : r in R}$
Now I can't really see why $R/R={0}$ with $R$ being the Real number in this definition. If someone could write out why the only coset of $R$ in $R$ is ${0}$ by the definition of cosets above I would be grateful.
My excuses if things seems unclear or not well defined or just not right, trying to learn some algebra on my own.
abstract-algebra ring-theory
edited Jan 26 at 11:19
Matt Samuel
38.9k63769
asked Feb 11 '15 at 16:08
Frederic DuryFrederic Dury
83
$endgroup$
add a comment |
$begingroup$
I know that for a Ring $R$, the Quotient Ring $R/I$ is defined as the set of all cosets of $I$ in $R$.
And the definition of a coset being ${r+I : r in R}$
Now I can't really see why $R/R={0}$ with $R$ being the Real number in this definition. If someone could write out why the only coset of $R$ in $R$ is ${0}$ by the definition of cosets above I would be grateful.
My excuses if things seems unclear or not well defined or just not right, trying to learn some algebra on my own.
abstract-algebra ring-theory
edited Jan 26 at 11:19
Matt Samuel
38.9k63769
asked Feb 11 '15 at 16:08
Frederic DuryFrederic Dury
83
$endgroup$
1
$begingroup$
There seems to be some confusion: ${r+Imid rin R}$ is not the definition of a coset but rather the set of all cosets: $R/I={r+Imid rin R}$; cosets are sets of the form $r+I$ for $rin R$ and $r+I$ is the set ${r+imid iin I}$. To your question: there is exactly one coset in $R/R$, namely $R=0+R=r+R$ for every $rin R$ and the quotient ring $R/R$ is a ring having only one element, that is, it is isomorphic to the ring ${0}$.
$endgroup$
– user 59363
Feb 11 '15 at 16:35
add a comment |
$begingroup$
I know that for a Ring $R$, the Quotient Ring $R/I$ is defined as the set of all cosets of $I$ in $R$.
And the definition of a coset being ${r+I : r in R}$
Now I can't really see why $R/R={0}$ with $R$ being the Real number in this definition. If someone could write out why the only coset of $R$ in $R$ is ${0}$ by the definition of cosets above I would be grateful.
My excuses if things seems unclear or not well defined or just not right, trying to learn some algebra on my own.
abstract-algebra ring-theory
edited Jan 26 at 11:19
Matt Samuel
38.9k63769
asked Feb 11 '15 at 16:08
Frederic DuryFrederic Dury
83
$endgroup$
I know that for a Ring $R$, the Quotient Ring $R/I$ is defined as the set of all cosets of $I$ in $R$.
And the definition of a coset being ${r+I : r in R}$
Now I can't really see why $R/R={0}$ with $R$ being the Real number in this definition. If someone could write out why the only coset of $R$ in $R$ is ${0}$ by the definition of cosets above I would be grateful.
My excuses if things seems unclear or not well defined or just not right, trying to learn some algebra on my own.
abstract-algebra ring-theory
abstract-algebra ring-theory
edited Jan 26 at 11:19
Matt Samuel
38.9k63769
asked Feb 11 '15 at 16:08
Frederic DuryFrederic Dury
83
edited Jan 26 at 11:19
Matt Samuel
38.9k63769
asked Feb 11 '15 at 16:08
Frederic DuryFrederic Dury
83
edited Jan 26 at 11:19
Matt Samuel
38.9k63769
edited Jan 26 at 11:19
Matt Samuel
38.9k63769
edited Jan 26 at 11:19
Matt Samuel
38.9k63769
38.9k63769
asked Feb 11 '15 at 16:08
Frederic DuryFrederic Dury
83
asked Feb 11 '15 at 16:08
Frederic DuryFrederic Dury
83
asked Feb 11 '15 at 16:08
Frederic DuryFrederic Dury
83
83
1
$begingroup$
There seems to be some confusion: ${r+Imid rin R}$ is not the definition of a coset but rather the set of all cosets: $R/I={r+Imid rin R}$; cosets are sets of the form $r+I$ for $rin R$ and $r+I$ is the set ${r+imid iin I}$. To your question: there is exactly one coset in $R/R$, namely $R=0+R=r+R$ for every $rin R$ and the quotient ring $R/R$ is a ring having only one element, that is, it is isomorphic to the ring ${0}$.
$endgroup$
– user 59363
Feb 11 '15 at 16:35
add a comment |
1
$begingroup$
There seems to be some confusion: ${r+Imid rin R}$ is not the definition of a coset but rather the set of all cosets: $R/I={r+Imid rin R}$; cosets are sets of the form $r+I$ for $rin R$ and $r+I$ is the set ${r+imid iin I}$. To your question: there is exactly one coset in $R/R$, namely $R=0+R=r+R$ for every $rin R$ and the quotient ring $R/R$ is a ring having only one element, that is, it is isomorphic to the ring ${0}$.
$endgroup$
– user 59363
Feb 11 '15 at 16:35
1
1
$begingroup$
There seems to be some confusion: ${r+Imid rin R}$ is not the definition of a coset but rather the set of all cosets: $R/I={r+Imid rin R}$; cosets are sets of the form $r+I$ for $rin R$ and $r+I$ is the set ${r+imid iin I}$. To your question: there is exactly one coset in $R/R$, namely $R=0+R=r+R$ for every $rin R$ and the quotient ring $R/R$ is a ring having only one element, that is, it is isomorphic to the ring ${0}$.
$endgroup$
– user 59363
Feb 11 '15 at 16:35
$begingroup$
There seems to be some confusion: ${r+Imid rin R}$ is not the definition of a coset but rather the set of all cosets: $R/I={r+Imid rin R}$; cosets are sets of the form $r+I$ for $rin R$ and $r+I$ is the set ${r+imid iin I}$. To your question: there is exactly one coset in $R/R$, namely $R=0+R=r+R$ for every $rin R$ and the quotient ring $R/R$ is a ring having only one element, that is, it is isomorphic to the ring ${0}$.
$endgroup$
– user 59363
Feb 11 '15 at 16:35
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Two elements $r,s$ are in the same coset of an ideal $I$ is $r-sin I$. If $I=R$, then $r-0in I$ for all $r$. Thus every element is in the same coset as 0.
answered Feb 11 '15 at 16:20
Matt SamuelMatt Samuel
38.9k63769
$endgroup$
add a comment |
$begingroup$
$x$ ~ $y$ for all $x, y in R$
Then R is a point. Say {0}
answered Feb 11 '15 at 16:37
Vinícius FerrazVinícius Ferraz
1214
$endgroup$
$begingroup$
Not $R$ is a point but rather $R/R$ is.
$endgroup$
– user 59363
Feb 11 '15 at 16:38
$begingroup$
Sure. R/I = R/R = R/~
$endgroup$
– Vinícius Ferraz
Feb 14 '15 at 1:15
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
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active
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active
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votes
$begingroup$
Two elements $r,s$ are in the same coset of an ideal $I$ is $r-sin I$. If $I=R$, then $r-0in I$ for all $r$. Thus every element is in the same coset as 0.
answered Feb 11 '15 at 16:20
Matt SamuelMatt Samuel
38.9k63769
$endgroup$
add a comment |
$begingroup$
Two elements $r,s$ are in the same coset of an ideal $I$ is $r-sin I$. If $I=R$, then $r-0in I$ for all $r$. Thus every element is in the same coset as 0.
answered Feb 11 '15 at 16:20
Matt SamuelMatt Samuel
38.9k63769
$endgroup$
add a comment |
$begingroup$
Two elements $r,s$ are in the same coset of an ideal $I$ is $r-sin I$. If $I=R$, then $r-0in I$ for all $r$. Thus every element is in the same coset as 0.
answered Feb 11 '15 at 16:20
Matt SamuelMatt Samuel
38.9k63769
$endgroup$
Two elements $r,s$ are in the same coset of an ideal $I$ is $r-sin I$. If $I=R$, then $r-0in I$ for all $r$. Thus every element is in the same coset as 0.
answered Feb 11 '15 at 16:20
Matt SamuelMatt Samuel
38.9k63769
answered Feb 11 '15 at 16:20
Matt SamuelMatt Samuel
38.9k63769
answered Feb 11 '15 at 16:20
Matt SamuelMatt Samuel
38.9k63769
answered Feb 11 '15 at 16:20
Matt SamuelMatt Samuel
38.9k63769
38.9k63769
add a comment |
add a comment |
$begingroup$
$x$ ~ $y$ for all $x, y in R$
Then R is a point. Say {0}
answered Feb 11 '15 at 16:37
Vinícius FerrazVinícius Ferraz
1214
$endgroup$
$begingroup$
Not $R$ is a point but rather $R/R$ is.
$endgroup$
– user 59363
Feb 11 '15 at 16:38
$begingroup$
Sure. R/I = R/R = R/~
$endgroup$
– Vinícius Ferraz
Feb 14 '15 at 1:15
add a comment |
$begingroup$
$x$ ~ $y$ for all $x, y in R$
Then R is a point. Say {0}
answered Feb 11 '15 at 16:37
Vinícius FerrazVinícius Ferraz
1214
$endgroup$
$begingroup$
Not $R$ is a point but rather $R/R$ is.
$endgroup$
– user 59363
Feb 11 '15 at 16:38
$begingroup$
Sure. R/I = R/R = R/~
$endgroup$
– Vinícius Ferraz
Feb 14 '15 at 1:15
add a comment |
$begingroup$
$x$ ~ $y$ for all $x, y in R$
Then R is a point. Say {0}
answered Feb 11 '15 at 16:37
Vinícius FerrazVinícius Ferraz
1214
$endgroup$
$x$ ~ $y$ for all $x, y in R$
Then R is a point. Say {0}
answered Feb 11 '15 at 16:37
Vinícius FerrazVinícius Ferraz
1214
answered Feb 11 '15 at 16:37
Vinícius FerrazVinícius Ferraz
1214
answered Feb 11 '15 at 16:37
Vinícius FerrazVinícius Ferraz
1214
answered Feb 11 '15 at 16:37
Vinícius FerrazVinícius Ferraz
1214
1214
$begingroup$
Not $R$ is a point but rather $R/R$ is.
$endgroup$
– user 59363
Feb 11 '15 at 16:38
$begingroup$
Sure. R/I = R/R = R/~
$endgroup$
– Vinícius Ferraz
Feb 14 '15 at 1:15
add a comment |
$begingroup$
Not $R$ is a point but rather $R/R$ is.
$endgroup$
– user 59363
Feb 11 '15 at 16:38
$begingroup$
Sure. R/I = R/R = R/~
$endgroup$
– Vinícius Ferraz
Feb 14 '15 at 1:15
$begingroup$
Not $R$ is a point but rather $R/R$ is.
$endgroup$
– user 59363
Feb 11 '15 at 16:38
$begingroup$
Not $R$ is a point but rather $R/R$ is.
$endgroup$
– user 59363
Feb 11 '15 at 16:38
$begingroup$
Sure. R/I = R/R = R/~
$endgroup$
– Vinícius Ferraz
Feb 14 '15 at 1:15
$begingroup$
Sure. R/I = R/R = R/~
$endgroup$
– Vinícius Ferraz
Feb 14 '15 at 1:15
add a comment |
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$begingroup$
There seems to be some confusion: ${r+Imid rin R}$ is not the definition of a coset but rather the set of all cosets: $R/I={r+Imid rin R}$; cosets are sets of the form $r+I$ for $rin R$ and $r+I$ is the set ${r+imid iin I}$. To your question: there is exactly one coset in $R/R$, namely $R=0+R=r+R$ for every $rin R$ and the quotient ring $R/R$ is a ring having only one element, that is, it is isomorphic to the ring ${0}$.
$endgroup$
– user 59363
Feb 11 '15 at 16:35