In $mathbb{Z}$ ring: if $D(I):={x in R,, |,, x+x in I }$ then $D(nmathbb{Z})=nmathbb{Z} iff n,,,,$is odd
$begingroup$
Let $R$ be a commutative ring and $I$ its ideal. Consider $D(I):={x in R,, |,, x+x in I }$, which is (easy to show) an ideal of $R$.
Assuming $R = mathbb{Z}$ and $n ge 2$, then $D(nmathbb{Z})=nmathbb{Z} iff n,,,,$is odd
Any tips to show that? I am sure that the solution is almost elementary, but there must be a relation that I am missing.
abstract-algebra ring-theory
asked Jan 11 at 15:24
F.incF.inc
403110
$endgroup$
add a comment |
$begingroup$
Let $R$ be a commutative ring and $I$ its ideal. Consider $D(I):={x in R,, |,, x+x in I }$, which is (easy to show) an ideal of $R$.
Assuming $R = mathbb{Z}$ and $n ge 2$, then $D(nmathbb{Z})=nmathbb{Z} iff n,,,,$is odd
Any tips to show that? I am sure that the solution is almost elementary, but there must be a relation that I am missing.
abstract-algebra ring-theory
asked Jan 11 at 15:24
F.incF.inc
403110
$endgroup$
add a comment |
$begingroup$
Let $R$ be a commutative ring and $I$ its ideal. Consider $D(I):={x in R,, |,, x+x in I }$, which is (easy to show) an ideal of $R$.
Assuming $R = mathbb{Z}$ and $n ge 2$, then $D(nmathbb{Z})=nmathbb{Z} iff n,,,,$is odd
Any tips to show that? I am sure that the solution is almost elementary, but there must be a relation that I am missing.
abstract-algebra ring-theory
asked Jan 11 at 15:24
F.incF.inc
403110
$endgroup$
Let $R$ be a commutative ring and $I$ its ideal. Consider $D(I):={x in R,, |,, x+x in I }$, which is (easy to show) an ideal of $R$.
Assuming $R = mathbb{Z}$ and $n ge 2$, then $D(nmathbb{Z})=nmathbb{Z} iff n,,,,$is odd
Any tips to show that? I am sure that the solution is almost elementary, but there must be a relation that I am missing.
abstract-algebra ring-theory
abstract-algebra ring-theory
asked Jan 11 at 15:24
F.incF.inc
403110
asked Jan 11 at 15:24
F.incF.inc
403110
asked Jan 11 at 15:24
F.incF.inc
403110
asked Jan 11 at 15:24
F.incF.inc
403110
asked Jan 11 at 15:24
F.incF.inc
403110
403110
add a comment |
add a comment |
1 Answer
1
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$begingroup$
Obviously you always have $D(I)supseteq I$.
When $R=mathbb Z$, the condition means that $D(I)={zin mathbb Zmid 2zin I}$.
If $n$ is even, for example $n=2$, then you obviously have that $1in D(2mathbb Z)supsetneq2mathbb Z$.
When $n$ is odd, you have to take $zinmathbb Z$ and consider what happens when $2zin nmathbb Z$.
Just look: $2z=nr$ with $n$ odd implies that $2|r$. Do you see how to finish?
answered Jan 11 at 15:56
rschwiebrschwieb
106k12101248
$endgroup$
$begingroup$
It's clear! Thank you!
$endgroup$
– F.inc
Jan 11 at 15:59
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Obviously you always have $D(I)supseteq I$.
When $R=mathbb Z$, the condition means that $D(I)={zin mathbb Zmid 2zin I}$.
If $n$ is even, for example $n=2$, then you obviously have that $1in D(2mathbb Z)supsetneq2mathbb Z$.
When $n$ is odd, you have to take $zinmathbb Z$ and consider what happens when $2zin nmathbb Z$.
Just look: $2z=nr$ with $n$ odd implies that $2|r$. Do you see how to finish?
answered Jan 11 at 15:56
rschwiebrschwieb
106k12101248
$endgroup$
$begingroup$
It's clear! Thank you!
$endgroup$
– F.inc
Jan 11 at 15:59
add a comment |
$begingroup$
Obviously you always have $D(I)supseteq I$.
When $R=mathbb Z$, the condition means that $D(I)={zin mathbb Zmid 2zin I}$.
If $n$ is even, for example $n=2$, then you obviously have that $1in D(2mathbb Z)supsetneq2mathbb Z$.
When $n$ is odd, you have to take $zinmathbb Z$ and consider what happens when $2zin nmathbb Z$.
Just look: $2z=nr$ with $n$ odd implies that $2|r$. Do you see how to finish?
answered Jan 11 at 15:56
rschwiebrschwieb
106k12101248
$endgroup$
$begingroup$
It's clear! Thank you!
$endgroup$
– F.inc
Jan 11 at 15:59
add a comment |
$begingroup$
Obviously you always have $D(I)supseteq I$.
When $R=mathbb Z$, the condition means that $D(I)={zin mathbb Zmid 2zin I}$.
If $n$ is even, for example $n=2$, then you obviously have that $1in D(2mathbb Z)supsetneq2mathbb Z$.
When $n$ is odd, you have to take $zinmathbb Z$ and consider what happens when $2zin nmathbb Z$.
Just look: $2z=nr$ with $n$ odd implies that $2|r$. Do you see how to finish?
answered Jan 11 at 15:56
rschwiebrschwieb
106k12101248
$endgroup$
Obviously you always have $D(I)supseteq I$.
When $R=mathbb Z$, the condition means that $D(I)={zin mathbb Zmid 2zin I}$.
If $n$ is even, for example $n=2$, then you obviously have that $1in D(2mathbb Z)supsetneq2mathbb Z$.
When $n$ is odd, you have to take $zinmathbb Z$ and consider what happens when $2zin nmathbb Z$.
Just look: $2z=nr$ with $n$ odd implies that $2|r$. Do you see how to finish?
answered Jan 11 at 15:56
rschwiebrschwieb
106k12101248
answered Jan 11 at 15:56
rschwiebrschwieb
106k12101248
answered Jan 11 at 15:56
rschwiebrschwieb
106k12101248
answered Jan 11 at 15:56
rschwiebrschwieb
106k12101248
106k12101248
$begingroup$
It's clear! Thank you!
$endgroup$
– F.inc
Jan 11 at 15:59
add a comment |
$begingroup$
It's clear! Thank you!
$endgroup$
– F.inc
Jan 11 at 15:59
$begingroup$
It's clear! Thank you!
$endgroup$
– F.inc
Jan 11 at 15:59
$begingroup$
It's clear! Thank you!
$endgroup$
– F.inc
Jan 11 at 15:59
add a comment |
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