Radical of an ideal equals the intersection of all prime ideals containing it.
$begingroup$
I am trying to prove that for a ring $R$ and an ideal $Ileq R$ we have
$$
sqrt{I}=bigcap_{Ileqmathfrak p}mathfrak p,
$$
the intersection of all prime ideals containing $I$.
The definition of the radical of an ideal I am working with is
$$
sqrt{I}={xin Rmidexists minmathbb{N} text{ with }x^min I}.
$$
I am a bit confused about the RHS in particular, since the intersection could theoretically be infinite, in which case we do not have primality of the ideal on the left. Is another assumption necessary, maybe that the ring is Noetherian?
abstract-algebra ring-theory ideals maximal-and-prime-ideals
edited Dec 18 '18 at 18:25
user26857
39.3k124183
asked Oct 1 '16 at 1:19
qbertqbert
22.1k32561
$endgroup$
add a comment |
$begingroup$
I am trying to prove that for a ring $R$ and an ideal $Ileq R$ we have
$$
sqrt{I}=bigcap_{Ileqmathfrak p}mathfrak p,
$$
the intersection of all prime ideals containing $I$.
The definition of the radical of an ideal I am working with is
$$
sqrt{I}={xin Rmidexists minmathbb{N} text{ with }x^min I}.
$$
I am a bit confused about the RHS in particular, since the intersection could theoretically be infinite, in which case we do not have primality of the ideal on the left. Is another assumption necessary, maybe that the ring is Noetherian?
abstract-algebra ring-theory ideals maximal-and-prime-ideals
edited Dec 18 '18 at 18:25
user26857
39.3k124183
asked Oct 1 '16 at 1:19
qbertqbert
22.1k32561
$endgroup$
$begingroup$
in which case we do not have primality of the ideal on the left. Radicals are not always prime, so that is totally expected. Why do you think otherwise? Is another assumption necessary, maybe that the Ring is Noetherian? For what? The equality? Then no.
$endgroup$
– rschwieb
Oct 1 '16 at 11:37
$begingroup$
@rschwieb I got the direction of implication mixed up, prime implies radical, but radical does not imply prime.
$endgroup$
– qbert
Oct 1 '16 at 19:37
add a comment |
$begingroup$
I am trying to prove that for a ring $R$ and an ideal $Ileq R$ we have
$$
sqrt{I}=bigcap_{Ileqmathfrak p}mathfrak p,
$$
the intersection of all prime ideals containing $I$.
The definition of the radical of an ideal I am working with is
$$
sqrt{I}={xin Rmidexists minmathbb{N} text{ with }x^min I}.
$$
I am a bit confused about the RHS in particular, since the intersection could theoretically be infinite, in which case we do not have primality of the ideal on the left. Is another assumption necessary, maybe that the ring is Noetherian?
abstract-algebra ring-theory ideals maximal-and-prime-ideals
edited Dec 18 '18 at 18:25
user26857
39.3k124183
asked Oct 1 '16 at 1:19
qbertqbert
22.1k32561
$endgroup$
I am trying to prove that for a ring $R$ and an ideal $Ileq R$ we have
$$
sqrt{I}=bigcap_{Ileqmathfrak p}mathfrak p,
$$
the intersection of all prime ideals containing $I$.
The definition of the radical of an ideal I am working with is
$$
sqrt{I}={xin Rmidexists minmathbb{N} text{ with }x^min I}.
$$
I am a bit confused about the RHS in particular, since the intersection could theoretically be infinite, in which case we do not have primality of the ideal on the left. Is another assumption necessary, maybe that the ring is Noetherian?
abstract-algebra ring-theory ideals maximal-and-prime-ideals
abstract-algebra ring-theory ideals maximal-and-prime-ideals
edited Dec 18 '18 at 18:25
user26857
39.3k124183
asked Oct 1 '16 at 1:19
qbertqbert
22.1k32561
edited Dec 18 '18 at 18:25
user26857
39.3k124183
asked Oct 1 '16 at 1:19
qbertqbert
22.1k32561
edited Dec 18 '18 at 18:25
user26857
39.3k124183
edited Dec 18 '18 at 18:25
user26857
39.3k124183
edited Dec 18 '18 at 18:25
user26857
39.3k124183
39.3k124183
asked Oct 1 '16 at 1:19
qbertqbert
22.1k32561
asked Oct 1 '16 at 1:19
qbertqbert
22.1k32561
asked Oct 1 '16 at 1:19
qbertqbert
22.1k32561
22.1k32561
$begingroup$
in which case we do not have primality of the ideal on the left. Radicals are not always prime, so that is totally expected. Why do you think otherwise? Is another assumption necessary, maybe that the Ring is Noetherian? For what? The equality? Then no.
$endgroup$
– rschwieb
Oct 1 '16 at 11:37
$begingroup$
@rschwieb I got the direction of implication mixed up, prime implies radical, but radical does not imply prime.
$endgroup$
– qbert
Oct 1 '16 at 19:37
add a comment |
$begingroup$
in which case we do not have primality of the ideal on the left. Radicals are not always prime, so that is totally expected. Why do you think otherwise? Is another assumption necessary, maybe that the Ring is Noetherian? For what? The equality? Then no.
$endgroup$
– rschwieb
Oct 1 '16 at 11:37
$begingroup$
@rschwieb I got the direction of implication mixed up, prime implies radical, but radical does not imply prime.
$endgroup$
– qbert
Oct 1 '16 at 19:37
$begingroup$
in which case we do not have primality of the ideal on the left. Radicals are not always prime, so that is totally expected. Why do you think otherwise? Is another assumption necessary, maybe that the Ring is Noetherian? For what? The equality? Then no.
$endgroup$
– rschwieb
Oct 1 '16 at 11:37
$begingroup$
in which case we do not have primality of the ideal on the left. Radicals are not always prime, so that is totally expected. Why do you think otherwise? Is another assumption necessary, maybe that the Ring is Noetherian? For what? The equality? Then no.
$endgroup$
– rschwieb
Oct 1 '16 at 11:37
$begingroup$
@rschwieb I got the direction of implication mixed up, prime implies radical, but radical does not imply prime.
$endgroup$
– qbert
Oct 1 '16 at 19:37
$begingroup$
@rschwieb I got the direction of implication mixed up, prime implies radical, but radical does not imply prime.
$endgroup$
– qbert
Oct 1 '16 at 19:37
add a comment |
1 Answer
1
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oldest
votes
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I'll let $N$ represent the intersection of all prime ideals of $R$ which contain $I$. Following through with definitions, you can prove $sqrt{I}subset N$. Proving the reverse containinment is not as simple. Given $xnotin sqrt{I}$, consider the collection $Omega$ of all ideals $J supset I$ such that for all $nin Bbb N$, $x^n notin J$. Partially order $Omega$ by inclusion, and show by Zorn's lemma that $Omega$ has a maximal element $frak{p}$. Since $mathfrak{p}supset I$, if you can show that $frak{p}$ is prime, then you can claim $xnotin N$ (since $xnotin frak{p}$). Take $a,bnotin frak{p}$, and using maximality of $mathfrak{p}$, show that $mathfrak{p} + (ab) notin Omega$; this will imply $abnotin mathfrak{p}$ and thus $mathfrak{p}$ is prime.
answered Oct 1 '16 at 4:08
kobekobe
34.9k22248
$endgroup$
$begingroup$
What justifies the application of Zorn's lemma? Why do we know that the chain of ideals must terminate?
$endgroup$
– qbert
Oct 10 '16 at 0:42
$begingroup$
Given a chain $mathcal C$ in $Omega$, the union of elements of $mathcal C$ is an upper bound of $mathcal C$ @qbert.
$endgroup$
– kobe
Oct 10 '16 at 15:03
add a comment |
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1 Answer
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1 Answer
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active
oldest
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oldest
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$begingroup$
I'll let $N$ represent the intersection of all prime ideals of $R$ which contain $I$. Following through with definitions, you can prove $sqrt{I}subset N$. Proving the reverse containinment is not as simple. Given $xnotin sqrt{I}$, consider the collection $Omega$ of all ideals $J supset I$ such that for all $nin Bbb N$, $x^n notin J$. Partially order $Omega$ by inclusion, and show by Zorn's lemma that $Omega$ has a maximal element $frak{p}$. Since $mathfrak{p}supset I$, if you can show that $frak{p}$ is prime, then you can claim $xnotin N$ (since $xnotin frak{p}$). Take $a,bnotin frak{p}$, and using maximality of $mathfrak{p}$, show that $mathfrak{p} + (ab) notin Omega$; this will imply $abnotin mathfrak{p}$ and thus $mathfrak{p}$ is prime.
answered Oct 1 '16 at 4:08
kobekobe
34.9k22248
$endgroup$
$begingroup$
What justifies the application of Zorn's lemma? Why do we know that the chain of ideals must terminate?
$endgroup$
– qbert
Oct 10 '16 at 0:42
$begingroup$
Given a chain $mathcal C$ in $Omega$, the union of elements of $mathcal C$ is an upper bound of $mathcal C$ @qbert.
$endgroup$
– kobe
Oct 10 '16 at 15:03
add a comment |
$begingroup$
I'll let $N$ represent the intersection of all prime ideals of $R$ which contain $I$. Following through with definitions, you can prove $sqrt{I}subset N$. Proving the reverse containinment is not as simple. Given $xnotin sqrt{I}$, consider the collection $Omega$ of all ideals $J supset I$ such that for all $nin Bbb N$, $x^n notin J$. Partially order $Omega$ by inclusion, and show by Zorn's lemma that $Omega$ has a maximal element $frak{p}$. Since $mathfrak{p}supset I$, if you can show that $frak{p}$ is prime, then you can claim $xnotin N$ (since $xnotin frak{p}$). Take $a,bnotin frak{p}$, and using maximality of $mathfrak{p}$, show that $mathfrak{p} + (ab) notin Omega$; this will imply $abnotin mathfrak{p}$ and thus $mathfrak{p}$ is prime.
answered Oct 1 '16 at 4:08
kobekobe
34.9k22248
$endgroup$
$begingroup$
What justifies the application of Zorn's lemma? Why do we know that the chain of ideals must terminate?
$endgroup$
– qbert
Oct 10 '16 at 0:42
$begingroup$
Given a chain $mathcal C$ in $Omega$, the union of elements of $mathcal C$ is an upper bound of $mathcal C$ @qbert.
$endgroup$
– kobe
Oct 10 '16 at 15:03
add a comment |
$begingroup$
I'll let $N$ represent the intersection of all prime ideals of $R$ which contain $I$. Following through with definitions, you can prove $sqrt{I}subset N$. Proving the reverse containinment is not as simple. Given $xnotin sqrt{I}$, consider the collection $Omega$ of all ideals $J supset I$ such that for all $nin Bbb N$, $x^n notin J$. Partially order $Omega$ by inclusion, and show by Zorn's lemma that $Omega$ has a maximal element $frak{p}$. Since $mathfrak{p}supset I$, if you can show that $frak{p}$ is prime, then you can claim $xnotin N$ (since $xnotin frak{p}$). Take $a,bnotin frak{p}$, and using maximality of $mathfrak{p}$, show that $mathfrak{p} + (ab) notin Omega$; this will imply $abnotin mathfrak{p}$ and thus $mathfrak{p}$ is prime.
answered Oct 1 '16 at 4:08
kobekobe
34.9k22248
$endgroup$
I'll let $N$ represent the intersection of all prime ideals of $R$ which contain $I$. Following through with definitions, you can prove $sqrt{I}subset N$. Proving the reverse containinment is not as simple. Given $xnotin sqrt{I}$, consider the collection $Omega$ of all ideals $J supset I$ such that for all $nin Bbb N$, $x^n notin J$. Partially order $Omega$ by inclusion, and show by Zorn's lemma that $Omega$ has a maximal element $frak{p}$. Since $mathfrak{p}supset I$, if you can show that $frak{p}$ is prime, then you can claim $xnotin N$ (since $xnotin frak{p}$). Take $a,bnotin frak{p}$, and using maximality of $mathfrak{p}$, show that $mathfrak{p} + (ab) notin Omega$; this will imply $abnotin mathfrak{p}$ and thus $mathfrak{p}$ is prime.
answered Oct 1 '16 at 4:08
kobekobe
34.9k22248
answered Oct 1 '16 at 4:08
kobekobe
34.9k22248
answered Oct 1 '16 at 4:08
kobekobe
34.9k22248
answered Oct 1 '16 at 4:08
kobekobe
34.9k22248
34.9k22248
$begingroup$
What justifies the application of Zorn's lemma? Why do we know that the chain of ideals must terminate?
$endgroup$
– qbert
Oct 10 '16 at 0:42
$begingroup$
Given a chain $mathcal C$ in $Omega$, the union of elements of $mathcal C$ is an upper bound of $mathcal C$ @qbert.
$endgroup$
– kobe
Oct 10 '16 at 15:03
add a comment |
$begingroup$
What justifies the application of Zorn's lemma? Why do we know that the chain of ideals must terminate?
$endgroup$
– qbert
Oct 10 '16 at 0:42
$begingroup$
Given a chain $mathcal C$ in $Omega$, the union of elements of $mathcal C$ is an upper bound of $mathcal C$ @qbert.
$endgroup$
– kobe
Oct 10 '16 at 15:03
$begingroup$
What justifies the application of Zorn's lemma? Why do we know that the chain of ideals must terminate?
$endgroup$
– qbert
Oct 10 '16 at 0:42
$begingroup$
What justifies the application of Zorn's lemma? Why do we know that the chain of ideals must terminate?
$endgroup$
– qbert
Oct 10 '16 at 0:42
$begingroup$
Given a chain $mathcal C$ in $Omega$, the union of elements of $mathcal C$ is an upper bound of $mathcal C$ @qbert.
$endgroup$
– kobe
Oct 10 '16 at 15:03
$begingroup$
Given a chain $mathcal C$ in $Omega$, the union of elements of $mathcal C$ is an upper bound of $mathcal C$ @qbert.
$endgroup$
– kobe
Oct 10 '16 at 15:03
add a comment |
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$begingroup$
in which case we do not have primality of the ideal on the left. Radicals are not always prime, so that is totally expected. Why do you think otherwise? Is another assumption necessary, maybe that the Ring is Noetherian? For what? The equality? Then no.
$endgroup$
– rschwieb
Oct 1 '16 at 11:37
$begingroup$
@rschwieb I got the direction of implication mixed up, prime implies radical, but radical does not imply prime.
$endgroup$
– qbert
Oct 1 '16 at 19:37