Nil-Radical equals Jacobson Radical even though not every prime ideal is maximal?If nilradical is equal to Jacobson radical, does every prime ideal be maximal?Examples of a commutative ring without an identity in which a maximal ideal is not a prime idealjudge if nilradical equals jacobson radicalOne-sided nilpotent ideal not in the Jacobson radical?Every prime is maximal in a Jacobson ring?The Jacobson radical of an IdealIs there an adjective for rings whose every non-zero prime ideal is maximal?A prime ideal which is not maximalJacobson radical and prime idealEvery maximal ideal is a prime idealA one-dimensional domain with infinitely many maximal ideals and non-zero Jacobson radical?
Justification failure in beamer enumerate list
Pre-Employment Background Check With Consent For Future Checks
Imaginary part of expression too difficult to calculate
Single word to change groups
When should a starting writer get his own webpage?
How can I create URL shortcuts/redirects for task/diff IDs in Phabricator?
Hackerrank All Women's Codesprint 2019: Name the Product
Have any astronauts/cosmonauts died in space?
Print a physical multiplication table
Do people actually use the word "kaputt" in conversation?
Do native speakers use "ultima" and "proxima" frequently in spoken English?
is this saw blade faulty?
Determine voltage drop over 10G resistors with cheap multimeter
Turning a hard to access nut?
What is it called when someone votes for an option that's not their first choice?
PTIJ: Which Dr. Seuss books should one obtain?
Why is "la Gestapo" feminine?
Air travel with refrigerated insulin
What are the rules for concealing thieves' tools (or items in general)?
Do I need to convey a moral for each of my blog post?
Logic with "co-relations" - sources?
Does convergence of polynomials imply that of its coefficients?
What is the difference between something being completely legal and being completely decriminalized?
Is "inadequate referencing" a euphemism for plagiarism?
Nil-Radical equals Jacobson Radical even though not every prime ideal is maximal?
If nilradical is equal to Jacobson radical, does every prime ideal be maximal?Examples of a commutative ring without an identity in which a maximal ideal is not a prime idealjudge if nilradical equals jacobson radicalOne-sided nilpotent ideal not in the Jacobson radical?Every prime is maximal in a Jacobson ring?The Jacobson radical of an IdealIs there an adjective for rings whose every non-zero prime ideal is maximal?A prime ideal which is not maximalJacobson radical and prime idealEvery maximal ideal is a prime idealA one-dimensional domain with infinitely many maximal ideals and non-zero Jacobson radical?
$begingroup$
Let's assume we have a commutative ring with identity. Can the Nil-Radical and the Jacobson Radical be equal in a non-trivial case (i.e. not every nonzero prime ideal in said ring is maximal)?
Are there any interesting examples of this case?
abstract-algebra commutative-algebra ring-theory
asked Oct 26 '10 at 19:33
PandoraPandora
2,70113157
$endgroup$
|
show 1 more comment
$begingroup$
Let's assume we have a commutative ring with identity. Can the Nil-Radical and the Jacobson Radical be equal in a non-trivial case (i.e. not every nonzero prime ideal in said ring is maximal)?
Are there any interesting examples of this case?
abstract-algebra commutative-algebra ring-theory
asked Oct 26 '10 at 19:33
PandoraPandora
2,70113157
$endgroup$
$begingroup$
@Natalia S: You tagged the question "commutative algebra", but you might want to be explicit in the question on whether you are considering only commutative rings or not (and while you are at it, whether your rings must have a unity or not).
$endgroup$
– Arturo Magidin
Oct 26 '10 at 19:36
4
$begingroup$
Since the concept of prime ideal is being used, it seems likely that the author of the question has the commutative case in mind. In this case, there is an important class of rings (including in particular the examples of Arturo Magidin's answer) for which the answer is yes, namely Jacobson rings: en.wikipedia.org/wiki/Jacobson_ring As that wikipedia entry (and Arturo Magidin's answer) make clear, this question is rather closely bound up with the ideas behind Hilbert's Nullstellensatz.
$endgroup$
– Matt E
Oct 26 '10 at 20:03
$begingroup$
@Arturo Magidin: Thanks for your comment, I edited the question.
$endgroup$
– Pandora
Oct 26 '10 at 20:03
$begingroup$
@Natalia S: I also added "nonzero"; otherwise, $mathbbZ$ itself would be an example.
$endgroup$
– Arturo Magidin
Oct 26 '10 at 20:06
3
$begingroup$
For any commutative ring R, R[x] has this property. This is one of the first exercises in Atiyah-Macdonald.
$endgroup$
– user641
Oct 27 '10 at 3:58
|
show 1 more comment
$begingroup$
Let's assume we have a commutative ring with identity. Can the Nil-Radical and the Jacobson Radical be equal in a non-trivial case (i.e. not every nonzero prime ideal in said ring is maximal)?
Are there any interesting examples of this case?
abstract-algebra commutative-algebra ring-theory
asked Oct 26 '10 at 19:33
PandoraPandora
2,70113157
$endgroup$
Let's assume we have a commutative ring with identity. Can the Nil-Radical and the Jacobson Radical be equal in a non-trivial case (i.e. not every nonzero prime ideal in said ring is maximal)?
Are there any interesting examples of this case?
abstract-algebra commutative-algebra ring-theory
abstract-algebra commutative-algebra ring-theory
asked Oct 26 '10 at 19:33
PandoraPandora
2,70113157
asked Oct 26 '10 at 19:33
PandoraPandora
2,70113157
edited Oct 26 '10 at 20:03
Pandora
asked Oct 26 '10 at 19:33
PandoraPandora
2,70113157
asked Oct 26 '10 at 19:33
PandoraPandora
2,70113157
asked Oct 26 '10 at 19:33
PandoraPandora
2,70113157
2,70113157
$begingroup$
@Natalia S: You tagged the question "commutative algebra", but you might want to be explicit in the question on whether you are considering only commutative rings or not (and while you are at it, whether your rings must have a unity or not).
$endgroup$
– Arturo Magidin
Oct 26 '10 at 19:36
4
$begingroup$
Since the concept of prime ideal is being used, it seems likely that the author of the question has the commutative case in mind. In this case, there is an important class of rings (including in particular the examples of Arturo Magidin's answer) for which the answer is yes, namely Jacobson rings: en.wikipedia.org/wiki/Jacobson_ring As that wikipedia entry (and Arturo Magidin's answer) make clear, this question is rather closely bound up with the ideas behind Hilbert's Nullstellensatz.
$endgroup$
– Matt E
Oct 26 '10 at 20:03
$begingroup$
@Arturo Magidin: Thanks for your comment, I edited the question.
$endgroup$
– Pandora
Oct 26 '10 at 20:03
$begingroup$
@Natalia S: I also added "nonzero"; otherwise, $mathbbZ$ itself would be an example.
$endgroup$
– Arturo Magidin
Oct 26 '10 at 20:06
3
$begingroup$
For any commutative ring R, R[x] has this property. This is one of the first exercises in Atiyah-Macdonald.
$endgroup$
– user641
Oct 27 '10 at 3:58
|
show 1 more comment
$begingroup$
@Natalia S: You tagged the question "commutative algebra", but you might want to be explicit in the question on whether you are considering only commutative rings or not (and while you are at it, whether your rings must have a unity or not).
$endgroup$
– Arturo Magidin
Oct 26 '10 at 19:36
4
$begingroup$
Since the concept of prime ideal is being used, it seems likely that the author of the question has the commutative case in mind. In this case, there is an important class of rings (including in particular the examples of Arturo Magidin's answer) for which the answer is yes, namely Jacobson rings: en.wikipedia.org/wiki/Jacobson_ring As that wikipedia entry (and Arturo Magidin's answer) make clear, this question is rather closely bound up with the ideas behind Hilbert's Nullstellensatz.
$endgroup$
– Matt E
Oct 26 '10 at 20:03
$begingroup$
@Arturo Magidin: Thanks for your comment, I edited the question.
$endgroup$
– Pandora
Oct 26 '10 at 20:03
$begingroup$
@Natalia S: I also added "nonzero"; otherwise, $mathbbZ$ itself would be an example.
$endgroup$
– Arturo Magidin
Oct 26 '10 at 20:06
3
$begingroup$
For any commutative ring R, R[x] has this property. This is one of the first exercises in Atiyah-Macdonald.
$endgroup$
– user641
Oct 27 '10 at 3:58
$begingroup$
@Natalia S: You tagged the question "commutative algebra", but you might want to be explicit in the question on whether you are considering only commutative rings or not (and while you are at it, whether your rings must have a unity or not).
$endgroup$
– Arturo Magidin
Oct 26 '10 at 19:36
$begingroup$
@Natalia S: You tagged the question "commutative algebra", but you might want to be explicit in the question on whether you are considering only commutative rings or not (and while you are at it, whether your rings must have a unity or not).
$endgroup$
– Arturo Magidin
Oct 26 '10 at 19:36
4
4
$begingroup$
Since the concept of prime ideal is being used, it seems likely that the author of the question has the commutative case in mind. In this case, there is an important class of rings (including in particular the examples of Arturo Magidin's answer) for which the answer is yes, namely Jacobson rings: en.wikipedia.org/wiki/Jacobson_ring As that wikipedia entry (and Arturo Magidin's answer) make clear, this question is rather closely bound up with the ideas behind Hilbert's Nullstellensatz.
$endgroup$
– Matt E
Oct 26 '10 at 20:03
$begingroup$
Since the concept of prime ideal is being used, it seems likely that the author of the question has the commutative case in mind. In this case, there is an important class of rings (including in particular the examples of Arturo Magidin's answer) for which the answer is yes, namely Jacobson rings: en.wikipedia.org/wiki/Jacobson_ring As that wikipedia entry (and Arturo Magidin's answer) make clear, this question is rather closely bound up with the ideas behind Hilbert's Nullstellensatz.
$endgroup$
– Matt E
Oct 26 '10 at 20:03
$begingroup$
@Arturo Magidin: Thanks for your comment, I edited the question.
$endgroup$
– Pandora
Oct 26 '10 at 20:03
$begingroup$
@Arturo Magidin: Thanks for your comment, I edited the question.
$endgroup$
– Pandora
Oct 26 '10 at 20:03
$begingroup$
@Natalia S: I also added "nonzero"; otherwise, $mathbbZ$ itself would be an example.
$endgroup$
– Arturo Magidin
Oct 26 '10 at 20:06
$begingroup$
@Natalia S: I also added "nonzero"; otherwise, $mathbbZ$ itself would be an example.
$endgroup$
– Arturo Magidin
Oct 26 '10 at 20:06
3
3
$begingroup$
For any commutative ring R, R[x] has this property. This is one of the first exercises in Atiyah-Macdonald.
$endgroup$
– user641
Oct 27 '10 at 3:58
$begingroup$
For any commutative ring R, R[x] has this property. This is one of the first exercises in Atiyah-Macdonald.
$endgroup$
– user641
Oct 27 '10 at 3:58
|
show 1 more comment
3 Answers
3
active
oldest
votes
$begingroup$
There are indeed very many rings in which the nilradical equals the Jacobson radical.
Consider the following property of a ring: every prime ideal is the intersection of the maximal ideals containing it. These rings are called Hilbert-Jacobson rings. In a Hilbert-Jacobson ring, the intersection of all the prime ideals is therefore also the intersection of all maximal ideals, i.e., the nilradical and Jacobson radical coincide.
And indeed there are very many Hilbert-Jacobson rings. One useful result in this direction is:
Theorem: Let $R$ be a Hilbert Jacobson ring, and let $S$ be a ring which is finitely generated as an $R$-algebra. Then $S$ is also a Hilbert-Jacobson ring.
Since any field is trivially a Hilbert-Jacobson ring, it follows that all algebras which are finitely generated over a field are Hilbert-Jacobson, as Arturo Magidin pointed out in his answer.
For some information on this subject, including a proof of the theorem, see these notes.
answered Oct 26 '10 at 21:29
Pete L. ClarkPete L. Clark
80.8k9162314
$endgroup$
add a comment |
$begingroup$
Let $k$ be a field, and let $R=k[x_1,ldots,x_n]/mathfrakI$, where $mathfrakI$ is an ideal of $k[x_1,ldots,x_n]$.
The nilradical of $R$ is $sqrtmathfrakI/mathfrakI$, and the Jacobson radical of $R$ is $mathfrakM/mathfrakI$, where $mathfrakM$ is the intersection of all maximal ideals of $k[x_1,ldots,x_n]$ that contain $mathfrakI$.
But by Hilbert's Nullstellensatz, $sqrtmathfrakI$, the radical of $mathfrakI$, is the intersection of all maximal ideals of $k[x_1,ldots,x_n]$ that contain $mathfrakI$, so the nilradical of $R$ equals the Jacobson radical of $R$.
However, in these rings there are generally prime ideals that are not maximal. For example, if $ngeq 3$ and you take $mathfrakI=(x_1)$, then $(x_1,x_2)+mathfrakI$ is prime but not maximal.
answered Oct 26 '10 at 19:49
Arturo MagidinArturo Magidin
265k34590919
$endgroup$
add a comment |
$begingroup$
Theorem 5.1 in T.Y. Lam's book "A First Course in Noncommutative Rings" states that every polynomial ring $R[T]$ over a commutative ring $R$ satisfies
$$rad ,, R[T] = Nil(R[T]) = (Nil ,,R)[T]$$
answered Apr 23 '12 at 10:14
CihanCihan
1,682916
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f7954%2fnil-radical-equals-jacobson-radical-even-though-not-every-prime-ideal-is-maximal%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
There are indeed very many rings in which the nilradical equals the Jacobson radical.
Consider the following property of a ring: every prime ideal is the intersection of the maximal ideals containing it. These rings are called Hilbert-Jacobson rings. In a Hilbert-Jacobson ring, the intersection of all the prime ideals is therefore also the intersection of all maximal ideals, i.e., the nilradical and Jacobson radical coincide.
And indeed there are very many Hilbert-Jacobson rings. One useful result in this direction is:
Theorem: Let $R$ be a Hilbert Jacobson ring, and let $S$ be a ring which is finitely generated as an $R$-algebra. Then $S$ is also a Hilbert-Jacobson ring.
Since any field is trivially a Hilbert-Jacobson ring, it follows that all algebras which are finitely generated over a field are Hilbert-Jacobson, as Arturo Magidin pointed out in his answer.
For some information on this subject, including a proof of the theorem, see these notes.
answered Oct 26 '10 at 21:29
Pete L. ClarkPete L. Clark
80.8k9162314
$endgroup$
add a comment |
$begingroup$
There are indeed very many rings in which the nilradical equals the Jacobson radical.
Consider the following property of a ring: every prime ideal is the intersection of the maximal ideals containing it. These rings are called Hilbert-Jacobson rings. In a Hilbert-Jacobson ring, the intersection of all the prime ideals is therefore also the intersection of all maximal ideals, i.e., the nilradical and Jacobson radical coincide.
And indeed there are very many Hilbert-Jacobson rings. One useful result in this direction is:
Theorem: Let $R$ be a Hilbert Jacobson ring, and let $S$ be a ring which is finitely generated as an $R$-algebra. Then $S$ is also a Hilbert-Jacobson ring.
Since any field is trivially a Hilbert-Jacobson ring, it follows that all algebras which are finitely generated over a field are Hilbert-Jacobson, as Arturo Magidin pointed out in his answer.
For some information on this subject, including a proof of the theorem, see these notes.
answered Oct 26 '10 at 21:29
Pete L. ClarkPete L. Clark
80.8k9162314
$endgroup$
add a comment |
$begingroup$
There are indeed very many rings in which the nilradical equals the Jacobson radical.
Consider the following property of a ring: every prime ideal is the intersection of the maximal ideals containing it. These rings are called Hilbert-Jacobson rings. In a Hilbert-Jacobson ring, the intersection of all the prime ideals is therefore also the intersection of all maximal ideals, i.e., the nilradical and Jacobson radical coincide.
And indeed there are very many Hilbert-Jacobson rings. One useful result in this direction is:
Theorem: Let $R$ be a Hilbert Jacobson ring, and let $S$ be a ring which is finitely generated as an $R$-algebra. Then $S$ is also a Hilbert-Jacobson ring.
Since any field is trivially a Hilbert-Jacobson ring, it follows that all algebras which are finitely generated over a field are Hilbert-Jacobson, as Arturo Magidin pointed out in his answer.
For some information on this subject, including a proof of the theorem, see these notes.
answered Oct 26 '10 at 21:29
Pete L. ClarkPete L. Clark
80.8k9162314
$endgroup$
There are indeed very many rings in which the nilradical equals the Jacobson radical.
Consider the following property of a ring: every prime ideal is the intersection of the maximal ideals containing it. These rings are called Hilbert-Jacobson rings. In a Hilbert-Jacobson ring, the intersection of all the prime ideals is therefore also the intersection of all maximal ideals, i.e., the nilradical and Jacobson radical coincide.
And indeed there are very many Hilbert-Jacobson rings. One useful result in this direction is:
Theorem: Let $R$ be a Hilbert Jacobson ring, and let $S$ be a ring which is finitely generated as an $R$-algebra. Then $S$ is also a Hilbert-Jacobson ring.
Since any field is trivially a Hilbert-Jacobson ring, it follows that all algebras which are finitely generated over a field are Hilbert-Jacobson, as Arturo Magidin pointed out in his answer.
For some information on this subject, including a proof of the theorem, see these notes.
answered Oct 26 '10 at 21:29
Pete L. ClarkPete L. Clark
80.8k9162314
edited Apr 11 '15 at 16:28
answered Oct 26 '10 at 21:29
Pete L. ClarkPete L. Clark
80.8k9162314
answered Oct 26 '10 at 21:29
Pete L. ClarkPete L. Clark
80.8k9162314
answered Oct 26 '10 at 21:29
Pete L. ClarkPete L. Clark
80.8k9162314
80.8k9162314
add a comment |
add a comment |
$begingroup$
Let $k$ be a field, and let $R=k[x_1,ldots,x_n]/mathfrakI$, where $mathfrakI$ is an ideal of $k[x_1,ldots,x_n]$.
The nilradical of $R$ is $sqrtmathfrakI/mathfrakI$, and the Jacobson radical of $R$ is $mathfrakM/mathfrakI$, where $mathfrakM$ is the intersection of all maximal ideals of $k[x_1,ldots,x_n]$ that contain $mathfrakI$.
But by Hilbert's Nullstellensatz, $sqrtmathfrakI$, the radical of $mathfrakI$, is the intersection of all maximal ideals of $k[x_1,ldots,x_n]$ that contain $mathfrakI$, so the nilradical of $R$ equals the Jacobson radical of $R$.
However, in these rings there are generally prime ideals that are not maximal. For example, if $ngeq 3$ and you take $mathfrakI=(x_1)$, then $(x_1,x_2)+mathfrakI$ is prime but not maximal.
answered Oct 26 '10 at 19:49
Arturo MagidinArturo Magidin
265k34590919
$endgroup$
add a comment |
$begingroup$
Let $k$ be a field, and let $R=k[x_1,ldots,x_n]/mathfrakI$, where $mathfrakI$ is an ideal of $k[x_1,ldots,x_n]$.
The nilradical of $R$ is $sqrtmathfrakI/mathfrakI$, and the Jacobson radical of $R$ is $mathfrakM/mathfrakI$, where $mathfrakM$ is the intersection of all maximal ideals of $k[x_1,ldots,x_n]$ that contain $mathfrakI$.
But by Hilbert's Nullstellensatz, $sqrtmathfrakI$, the radical of $mathfrakI$, is the intersection of all maximal ideals of $k[x_1,ldots,x_n]$ that contain $mathfrakI$, so the nilradical of $R$ equals the Jacobson radical of $R$.
However, in these rings there are generally prime ideals that are not maximal. For example, if $ngeq 3$ and you take $mathfrakI=(x_1)$, then $(x_1,x_2)+mathfrakI$ is prime but not maximal.
answered Oct 26 '10 at 19:49
Arturo MagidinArturo Magidin
265k34590919
$endgroup$
add a comment |
$begingroup$
Let $k$ be a field, and let $R=k[x_1,ldots,x_n]/mathfrakI$, where $mathfrakI$ is an ideal of $k[x_1,ldots,x_n]$.
The nilradical of $R$ is $sqrtmathfrakI/mathfrakI$, and the Jacobson radical of $R$ is $mathfrakM/mathfrakI$, where $mathfrakM$ is the intersection of all maximal ideals of $k[x_1,ldots,x_n]$ that contain $mathfrakI$.
But by Hilbert's Nullstellensatz, $sqrtmathfrakI$, the radical of $mathfrakI$, is the intersection of all maximal ideals of $k[x_1,ldots,x_n]$ that contain $mathfrakI$, so the nilradical of $R$ equals the Jacobson radical of $R$.
However, in these rings there are generally prime ideals that are not maximal. For example, if $ngeq 3$ and you take $mathfrakI=(x_1)$, then $(x_1,x_2)+mathfrakI$ is prime but not maximal.
answered Oct 26 '10 at 19:49
Arturo MagidinArturo Magidin
265k34590919
$endgroup$
Let $k$ be a field, and let $R=k[x_1,ldots,x_n]/mathfrakI$, where $mathfrakI$ is an ideal of $k[x_1,ldots,x_n]$.
The nilradical of $R$ is $sqrtmathfrakI/mathfrakI$, and the Jacobson radical of $R$ is $mathfrakM/mathfrakI$, where $mathfrakM$ is the intersection of all maximal ideals of $k[x_1,ldots,x_n]$ that contain $mathfrakI$.
But by Hilbert's Nullstellensatz, $sqrtmathfrakI$, the radical of $mathfrakI$, is the intersection of all maximal ideals of $k[x_1,ldots,x_n]$ that contain $mathfrakI$, so the nilradical of $R$ equals the Jacobson radical of $R$.
However, in these rings there are generally prime ideals that are not maximal. For example, if $ngeq 3$ and you take $mathfrakI=(x_1)$, then $(x_1,x_2)+mathfrakI$ is prime but not maximal.
answered Oct 26 '10 at 19:49
Arturo MagidinArturo Magidin
265k34590919
answered Oct 26 '10 at 19:49
Arturo MagidinArturo Magidin
265k34590919
answered Oct 26 '10 at 19:49
Arturo MagidinArturo Magidin
265k34590919
answered Oct 26 '10 at 19:49
Arturo MagidinArturo Magidin
265k34590919
265k34590919
add a comment |
add a comment |
$begingroup$
Theorem 5.1 in T.Y. Lam's book "A First Course in Noncommutative Rings" states that every polynomial ring $R[T]$ over a commutative ring $R$ satisfies
$$rad ,, R[T] = Nil(R[T]) = (Nil ,,R)[T]$$
answered Apr 23 '12 at 10:14
CihanCihan
1,682916
$endgroup$
add a comment |
$begingroup$
Theorem 5.1 in T.Y. Lam's book "A First Course in Noncommutative Rings" states that every polynomial ring $R[T]$ over a commutative ring $R$ satisfies
$$rad ,, R[T] = Nil(R[T]) = (Nil ,,R)[T]$$
answered Apr 23 '12 at 10:14
CihanCihan
1,682916
$endgroup$
add a comment |
$begingroup$
Theorem 5.1 in T.Y. Lam's book "A First Course in Noncommutative Rings" states that every polynomial ring $R[T]$ over a commutative ring $R$ satisfies
$$rad ,, R[T] = Nil(R[T]) = (Nil ,,R)[T]$$
answered Apr 23 '12 at 10:14
CihanCihan
1,682916
$endgroup$
Theorem 5.1 in T.Y. Lam's book "A First Course in Noncommutative Rings" states that every polynomial ring $R[T]$ over a commutative ring $R$ satisfies
$$rad ,, R[T] = Nil(R[T]) = (Nil ,,R)[T]$$
answered Apr 23 '12 at 10:14
CihanCihan
1,682916
answered Apr 23 '12 at 10:14
CihanCihan
1,682916
answered Apr 23 '12 at 10:14
CihanCihan
1,682916
answered Apr 23 '12 at 10:14
CihanCihan
1,682916
1,682916
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f7954%2fnil-radical-equals-jacobson-radical-even-though-not-every-prime-ideal-is-maximal%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
@Natalia S: You tagged the question "commutative algebra", but you might want to be explicit in the question on whether you are considering only commutative rings or not (and while you are at it, whether your rings must have a unity or not).
$endgroup$
– Arturo Magidin
Oct 26 '10 at 19:36
4
$begingroup$
Since the concept of prime ideal is being used, it seems likely that the author of the question has the commutative case in mind. In this case, there is an important class of rings (including in particular the examples of Arturo Magidin's answer) for which the answer is yes, namely Jacobson rings: en.wikipedia.org/wiki/Jacobson_ring As that wikipedia entry (and Arturo Magidin's answer) make clear, this question is rather closely bound up with the ideas behind Hilbert's Nullstellensatz.
$endgroup$
– Matt E
Oct 26 '10 at 20:03
$begingroup$
@Arturo Magidin: Thanks for your comment, I edited the question.
$endgroup$
– Pandora
Oct 26 '10 at 20:03
$begingroup$
@Natalia S: I also added "nonzero"; otherwise, $mathbbZ$ itself would be an example.
$endgroup$
– Arturo Magidin
Oct 26 '10 at 20:06
3
$begingroup$
For any commutative ring R, R[x] has this property. This is one of the first exercises in Atiyah-Macdonald.
$endgroup$
– user641
Oct 27 '10 at 3:58