Radical of an ideal (properties)
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How do I prove that for $I$ an ideal of a ring $A$ the radical of $I$ is the intersection of all prime ideals of $A$ that are minimal among those containing $I$?
abstract-algebra ideals
asked Nov 20 at 16:19
Gentiana
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add a comment |
up vote
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How do I prove that for $I$ an ideal of a ring $A$ the radical of $I$ is the intersection of all prime ideals of $A$ that are minimal among those containing $I$?
abstract-algebra ideals
asked Nov 20 at 16:19
Gentiana
304
math.stackexchange.com/questions/1948757/…
– Paul K
Nov 20 at 16:40
Thank you very much!
– Gentiana
Nov 20 at 16:41
@PaulK That question and its solutions say nothing about the equivalence of the definition using minimal primes.
– rschwieb
Nov 20 at 17:46
@rschwieb that step is like one line.
– Paul K
Nov 20 at 18:06
@PaulK OK, yes, depending on what we're guessing the poster's definition to be, and other things known about the poset of prime ideals. You could demonstrate by including that one line in a comment, perhaps?
– rschwieb
Nov 20 at 18:14
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
How do I prove that for $I$ an ideal of a ring $A$ the radical of $I$ is the intersection of all prime ideals of $A$ that are minimal among those containing $I$?
abstract-algebra ideals
asked Nov 20 at 16:19
Gentiana
304
How do I prove that for $I$ an ideal of a ring $A$ the radical of $I$ is the intersection of all prime ideals of $A$ that are minimal among those containing $I$?
abstract-algebra ideals
abstract-algebra ideals
asked Nov 20 at 16:19
Gentiana
304
asked Nov 20 at 16:19
Gentiana
304
asked Nov 20 at 16:19
Gentiana
304
asked Nov 20 at 16:19
Gentiana
304
asked Nov 20 at 16:19
Gentiana
304
304
math.stackexchange.com/questions/1948757/…
– Paul K
Nov 20 at 16:40
Thank you very much!
– Gentiana
Nov 20 at 16:41
@PaulK That question and its solutions say nothing about the equivalence of the definition using minimal primes.
– rschwieb
Nov 20 at 17:46
@rschwieb that step is like one line.
– Paul K
Nov 20 at 18:06
@PaulK OK, yes, depending on what we're guessing the poster's definition to be, and other things known about the poset of prime ideals. You could demonstrate by including that one line in a comment, perhaps?
– rschwieb
Nov 20 at 18:14
add a comment |
math.stackexchange.com/questions/1948757/…
– Paul K
Nov 20 at 16:40
Thank you very much!
– Gentiana
Nov 20 at 16:41
@PaulK That question and its solutions say nothing about the equivalence of the definition using minimal primes.
– rschwieb
Nov 20 at 17:46
@rschwieb that step is like one line.
– Paul K
Nov 20 at 18:06
@PaulK OK, yes, depending on what we're guessing the poster's definition to be, and other things known about the poset of prime ideals. You could demonstrate by including that one line in a comment, perhaps?
– rschwieb
Nov 20 at 18:14
math.stackexchange.com/questions/1948757/…
– Paul K
Nov 20 at 16:40
math.stackexchange.com/questions/1948757/…
– Paul K
Nov 20 at 16:40
Thank you very much!
– Gentiana
Nov 20 at 16:41
Thank you very much!
– Gentiana
Nov 20 at 16:41
@PaulK That question and its solutions say nothing about the equivalence of the definition using minimal primes.
– rschwieb
Nov 20 at 17:46
@PaulK That question and its solutions say nothing about the equivalence of the definition using minimal primes.
– rschwieb
Nov 20 at 17:46
@rschwieb that step is like one line.
– Paul K
Nov 20 at 18:06
@rschwieb that step is like one line.
– Paul K
Nov 20 at 18:06
@PaulK OK, yes, depending on what we're guessing the poster's definition to be, and other things known about the poset of prime ideals. You could demonstrate by including that one line in a comment, perhaps?
– rschwieb
Nov 20 at 18:14
@PaulK OK, yes, depending on what we're guessing the poster's definition to be, and other things known about the poset of prime ideals. You could demonstrate by including that one line in a comment, perhaps?
– rschwieb
Nov 20 at 18:14
add a comment |
2 Answers
2
active
oldest
votes
up vote
1
down vote
If you're don't already know that the radical of $I$ is the intersection of all prime ideals over $I$, then see this and come back.
The containment $bigcap_{Ptext{ prime over } I}Psubseteq bigcap_{Ptext{ prime, minimal over } I}P$ is trivial.
Suppose the containment was strict, i.e. that there exists an element $xin bigcap_{Ptext{ prime, minimal over }I}P$ that fails to be in some prime ideal $Q$ containing $I$.
So the next step that suggests itself would be to demonstrate that $Q$ has to contain a minimal prime ideal $Q'$ of $A$ that contains $I$, and then we would have arrived at a contradiction (since $xin Q'subseteq Q$.)
There is probably more than one way to do that, but here's one way:
There is a well-known exercise that every ring with identity contains minimal primes. To prove it, one usually notes that Zorn's Lemma applies to the poset of prime ideals (order by reverse-inclusion). Now, you can repeat the argument with the poset of prime ideals between $I$ and $Q$, including $Q$, and you will know there is a prime ideal of $A$ within $Q$ and minimal over $I$.
answered Nov 20 at 18:12
rschwieb
104k1299238
Another way would be to localize at Q and then use the usual existence of minimal prime ideals and the correspondence between prime ideals in a ring and its localization!
– Paul K
Nov 20 at 18:22
1
@PaulK Indeed, although the explanation somewhat complicated by $I$. Perhaps it would be even better to start afresh and say "WLOG we replace $A$ with $A/I$." and that would pave the way for the second explanation to be easier.
– rschwieb
Nov 20 at 18:30
add a comment |
up vote
0
down vote
I will give a less elementary proof then the proof given in the link above.
The other answer shows how to obtain the equality of the intersection of all primes containing $I$ and the intersection of minimal primes containing $I$. (A full proof is given here: Existence of minimal prime ideal contained in given prime ideal and containing a given subset)
So, let now $I$ be an ideal. By looking at $A / I$ and observing that the nilradical, i.e. the radical of $(0)$, corresponds to the radical of $I$. So it is sufficient to prove the assertion for $I = (0)$. The direction $text{rad}(I) subseteq bigcap_{P supseteq I} I$ is easy to see. For the converse direction let $x notin text{rad}(I)$. Then
$$S = {x^n mid n geq 0}$$
is multiplicative and we can look at the localization $A_x = S^{-1} A$. For localizations there is a $1:1$-correspondence between prime ideals in $S^{-1} A$ and prime ideals which have empty intersection with $S$. Now, by existence of maximal ideals, $A_x$ has a maximal ideal which then corresponds to a prime ideal $P$ in $A$ which has empty intersection with $S$, in particular $x notin P$. Therefore $x notin bigcap_{P supseteq I} P$.
answered Nov 21 at 8:10
Paul K
2,685315
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
If you're don't already know that the radical of $I$ is the intersection of all prime ideals over $I$, then see this and come back.
The containment $bigcap_{Ptext{ prime over } I}Psubseteq bigcap_{Ptext{ prime, minimal over } I}P$ is trivial.
Suppose the containment was strict, i.e. that there exists an element $xin bigcap_{Ptext{ prime, minimal over }I}P$ that fails to be in some prime ideal $Q$ containing $I$.
So the next step that suggests itself would be to demonstrate that $Q$ has to contain a minimal prime ideal $Q'$ of $A$ that contains $I$, and then we would have arrived at a contradiction (since $xin Q'subseteq Q$.)
There is probably more than one way to do that, but here's one way:
There is a well-known exercise that every ring with identity contains minimal primes. To prove it, one usually notes that Zorn's Lemma applies to the poset of prime ideals (order by reverse-inclusion). Now, you can repeat the argument with the poset of prime ideals between $I$ and $Q$, including $Q$, and you will know there is a prime ideal of $A$ within $Q$ and minimal over $I$.
answered Nov 20 at 18:12
rschwieb
104k1299238
Another way would be to localize at Q and then use the usual existence of minimal prime ideals and the correspondence between prime ideals in a ring and its localization!
– Paul K
Nov 20 at 18:22
1
@PaulK Indeed, although the explanation somewhat complicated by $I$. Perhaps it would be even better to start afresh and say "WLOG we replace $A$ with $A/I$." and that would pave the way for the second explanation to be easier.
– rschwieb
Nov 20 at 18:30
add a comment |
up vote
1
down vote
If you're don't already know that the radical of $I$ is the intersection of all prime ideals over $I$, then see this and come back.
The containment $bigcap_{Ptext{ prime over } I}Psubseteq bigcap_{Ptext{ prime, minimal over } I}P$ is trivial.
Suppose the containment was strict, i.e. that there exists an element $xin bigcap_{Ptext{ prime, minimal over }I}P$ that fails to be in some prime ideal $Q$ containing $I$.
So the next step that suggests itself would be to demonstrate that $Q$ has to contain a minimal prime ideal $Q'$ of $A$ that contains $I$, and then we would have arrived at a contradiction (since $xin Q'subseteq Q$.)
There is probably more than one way to do that, but here's one way:
There is a well-known exercise that every ring with identity contains minimal primes. To prove it, one usually notes that Zorn's Lemma applies to the poset of prime ideals (order by reverse-inclusion). Now, you can repeat the argument with the poset of prime ideals between $I$ and $Q$, including $Q$, and you will know there is a prime ideal of $A$ within $Q$ and minimal over $I$.
answered Nov 20 at 18:12
rschwieb
104k1299238
Another way would be to localize at Q and then use the usual existence of minimal prime ideals and the correspondence between prime ideals in a ring and its localization!
– Paul K
Nov 20 at 18:22
1
@PaulK Indeed, although the explanation somewhat complicated by $I$. Perhaps it would be even better to start afresh and say "WLOG we replace $A$ with $A/I$." and that would pave the way for the second explanation to be easier.
– rschwieb
Nov 20 at 18:30
add a comment |
up vote
1
down vote
up vote
1
down vote
If you're don't already know that the radical of $I$ is the intersection of all prime ideals over $I$, then see this and come back.
The containment $bigcap_{Ptext{ prime over } I}Psubseteq bigcap_{Ptext{ prime, minimal over } I}P$ is trivial.
Suppose the containment was strict, i.e. that there exists an element $xin bigcap_{Ptext{ prime, minimal over }I}P$ that fails to be in some prime ideal $Q$ containing $I$.
So the next step that suggests itself would be to demonstrate that $Q$ has to contain a minimal prime ideal $Q'$ of $A$ that contains $I$, and then we would have arrived at a contradiction (since $xin Q'subseteq Q$.)
There is probably more than one way to do that, but here's one way:
There is a well-known exercise that every ring with identity contains minimal primes. To prove it, one usually notes that Zorn's Lemma applies to the poset of prime ideals (order by reverse-inclusion). Now, you can repeat the argument with the poset of prime ideals between $I$ and $Q$, including $Q$, and you will know there is a prime ideal of $A$ within $Q$ and minimal over $I$.
answered Nov 20 at 18:12
rschwieb
104k1299238
If you're don't already know that the radical of $I$ is the intersection of all prime ideals over $I$, then see this and come back.
The containment $bigcap_{Ptext{ prime over } I}Psubseteq bigcap_{Ptext{ prime, minimal over } I}P$ is trivial.
Suppose the containment was strict, i.e. that there exists an element $xin bigcap_{Ptext{ prime, minimal over }I}P$ that fails to be in some prime ideal $Q$ containing $I$.
So the next step that suggests itself would be to demonstrate that $Q$ has to contain a minimal prime ideal $Q'$ of $A$ that contains $I$, and then we would have arrived at a contradiction (since $xin Q'subseteq Q$.)
There is probably more than one way to do that, but here's one way:
There is a well-known exercise that every ring with identity contains minimal primes. To prove it, one usually notes that Zorn's Lemma applies to the poset of prime ideals (order by reverse-inclusion). Now, you can repeat the argument with the poset of prime ideals between $I$ and $Q$, including $Q$, and you will know there is a prime ideal of $A$ within $Q$ and minimal over $I$.
answered Nov 20 at 18:12
rschwieb
104k1299238
answered Nov 20 at 18:12
rschwieb
104k1299238
answered Nov 20 at 18:12
rschwieb
104k1299238
answered Nov 20 at 18:12
rschwieb
104k1299238
104k1299238
Another way would be to localize at Q and then use the usual existence of minimal prime ideals and the correspondence between prime ideals in a ring and its localization!
– Paul K
Nov 20 at 18:22
1
@PaulK Indeed, although the explanation somewhat complicated by $I$. Perhaps it would be even better to start afresh and say "WLOG we replace $A$ with $A/I$." and that would pave the way for the second explanation to be easier.
– rschwieb
Nov 20 at 18:30
add a comment |
Another way would be to localize at Q and then use the usual existence of minimal prime ideals and the correspondence between prime ideals in a ring and its localization!
– Paul K
Nov 20 at 18:22
1
@PaulK Indeed, although the explanation somewhat complicated by $I$. Perhaps it would be even better to start afresh and say "WLOG we replace $A$ with $A/I$." and that would pave the way for the second explanation to be easier.
– rschwieb
Nov 20 at 18:30
Another way would be to localize at Q and then use the usual existence of minimal prime ideals and the correspondence between prime ideals in a ring and its localization!
– Paul K
Nov 20 at 18:22
Another way would be to localize at Q and then use the usual existence of minimal prime ideals and the correspondence between prime ideals in a ring and its localization!
– Paul K
Nov 20 at 18:22
1
1
@PaulK Indeed, although the explanation somewhat complicated by $I$. Perhaps it would be even better to start afresh and say "WLOG we replace $A$ with $A/I$." and that would pave the way for the second explanation to be easier.
– rschwieb
Nov 20 at 18:30
@PaulK Indeed, although the explanation somewhat complicated by $I$. Perhaps it would be even better to start afresh and say "WLOG we replace $A$ with $A/I$." and that would pave the way for the second explanation to be easier.
– rschwieb
Nov 20 at 18:30
add a comment |
up vote
0
down vote
I will give a less elementary proof then the proof given in the link above.
The other answer shows how to obtain the equality of the intersection of all primes containing $I$ and the intersection of minimal primes containing $I$. (A full proof is given here: Existence of minimal prime ideal contained in given prime ideal and containing a given subset)
So, let now $I$ be an ideal. By looking at $A / I$ and observing that the nilradical, i.e. the radical of $(0)$, corresponds to the radical of $I$. So it is sufficient to prove the assertion for $I = (0)$. The direction $text{rad}(I) subseteq bigcap_{P supseteq I} I$ is easy to see. For the converse direction let $x notin text{rad}(I)$. Then
$$S = {x^n mid n geq 0}$$
is multiplicative and we can look at the localization $A_x = S^{-1} A$. For localizations there is a $1:1$-correspondence between prime ideals in $S^{-1} A$ and prime ideals which have empty intersection with $S$. Now, by existence of maximal ideals, $A_x$ has a maximal ideal which then corresponds to a prime ideal $P$ in $A$ which has empty intersection with $S$, in particular $x notin P$. Therefore $x notin bigcap_{P supseteq I} P$.
answered Nov 21 at 8:10
Paul K
2,685315
add a comment |
up vote
0
down vote
I will give a less elementary proof then the proof given in the link above.
The other answer shows how to obtain the equality of the intersection of all primes containing $I$ and the intersection of minimal primes containing $I$. (A full proof is given here: Existence of minimal prime ideal contained in given prime ideal and containing a given subset)
So, let now $I$ be an ideal. By looking at $A / I$ and observing that the nilradical, i.e. the radical of $(0)$, corresponds to the radical of $I$. So it is sufficient to prove the assertion for $I = (0)$. The direction $text{rad}(I) subseteq bigcap_{P supseteq I} I$ is easy to see. For the converse direction let $x notin text{rad}(I)$. Then
$$S = {x^n mid n geq 0}$$
is multiplicative and we can look at the localization $A_x = S^{-1} A$. For localizations there is a $1:1$-correspondence between prime ideals in $S^{-1} A$ and prime ideals which have empty intersection with $S$. Now, by existence of maximal ideals, $A_x$ has a maximal ideal which then corresponds to a prime ideal $P$ in $A$ which has empty intersection with $S$, in particular $x notin P$. Therefore $x notin bigcap_{P supseteq I} P$.
answered Nov 21 at 8:10
Paul K
2,685315
add a comment |
up vote
0
down vote
up vote
0
down vote
I will give a less elementary proof then the proof given in the link above.
The other answer shows how to obtain the equality of the intersection of all primes containing $I$ and the intersection of minimal primes containing $I$. (A full proof is given here: Existence of minimal prime ideal contained in given prime ideal and containing a given subset)
So, let now $I$ be an ideal. By looking at $A / I$ and observing that the nilradical, i.e. the radical of $(0)$, corresponds to the radical of $I$. So it is sufficient to prove the assertion for $I = (0)$. The direction $text{rad}(I) subseteq bigcap_{P supseteq I} I$ is easy to see. For the converse direction let $x notin text{rad}(I)$. Then
$$S = {x^n mid n geq 0}$$
is multiplicative and we can look at the localization $A_x = S^{-1} A$. For localizations there is a $1:1$-correspondence between prime ideals in $S^{-1} A$ and prime ideals which have empty intersection with $S$. Now, by existence of maximal ideals, $A_x$ has a maximal ideal which then corresponds to a prime ideal $P$ in $A$ which has empty intersection with $S$, in particular $x notin P$. Therefore $x notin bigcap_{P supseteq I} P$.
answered Nov 21 at 8:10
Paul K
2,685315
I will give a less elementary proof then the proof given in the link above.
The other answer shows how to obtain the equality of the intersection of all primes containing $I$ and the intersection of minimal primes containing $I$. (A full proof is given here: Existence of minimal prime ideal contained in given prime ideal and containing a given subset)
So, let now $I$ be an ideal. By looking at $A / I$ and observing that the nilradical, i.e. the radical of $(0)$, corresponds to the radical of $I$. So it is sufficient to prove the assertion for $I = (0)$. The direction $text{rad}(I) subseteq bigcap_{P supseteq I} I$ is easy to see. For the converse direction let $x notin text{rad}(I)$. Then
$$S = {x^n mid n geq 0}$$
is multiplicative and we can look at the localization $A_x = S^{-1} A$. For localizations there is a $1:1$-correspondence between prime ideals in $S^{-1} A$ and prime ideals which have empty intersection with $S$. Now, by existence of maximal ideals, $A_x$ has a maximal ideal which then corresponds to a prime ideal $P$ in $A$ which has empty intersection with $S$, in particular $x notin P$. Therefore $x notin bigcap_{P supseteq I} P$.
answered Nov 21 at 8:10
Paul K
2,685315
edited Nov 21 at 8:20
answered Nov 21 at 8:10
Paul K
2,685315
answered Nov 21 at 8:10
Paul K
2,685315
answered Nov 21 at 8:10
Paul K
2,685315
2,685315
add a comment |
add a comment |
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math.stackexchange.com/questions/1948757/…
– Paul K
Nov 20 at 16:40
Thank you very much!
– Gentiana
Nov 20 at 16:41
@PaulK That question and its solutions say nothing about the equivalence of the definition using minimal primes.
– rschwieb
Nov 20 at 17:46
@rschwieb that step is like one line.
– Paul K
Nov 20 at 18:06
@PaulK OK, yes, depending on what we're guessing the poster's definition to be, and other things known about the poset of prime ideals. You could demonstrate by including that one line in a comment, perhaps?
– rschwieb
Nov 20 at 18:14