An ideal is radical if and only if $x^2in IRightarrow xin I$ The 2019 Stack Overflow Developer Survey Results Are InRadical/Prime/Maximal ideals under quotient mapsAn ideal whose radical is maximal is primaryRadical ideals of $mathbbZ$?Let I be an unmixed radical ideal of R. then (I:x) is unmixedA radical ideal in a commutative ring is prime if and only if it is not an intersection of two radical ideals properly containing it?Radical of an ideal in $R [x]$Radical of an ideal in the ring $mathbbZ^mathbbN$Proof-verification: every prime ideal is radicalProperties of radical idealsRadical of an ideal (properties)
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An ideal is radical if and only if $x^2in IRightarrow xin I$
The 2019 Stack Overflow Developer Survey Results Are InRadical/Prime/Maximal ideals under quotient mapsAn ideal whose radical is maximal is primaryRadical ideals of $mathbbZ$?Let I be an unmixed radical ideal of R. then (I:x) is unmixedA radical ideal in a commutative ring is prime if and only if it is not an intersection of two radical ideals properly containing it?Radical of an ideal in $R [x]$Radical of an ideal in the ring $mathbbZ^mathbbN$Proof-verification: every prime ideal is radicalProperties of radical idealsRadical of an ideal (properties)
$begingroup$
I was trying to prove the following:
Let $R$ be a commutative ring. Suppose $I$ is an ideal of $R$. Then $I$ is a radical ideal if and only if for $xin R$ such that $x^2in I$ then $xin I$.
One direction is very easy. If you suppose $I$ is radical then the result is immediate.
The other direction is not so easy. We have to prove that for any $n$, $x^nin IRightarrow xin I$. I tried to prove it by induction, but there is some trouble when $n$ is odd. I think the problem can be reduced to prove it when $n$ is an odd prime.
But I don't know if this is the best approach. Perhaps it's better to suppose some $x$ is in all prime ideals containing $I$ and try to prove $xin I$ but at first sight it seems more convoluted to do it that way. So, how can I proceed?
commutative-algebra ideals
edited Mar 24 at 7:06
user26857
39.5k124284
asked Mar 24 at 5:58
NellNell
40429
$endgroup$
add a comment |
$begingroup$
I was trying to prove the following:
Let $R$ be a commutative ring. Suppose $I$ is an ideal of $R$. Then $I$ is a radical ideal if and only if for $xin R$ such that $x^2in I$ then $xin I$.
One direction is very easy. If you suppose $I$ is radical then the result is immediate.
The other direction is not so easy. We have to prove that for any $n$, $x^nin IRightarrow xin I$. I tried to prove it by induction, but there is some trouble when $n$ is odd. I think the problem can be reduced to prove it when $n$ is an odd prime.
But I don't know if this is the best approach. Perhaps it's better to suppose some $x$ is in all prime ideals containing $I$ and try to prove $xin I$ but at first sight it seems more convoluted to do it that way. So, how can I proceed?
commutative-algebra ideals
edited Mar 24 at 7:06
user26857
39.5k124284
asked Mar 24 at 5:58
NellNell
40429
$endgroup$
add a comment |
$begingroup$
I was trying to prove the following:
Let $R$ be a commutative ring. Suppose $I$ is an ideal of $R$. Then $I$ is a radical ideal if and only if for $xin R$ such that $x^2in I$ then $xin I$.
One direction is very easy. If you suppose $I$ is radical then the result is immediate.
The other direction is not so easy. We have to prove that for any $n$, $x^nin IRightarrow xin I$. I tried to prove it by induction, but there is some trouble when $n$ is odd. I think the problem can be reduced to prove it when $n$ is an odd prime.
But I don't know if this is the best approach. Perhaps it's better to suppose some $x$ is in all prime ideals containing $I$ and try to prove $xin I$ but at first sight it seems more convoluted to do it that way. So, how can I proceed?
commutative-algebra ideals
edited Mar 24 at 7:06
user26857
39.5k124284
asked Mar 24 at 5:58
NellNell
40429
$endgroup$
I was trying to prove the following:
Let $R$ be a commutative ring. Suppose $I$ is an ideal of $R$. Then $I$ is a radical ideal if and only if for $xin R$ such that $x^2in I$ then $xin I$.
One direction is very easy. If you suppose $I$ is radical then the result is immediate.
The other direction is not so easy. We have to prove that for any $n$, $x^nin IRightarrow xin I$. I tried to prove it by induction, but there is some trouble when $n$ is odd. I think the problem can be reduced to prove it when $n$ is an odd prime.
But I don't know if this is the best approach. Perhaps it's better to suppose some $x$ is in all prime ideals containing $I$ and try to prove $xin I$ but at first sight it seems more convoluted to do it that way. So, how can I proceed?
commutative-algebra ideals
commutative-algebra ideals
edited Mar 24 at 7:06
user26857
39.5k124284
asked Mar 24 at 5:58
NellNell
40429
edited Mar 24 at 7:06
user26857
39.5k124284
asked Mar 24 at 5:58
NellNell
40429
edited Mar 24 at 7:06
user26857
39.5k124284
edited Mar 24 at 7:06
user26857
39.5k124284
edited Mar 24 at 7:06
user26857
39.5k124284
39.5k124284
asked Mar 24 at 5:58
NellNell
40429
asked Mar 24 at 5:58
NellNell
40429
asked Mar 24 at 5:58
NellNell
40429
40429
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Assume your condition $x^2in Iimplies xin I$ holds.
Suppose $x^3in I$. Then $x^4in I$ since $I$ is an ideal, and $x^4=xx^3$.
Therefore $x^2in I$ (as $(x^2)^2in I$) and so $xin I$.
See above for the $x^4in I$ case.
Suppose $x^5in I$. Then $x^6in I$ since $I$ is an ideal, and $x^6=xx^5$.
Therefore $x^3in I$ (as $(x^3)^2in I$) and we have seen that implies
that $xin I$.
And so on...
answered Mar 24 at 6:10
Lord Shark the UnknownLord Shark the Unknown
108k1162136
$endgroup$
1
$begingroup$
Nice answer. To give a unified presentation to this argument: it is easy to prove (e.g., by induction) that if $x^2 in I implies x in I$, then $x^2^n in I implies x in I$ for all $n in mathbbN$. Hence, if $x^k in I$, and $t_k$ is the smallest power of $2$ bigger than $k$, then $x^t_k = x^t_k-k cdot x^k in I$, whence $x in I$.
$endgroup$
– Alex Wertheim
Mar 24 at 7:42
add a comment |
Your Answer
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1 Answer
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1 Answer
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oldest
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$begingroup$
Assume your condition $x^2in Iimplies xin I$ holds.
Suppose $x^3in I$. Then $x^4in I$ since $I$ is an ideal, and $x^4=xx^3$.
Therefore $x^2in I$ (as $(x^2)^2in I$) and so $xin I$.
See above for the $x^4in I$ case.
Suppose $x^5in I$. Then $x^6in I$ since $I$ is an ideal, and $x^6=xx^5$.
Therefore $x^3in I$ (as $(x^3)^2in I$) and we have seen that implies
that $xin I$.
And so on...
answered Mar 24 at 6:10
Lord Shark the UnknownLord Shark the Unknown
108k1162136
$endgroup$
1
$begingroup$
Nice answer. To give a unified presentation to this argument: it is easy to prove (e.g., by induction) that if $x^2 in I implies x in I$, then $x^2^n in I implies x in I$ for all $n in mathbbN$. Hence, if $x^k in I$, and $t_k$ is the smallest power of $2$ bigger than $k$, then $x^t_k = x^t_k-k cdot x^k in I$, whence $x in I$.
$endgroup$
– Alex Wertheim
Mar 24 at 7:42
add a comment |
$begingroup$
Assume your condition $x^2in Iimplies xin I$ holds.
Suppose $x^3in I$. Then $x^4in I$ since $I$ is an ideal, and $x^4=xx^3$.
Therefore $x^2in I$ (as $(x^2)^2in I$) and so $xin I$.
See above for the $x^4in I$ case.
Suppose $x^5in I$. Then $x^6in I$ since $I$ is an ideal, and $x^6=xx^5$.
Therefore $x^3in I$ (as $(x^3)^2in I$) and we have seen that implies
that $xin I$.
And so on...
answered Mar 24 at 6:10
Lord Shark the UnknownLord Shark the Unknown
108k1162136
$endgroup$
1
$begingroup$
Nice answer. To give a unified presentation to this argument: it is easy to prove (e.g., by induction) that if $x^2 in I implies x in I$, then $x^2^n in I implies x in I$ for all $n in mathbbN$. Hence, if $x^k in I$, and $t_k$ is the smallest power of $2$ bigger than $k$, then $x^t_k = x^t_k-k cdot x^k in I$, whence $x in I$.
$endgroup$
– Alex Wertheim
Mar 24 at 7:42
add a comment |
$begingroup$
Assume your condition $x^2in Iimplies xin I$ holds.
Suppose $x^3in I$. Then $x^4in I$ since $I$ is an ideal, and $x^4=xx^3$.
Therefore $x^2in I$ (as $(x^2)^2in I$) and so $xin I$.
See above for the $x^4in I$ case.
Suppose $x^5in I$. Then $x^6in I$ since $I$ is an ideal, and $x^6=xx^5$.
Therefore $x^3in I$ (as $(x^3)^2in I$) and we have seen that implies
that $xin I$.
And so on...
answered Mar 24 at 6:10
Lord Shark the UnknownLord Shark the Unknown
108k1162136
$endgroup$
Assume your condition $x^2in Iimplies xin I$ holds.
Suppose $x^3in I$. Then $x^4in I$ since $I$ is an ideal, and $x^4=xx^3$.
Therefore $x^2in I$ (as $(x^2)^2in I$) and so $xin I$.
See above for the $x^4in I$ case.
Suppose $x^5in I$. Then $x^6in I$ since $I$ is an ideal, and $x^6=xx^5$.
Therefore $x^3in I$ (as $(x^3)^2in I$) and we have seen that implies
that $xin I$.
And so on...
answered Mar 24 at 6:10
Lord Shark the UnknownLord Shark the Unknown
108k1162136
answered Mar 24 at 6:10
Lord Shark the UnknownLord Shark the Unknown
108k1162136
answered Mar 24 at 6:10
Lord Shark the UnknownLord Shark the Unknown
108k1162136
answered Mar 24 at 6:10
Lord Shark the UnknownLord Shark the Unknown
108k1162136
108k1162136
1
$begingroup$
Nice answer. To give a unified presentation to this argument: it is easy to prove (e.g., by induction) that if $x^2 in I implies x in I$, then $x^2^n in I implies x in I$ for all $n in mathbbN$. Hence, if $x^k in I$, and $t_k$ is the smallest power of $2$ bigger than $k$, then $x^t_k = x^t_k-k cdot x^k in I$, whence $x in I$.
$endgroup$
– Alex Wertheim
Mar 24 at 7:42
add a comment |
1
$begingroup$
Nice answer. To give a unified presentation to this argument: it is easy to prove (e.g., by induction) that if $x^2 in I implies x in I$, then $x^2^n in I implies x in I$ for all $n in mathbbN$. Hence, if $x^k in I$, and $t_k$ is the smallest power of $2$ bigger than $k$, then $x^t_k = x^t_k-k cdot x^k in I$, whence $x in I$.
$endgroup$
– Alex Wertheim
Mar 24 at 7:42
1
1
$begingroup$
Nice answer. To give a unified presentation to this argument: it is easy to prove (e.g., by induction) that if $x^2 in I implies x in I$, then $x^2^n in I implies x in I$ for all $n in mathbbN$. Hence, if $x^k in I$, and $t_k$ is the smallest power of $2$ bigger than $k$, then $x^t_k = x^t_k-k cdot x^k in I$, whence $x in I$.
$endgroup$
– Alex Wertheim
Mar 24 at 7:42
$begingroup$
Nice answer. To give a unified presentation to this argument: it is easy to prove (e.g., by induction) that if $x^2 in I implies x in I$, then $x^2^n in I implies x in I$ for all $n in mathbbN$. Hence, if $x^k in I$, and $t_k$ is the smallest power of $2$ bigger than $k$, then $x^t_k = x^t_k-k cdot x^k in I$, whence $x in I$.
$endgroup$
– Alex Wertheim
Mar 24 at 7:42
add a comment |
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